- Identify the base, exponent, and result. - The power exponent becomes the radical index. - A square-root index of \(2\) is usually omitted.

a) \(13^2=169\) becomes \(\sqrt{169}=13\). b) \(\sqrt[3]{216}=6\) becomes \(6^3=216\). c) \(0.2^5=0.00032\) becomes \(\sqrt[5]{0.00032}=0.2\). d) \(\sqrt[4]{\frac{16}{81}}=\frac{2}{3}\) becomes \(\left(\frac{2}{3}\right)^4=\frac{16}{81}\).

a) \(\sqrt{169}=13\) b) \(6^3=216\) c) \(\sqrt[5]{0.00032}=0.2\) d) \(\left(\frac{2}{3}\right)^4=\frac{16}{81}\)

- Look separately at the numerator and denominator of the rational exponent. - The denominator tells which root is involved. - The numerator tells which power is involved.

The denominator \(3\) indicates a cube root. The numerator \(2\) indicates a square.

Cube root, then square.

- Recall what the denominator of a unit fraction exponent represents. - Connect the notation to the inverse operation of raising a positive number to the fourth power.

1. An exponent of \(\frac{1}{4}\) denotes the principal fourth root, so \(x^{\frac{1}{4}}=\sqrt[4]{x}\). 2. By definition, \(\left(\sqrt[4]{x}\right)^4=x\).

\(x^{\frac{1}{4}}=\sqrt[4]{x}\), and the fourth power of that value equals \(x\).

- Which positive whole numbers have squares close to \(81\)? - Which root value follows the convention of using the nonnegative answer?

1. An exponent of \(\frac{1}{2}\) means the principal square root. 2. Since \(9^2=81\), the value is \(9\).

\(81^{\frac{1}{2}}=\sqrt{81}=9\).

- A cube root asks for a number whose third power gives the base. - What happens when you cube a negative number?

1. The exponent means the cube root of \(-64\). 2. Since \((-4)^3=-64\), its cube root is \(-4\).

\((-64)^{\frac{1}{3}}=-4\), because \((-4)^3=-64\).

- Compare \(\sqrt[3]{z}\) and \(z^{\frac{1}{3}}\). What stays the same? - Test your proposed notation with \(u=2\) to see if it describes the same operations.

1. A fifth root corresponds to a denominator of \(5\). 2. The square on \(u\) gives numerator \(2\), so the form is \(u^{\frac{2}{5}}\).

\(u^{\frac{2}{5}}\).

- How would you write \(8^{\frac{1}{3}}\) without an exponent? - Would splitting the sum change the base before the operation is carried out?

1. An exponent of one-third means cube root. 2. The whole sum \(p+q\) is the base, so the radical form is \(\sqrt[3]{p+q}\).

\(\sqrt[3]{p+q}\); the entire sum \(p+q\) is inside the radical.

- What happens when you divide the same nonzero power by itself? - Could that division be used with a zero base?

1. For any nonzero base, the zero-exponent rule gives \(a^0=1\). 2. For \(a\ne0\), the quotient \(a^3\div a^3\) equals both \(1\) and \(a^{3-3}=a^0\). This argument fails at \(a=0\) because it would divide by zero.

\(a^0=1\) for \(a\ne0\). The rule comes from dividing a nonzero power by itself, which cannot be done with base \(0\).

- A negative exponent does not make the value negative. - Think about what multiplied by \(5\) gives \(1\).

1. A negative exponent means take the reciprocal of the corresponding positive power. 2. Therefore \(5^{-1}=\frac{1}{5}\).

\(\frac{1}{5}\); it means take the reciprocal.

- What kind of root does the denominator \(2\) name? - Can squaring any real number give a negative result?

1. This expression asks for the real square root of \(-16\). 2. The square of every real number is nonnegative, so no real number squares to \(-16\).

No. \((-16)^{\frac{1}{2}}\) is not real: the square of a real number cannot be negative.

- Follow the vertical grid line at the stated input to the curve. - Evaluate the two possible powers at the same input and compare with your reading.

1. Curve r has height \(2\) at \(x=4\). 2. Since \(4^{\frac{1}{2}}=\sqrt{4}=2\), this rule matches the graph. 3. The other rule gives \(4^2=16\), which does not match the graph at \(x=4\).

At \(x=4\), the graph has height \(2\). Since \(4^{\frac{1}{2}}=2\) but \(4^2=16\), curve r represents \(y=x^{\frac{1}{2}}\).

- Rewrite every radicand as a power of \(10\). - Numbers between \(0\) and \(1\) use negative powers of \(10\). - Subtract exponents before taking the square root in part c).

The identity follows from \(\sqrt{10^{2n}}=(10^{2n})^{\frac{1}{2}}=10^n\). a) \(100{,}000{,}000=10^8\), so \(\sqrt{10^8}=10^4=10{,}000\). b) \(0.000001=10^{-6}\), so \(\sqrt{10^{-6}}=10^{-3}=0.001\). c) \(\frac{10^3}{10^7}=10^{-4}\), so \(\sqrt{10^{-4}}=10^{-2}=0.01\).

a) \(10{,}000\) b) \(0.001\) c) \(0.01\)

- Prove the general identity before evaluating the examples. - A negative exponent creates a reciprocal. - Rewrite the last radicand as a power of \(5\).

1. Rewrite the square root as a rational exponent: \(\sqrt{5^{2k}}=(5^{2k})^{\frac{1}{2}}=5^k\). 2. \(\sqrt{5^6}=5^3=125\). 3. \(\sqrt{5^{-4}}=5^{-2}=\frac{1}{25}\). 4. \(\sqrt{\frac{1}{5^2}}=\sqrt{5^{-2}}=5^{-1}=\frac{1}{5}\).

\(\sqrt{5^{2k}}=(5^{2k})^{\frac{1}{2}}=5^k\) \(\sqrt{5^6}=125\) \(\sqrt{5^{-4}}=\frac{1}{25}\) \(\sqrt{\frac{1}{5^2}}=\frac{1}{5}\)

- Rewrite a square root as a power with exponent \(\frac{1}{2}\). - Use the power-of-a-power property. - Recognize \(0.49\) as a perfect-square decimal. - Apply the product property when a product appears under the radical.

a) \(\sqrt{5^4}=5^{\frac{4}{2}}=5^2=25\). b) \(\sqrt{2^{10}}=2^{\frac{10}{2}}=2^5=32\). c) \(\sqrt{10^{-6}}=10^{-3}=0.001\). d) \(\sqrt{0.49\cdot10^8}=\sqrt{0.49}\cdot\sqrt{10^8}=0.7\cdot10^4=7000\).

a) \(25\) b) \(32\) c) \(0.001\) d) \(7000\)

- Rewrite decimal exponents as fractions. - Take the root before the remaining power when convenient. - Look for familiar perfect powers.

a) \(27^{\frac{4}{3}}=(\sqrt[3]{27})^4=3^4=81\). b) \(\sqrt[3]{10^9}=(10^9)^{\frac{1}{3}}=10^3=1000\). c) \(0.09^{1.5}=0.09^{\frac{3}{2}}=(\sqrt{0.09})^3=(0.3)^3=0.027\). d) \(\sqrt[4]{0.0016}=0.0016^{\frac{1}{4}}=0.2\), since \((0.2)^4=0.0016\).

a) \(81\) b) \(1000\) c) \(0.027\) d) \(0.2\)

- First identify whether the exponent is even or odd. - Taking an \(n\)th root corresponds to using an exponent of \(\frac{1}{n}\). - Consider the sign of the number on the right side. - An even power can never be negative over the real numbers.

a) Take both square roots: \(x=\pm\sqrt{144}=\pm144^{\frac{1}{2}}=\pm12\). There are two solutions. b) Take the cube root: \(x=\sqrt[3]{-125}=-5\). There is one real solution. c) The only solution is \(x=0\). There is one real solution. d) No real solution exists because an even power cannot be negative.

a) \(\{-12,12\}\); two solutions b) \(\{-5\}\); one solution c) \(\{0\}\); one solution d) \(\varnothing\); no real solution

- Convert decimal exponents to fractions. - Use \(x^{\frac{m}{n}}=\sqrt[n]{x^m}\). - Combine square roots in a product when possible.

1. \(e=0\). 2. Since \(0.25=\frac{1}{4}\), \(d=\sqrt[4]{16}=2\). 3. \(a=\sqrt[4]{81}=3\). 4. Since \(0.4=\frac{2}{5}\), \(b=32^{\frac{2}{5}}=(\sqrt[5]{32})^2=4\). 5. \(c=\sqrt{2\cdot18}=\sqrt{36}=6\). 6. Therefore, \(0<2<3<4<6\), so the order is \(e,d,a,b,c\).

\(e<d<a<b<c\), with values \(0<2<3<4<6\)

- Determine the number of real solutions from the exponent's parity and the right side's sign. - An \(n\)th root can be written using an exponent of \(\frac{1}{n}\). - Include both signs for an even power with a positive result. - Round only after finding the exact form.

a) Since \(x^4=150\) has a positive right side and an even exponent, \(x=\pm150^{\frac{1}{4}}=\pm\sqrt[4]{150}\approx\pm3.50\). b) Since the exponent is odd, \(x^3=-200\) has one real solution: \(x=-200^{\frac{1}{3}}=-\sqrt[3]{200}\approx-5.85\). c) The equation \(x^6=-10\) has no real solution because an even power is never negative. d) \(x^5=32\) has the exact solution \(x=32^{\frac{1}{5}}=\sqrt[5]{32}=2\).

a) \(x=\pm150^{\frac{1}{4}}=\pm\sqrt[4]{150}\approx\pm3.50\) b) \(x=-200^{\frac{1}{3}}=-\sqrt[3]{200}\approx-5.85\) c) No real solution d) \(x=32^{\frac{1}{5}}=2\)

- Check a proposed radical value by raising it to the radical index. - Odd powers preserve the sign of a negative base. - A principal even root is never negative.

1. True, because \((-0.5)^3=-0.125\). 2. True, because \((0.08)^2=0.0064\). 3. False, because \(5^4=625\), so \(\sqrt[4]{625}=5\). 4. False, because a principal square root is nonnegative: \(\sqrt{(-9)^2}=\sqrt{81}=9\). 5. True, because \((-2)^5=-32\).

1. True 2. True 3. False 4. False 5. True

- Evaluate each expression first. - Notice whether a negative sign is inside or outside a radical. - A principal square root is nonnegative.

1. \(a=-2\). 2. \(b=0.5\). 3. \(c=-3\). 4. \(d=3\), because a principal square root is nonnegative. 5. \(e=2\). 6. Therefore, \(-3<-2<0.5<2<3\), so \(c<a<b<e<d\).

\(c<a<b<e<d\)

- Decide whether each exponent is even or odd. - An even power with a positive result can have two real bases. - An odd power has exactly one real base. - Use familiar integer powers to evaluate the roots.

a) Since \(5^4=625\) and the exponent is even, \(x=\pm5\). b) Since \((-4)^3=-64\), \(x=-4\). c) Both \(1^8=1\) and \((-1)^8=1\), so \(x=\pm1\). d) The only number whose fifth power is \(0\) is \(0\), so \(x=0\).

a) \(\{-5,5\}\) b) \(\{-4\}\) c) \(\{-1,1\}\) d) \(\{0\}\)

- Rewrite each radicand as a familiar power. - For fractions, take the root of the numerator and denominator. - A negative exponent represents a reciprocal.

a) Since \(6^3=216\), \(\sqrt[3]{216}=6\). b) Since \((0.5)^4=0.0625\), \(\sqrt[4]{0.0625}=0.5\). c) Since \(32^{-1}=2^{-5}=(0.5)^5\), \(\sqrt[5]{32^{-1}}=0.5\). d) \(\sqrt[3]{\frac{27}{1000}}=\frac{3}{10}=0.3\).

a) \(6\) b) \(0.5\) c) \(0.5\) d) \(0.3\)

- Convert decimal exponents to fractions. - A negative exponent takes the reciprocal. - Evaluate the root before the remaining power when convenient.

a) \(144^{0.5}=144^{\frac{1}{2}}=\sqrt{144}=12\). b) \(\left(\frac{1}{27}\right)^{-\frac{1}{3}}=27^{\frac{1}{3}}=\sqrt[3]{27}=3\). c) \(32^{0.4}=32^{\frac{2}{5}}=(\sqrt[5]{32})^2=4\). d) \(1000^{-\frac{2}{3}}=\frac{1}{(\sqrt[3]{1000})^2}=\frac{1}{100}=0.01\).

a) \(12\) b) \(3\) c) \(4\) d) \(0.01\)

- Convert decimal exponents to fractions. - A negative exponent places the positive power in the denominator. - The denominator of a rational exponent gives the radical index.

a) \(x^{\frac{2}{3}}=\sqrt[3]{x^2}\). b) Since \(1.5=\frac{3}{2}\), \(y^{1.5}=\sqrt{y^3}\). c) \(z^{-\frac{1}{2}}=\frac{1}{\sqrt z}\). d) \((4a)^{\frac{3}{4}}=\sqrt[4]{(4a)^3}=\sqrt[4]{64a^3}\).

a) \(\sqrt[3]{x^2}\) b) \(\sqrt{y^3}\) c) \(\frac{1}{\sqrt z}\) d) \(\sqrt[4]{64a^3}\)

- Write \(64\) and \(8\) as powers of \(2\). - Rewrite each radical as a rational exponent. - Compare the resulting powers of \(2\).

1. \(\sqrt[6]{64}=\sqrt[6]{2^6}=2^1\). 2. \(\sqrt[3]{8}=\sqrt[3]{2^3}=2^1\). 3. Both statements are correct because all three expressions equal \(2\).

Both students are correct: \(\sqrt[6]{64}=\sqrt[3]{8}=2^1=2\).

- Write each radical as an exponent fraction. - Reduce the fraction just as you would any other fraction. - A negative exponent creates a reciprocal.

a) \(\sqrt[12]{b^3}=b^{\frac{3}{12}}=b^{\frac{1}{4}}=\sqrt[4]{b}\). b) \(\sqrt[10]{z^{15}}=z^{\frac{15}{10}}=z^{\frac{3}{2}}=\sqrt{z^3}\). c) \(\frac{1}{\sqrt[6]{y^4}}=y^{-\frac{4}{6}}=y^{-\frac{2}{3}}=\frac{1}{\sqrt[3]{y^2}}\).

a) \(\sqrt[4]{b}\) b) \(\sqrt{z^3}\) c) \(\frac{1}{\sqrt[3]{y^2}}\)

- Think about how a whole-number root can be written as a power. - How does writing a reciprocal affect the exponent? - Before calculating with a negative base, decide whether the required root is odd or even.

1. \(\sqrt[4]{20}=20^{\frac{1}{4}}\approx2.115\). 2. \(\sqrt[3]{5^2}=5^{\frac{2}{3}}\approx2.924\). 3. \(\frac{1}{\sqrt[5]{10}}=10^{-\frac{1}{5}}\approx0.631\). 4. \(\sqrt[3]{-27}=(-27)^{\frac{1}{3}}=-3\).

a) \(20^{\frac{1}{4}}\approx2.115\) b) \(5^{\frac{2}{3}}\approx2.924\) c) \(10^{-\frac{1}{5}}\approx0.631\) d) \((-27)^{\frac{1}{3}}=-3\)

- Rewrite every base as a power of \(2\). - Convert the decimal exponent to a fraction. - Apply the power-of-a-power property.

1. \(A=(2^4)^{\frac{3}{4}}=2^3=8\). 2. \(B=\sqrt[3]{2^6}=2^2=4\). 3. \(C=\left(\frac{1}{4}\right)^{-\frac{3}{2}}=4^{\frac{3}{2}}=8\). 4. \(D=(2^6)^{\frac{1}{2}}=2^3=8\). 5. Therefore, \(A=C=D\), while \(B=4\).

\(A=8\), \(B=4\), \(C=8\), and \(D=8\). Thus \(A=C=D\).

- Rewrite each radical using a rational exponent. - Express \(25\) and \(125\) as powers of \(5\). - Apply the power-of-a-power property.

1. \(T_1=\sqrt[4]{5^6}=5^{\frac{6}{4}}=5^{\frac{3}{2}}\). 2. \(T_2=\sqrt{5^3}=5^{\frac{3}{2}}\). 3. Since \(25=5^2\), \(T_3=(5^2)^{\frac{3}{4}}=5^{\frac{6}{4}}=5^{\frac{3}{2}}\). 4. Since \(125=5^3\), \(T_4=(5^3)^{\frac{1}{2}}=5^{\frac{3}{2}}\). 5. All four expressions equal \(5^{\frac{3}{2}}\), so they have the same value.

\(T_1=T_2=T_3=T_4=5^{\frac{3}{2}}\).

- Rewrite an \(n\)th root as an exponent of \(\frac{1}{n}\). - Apply the exponent to each factor in a product. - A factor moved from the denominator to the numerator has the opposite exponent. - Reduce each exponent fraction.

a) \(\sqrt[3]{x^9}=x^{\frac{9}{3}}=x^3\). b) \(\sqrt[5]{a^{10}b^{20}}=a^{\frac{10}{5}}b^{\frac{20}{5}}=a^2b^4\). c) \(\frac{1}{\sqrt[4]{y^{12}}}=\frac{1}{y^{\frac{12}{4}}}=y^{-3}\). d) \(\sqrt[k]{z^{5k}}=z^{\frac{5k}{k}}=z^5\).

a) \(x^3\) b) \(a^2b^4\) c) \(y^{-3}\) d) \(z^5\)

- Add exponents when multiplying powers with the same base. - Subtract exponents when dividing powers with the same base. - Distribute an exponent over a product. - Rewrite each radical as a rational exponent.

a) \(3^{\frac{1}{4}}\cdot3^{\frac{3}{4}}=3^{\frac{1}{4}+\frac{3}{4}}=3\). b) \(x^{\frac{5}{6}}\div x^{\frac{1}{3}}=x^{\frac{5}{6}-\frac{2}{6}}=x^{\frac{1}{2}}\). c) \((4y)^{\frac{1}{2}}\cdot y^{\frac{1}{2}}=4^{\frac{1}{2}}y^{\frac{1}{2}}y^{\frac{1}{2}}=2y\). d) \(\sqrt[5]{a^2}\cdot\sqrt[5]{a^3}=a^{\frac{2}{5}}a^{\frac{3}{5}}=a\).

a) \(3\) b) \(x^{\frac{1}{2}}\) c) \(2y\) d) \(a\)

- Rewrite an \(n\)th root as a power with exponent \(\frac{1}{n}\). - Apply the power-of-a-power property. - Express \(27\) as a power of \(3\). - Multiply powers with the same base by adding exponents.

a) \(A=(27^2)^{\frac{1}{3}}=27^{\frac{2}{3}}=(3^3)^{\frac{2}{3}}=3^2=9\). Also, \(B=(27^{\frac{1}{3}})^2=27^{\frac{2}{3}}=9\). Therefore, \(A\) and \(B\) are equivalent. b) \(C=(x^{12})^{\frac{1}{4}}=x^3\). Since \(x>0\), \(D=x^2(x^2)^{\frac{1}{2}}=x^2\cdot x=x^3\). Therefore, \(C\) and \(D\) are equivalent.

a) Yes. Both expressions equal \(9\). b) Yes. Both expressions simplify to \(x^3\).

- Rewrite radicals as powers with rational exponents. - Add or subtract exponents when the bases are the same. - Reduce exponent fractions before combining them. - Use the difference-of-squares pattern in part c).

a) \(\sqrt[3]{x^2}\sqrt[6]{x^2}=x^{\frac{2}{3}}x^{\frac{2}{6}}=x^{\frac{2}{3}+\frac{1}{3}}=x\). b) \(\frac{\sqrt{a^5}}{\sqrt[4]{a^2}}=a^{\frac{5}{2}-\frac{2}{4}}=a^{\frac{5}{2}-\frac{1}{2}}=a^2\). c) Apply the difference-of-squares pattern: \((\sqrt b+\sqrt a)(\sqrt b-\sqrt a)=(\sqrt b)^2-(\sqrt a)^2=b-a\).

a) \(x\) b) \(a^2\) c) \(b-a\)

- Rewrite the fourth root as a rational exponent. - Add exponents when multiplying powers with the same base. - Reduce the resulting exponent.

1. Rewrite the radical: \(\sqrt[4]{x^6}=x^{\frac{6}{4}}=x^{\frac{3}{2}}\). 2. Multiply powers with the same base by adding exponents: \(x^{-\frac{1}{2}}x^{\frac{3}{2}}=x^{-\frac{1}{2}+\frac{3}{2}}=x^1=x\).

\(x\)

- Substitute the surface area and simplify inside the parentheses. - Rewrite \(1.5\) as a fraction. - An exponent of \(\frac{1}{2}\) represents a square root.

1. Substitute \(S=54\): \(V=\left(\frac{54}{6}\right)^{1.5}=9^{1.5}\). 2. Rewrite the decimal exponent: \(1.5=\frac{3}{2}\), so \(V=9^{\frac{3}{2}}\). 3. Evaluate using the square root first: \(9^{\frac{3}{2}}=(\sqrt9)^3=3^3=27\). 4. The volume is \(27\,\text{cm}^3\).

\(27\,\text{cm}^3\)

- Rewrite each decimal as a familiar power. - Check how decimal places change when a decimal is squared, cubed, or raised to the fourth power.

1. \(\sqrt{1.44}=1.2\). 2. \(\sqrt[3]{0.008}=0.2\). 3. \(\sqrt[4]{0.0001}=0.1\). 4. Therefore, \(1.2+0.2-0.1=1.3\).

\(1.3\)

- Rewrite the cube root as a rational exponent. - Add exponents in the numerator. - Subtract the denominator exponent, including its negative sign. - Use a common denominator for the exponent fractions.

1. Rewrite the radical: \(\sqrt[3]{a}=a^{\frac{1}{3}}\). 2. Combine the numerator powers: \(a^{\frac{5}{6}}a^{\frac{1}{3}}=a^{\frac{5}{6}+\frac{2}{6}}=a^{\frac{7}{6}}\). 3. Divide by subtracting exponents: \(a^{\frac{7}{6}-(-\frac{1}{2})}=a^{\frac{7}{6}+\frac{3}{6}}=a^{\frac{10}{6}}=a^{\frac{5}{3}}\).

\(a^{\frac{5}{3}}\)

- Rewrite the equation in the form \(x^n=c\) before substituting values for \(n\). - Use a rational exponent to undo each power, then account for whether the exponent is odd or even. - Compare what happens when a positive value is raised to an odd power and to an even power.

First rewrite the equation: \(2(x^n-4)=120\), so \(x^n-4=60\) and \(x^n=64\). a) For \(n=3\), \(x^3=64\), so \(x=64^{\frac{1}{3}}=4\). An odd power gives one real solution. b) For \(n=6\), \(x^6=64\), so \(x=\pm64^{\frac{1}{6}}=\pm2\). An even power equal to a positive number gives two real solutions.

a) \(x=4\) b) \(x=\pm2\) The odd exponent gives one real solution, while the even exponent gives two real solutions because the right side is positive.

- Substitute the point into the function rule. - Isolate the exponential expression before identifying the exponent. - A cube root can be written using an exponent of \(\frac{1}{3}\).

a) Substitute \((4,32)\): \(32=\frac{1}{2}4^n\). Multiply by \(2\): \(64=4^n\). Since \(4^3=64\), \(n=3\). b) Use \(n=3\): \(\frac{1}{2}a^3=4\). Multiply by \(2\): \(a^3=8\), so \(a=8^{\frac{1}{3}}=\sqrt[3]{8}=2\).

a) \(n=3\) b) \(a=8^{\frac{1}{3}}=2\)

- Rewrite each radicand as a familiar power. - Odd roots of negative numbers are negative. - Convert fractions to decimals only if it helps compare the values.

a) \(\sqrt[3]{-0.125}=-0.5\). b) \(\sqrt[5]{243}=3\). c) \(\sqrt[3]{-\frac{27}{64}}=-\frac{3}{4}=-0.75\). d) \(\sqrt[4]{0.0001}=0.1\). Therefore, \(-0.75<-0.5<0.1<3\), so the order is \(c<a<d<b\).

a) \(-0.5\) b) \(3\) c) \(-0.75\) d) \(0.1\) Order: \(c<a<d<b\)

- Rewrite each radicand as a familiar power. - Interpret a rational exponent as a root followed by a power. - Take the square root of a fraction by taking the square root of its numerator and denominator.

1) Since \((0.1)^3=0.001\), the value is \(0.1\). 2) \(16^{\frac{3}{4}}=(\sqrt[4]{16})^3=2^3=8\). 3) \(\sqrt{\frac{25}{144}}=\frac{5}{12}\). 4) Since \(5^4=625\), the value is \(5\).

1) \(0.1\) 2) \(8\) 3) \(\frac{5}{12}\) 4) \(5\)

- Evaluate each root or rational exponent separately. - Check a proposed root by raising it to the radical index. - For a nested square root, work from the inside out.

a) True. \(\sqrt[3]{8}\cdot\sqrt[3]{27}=2\cdot3=6\), and \(\sqrt[3]{216}=6\). b) False. \(100^{\frac{1}{2}}-64^{\frac{1}{3}}=10-4=6\). c) True. \(\sqrt{\sqrt{16}}=\sqrt{4}=2\). d) False. Since \((0.1)^4=0.0001\), \(\sqrt[4]{0.0001}=0.1\).

a) True b) False; the correct value is \(6\). c) True d) False; the correct value is \(0.1\).

- Use a \(\frac{1}{3}\) power, equivalently a cube root, to find an edge length from a volume. - The volume of a cube with edge length \(s\) is \(s^3\). - Compare how multiplying an edge length affects the volume.

a) For the first tank, \(s^3=216\), so \(s=216^{\frac{1}{3}}=\sqrt[3]{216}=6\,\text{dm}\). b) For the larger tank, \(s^3=1728\), so \(s=1728^{\frac{1}{3}}=\sqrt[3]{1728}=12\,\text{dm}\). c) The edge-length ratio is \(12:6=2:1\). The volume ratio is \(1728:216=8:1\), which agrees with \(2^3=8\).

a) \(6\,\text{dm}\) b) \(12\,\text{dm}\) c) The larger tank has twice the edge length and eight times the volume.

- An odd root of a negative number can be real, but an even root cannot. - Check exact values by raising the proposed result to the radical index. - For an even index, \(\sqrt[n]{a^n}=|a|\).

1) \(\sqrt[4]{-81}\) has no real value because an even root of a negative number is not real. 2) \(\sqrt[5]{-32}=-2\) because \((-2)^5=-32\). 3) \(\sqrt[6]{(-2)^6}=|-2|=2\). 4) \(\sqrt[3]{0.008}=0.2\) because \((0.2)^3=0.008\). 5) \(\sqrt[10]{-1}\) has no real value because an even root of a negative number is not real.

1) Not real 2) \(-2\) 3) \(2\) 4) \(0.2\) 5) Not real

- Distinguish between even-index and odd-index radicals. - Test a negative number raised to an even power. What sign has its even root? - Use each given sign condition to decide whether the expression inside the power is positive or negative.

1. \(\sqrt{(a-8)^2}=|a-8|\). Since \(a-8\leq0\), the result is \(8-a\). 2. For an odd index, \(\sqrt[3]{(b+5)^3}=b+5\) for every real \(b\). 3. \(\sqrt[4]{(c-2)^4}=|c-2|\). Since \(c-2\leq0\), the result is \(2-c\). 4. \(\sqrt[6]{(1-x)^6}=|1-x|\). Since \(1-x\leq0\), the result is \(x-1\).

a) \(8-a\) b) \(b+5\) c) \(2-c\) d) \(x-1\)

- Rewrite each radical as a rational exponent and reduce the exponent. - Identify a familiar square, cube, or fourth power after simplifying. - Check each result by raising it to the appropriate power.

a) \(\sqrt[6]{125^2}=125^{\frac{2}{6}}=125^{\frac{1}{3}}=\sqrt[3]{125}=5\). b) \(\sqrt[4]{2.25^2}=2.25^{\frac{2}{4}}=\sqrt{2.25}=1.5\). c) \(\sqrt[8]{16^2}=16^{\frac{2}{8}}=16^{\frac{1}{4}}=\sqrt[4]{16}=2\). d) \(\sqrt[6]{0.001^2}=0.001^{\frac{2}{6}}=0.001^{\frac{1}{3}}=\sqrt[3]{0.001}=0.1\).

a) \(5\) b) \(1.5\) c) \(2\) d) \(0.1\)

- Find two consecutive integers whose fifth powers bracket \(100\). - Then test consecutive tenths. - Compare with the midpoint between the two tenths to determine the rounded value.

1. Since \(2^5=32\) and \(3^5=243\), the value lies between \(2\) and \(3\). 2. Test consecutive tenths: \((2.5)^5=97.65625\) and \((2.6)^5=118.81376\). Therefore \(2.5<\sqrt[5]{100}<2.6\). 3. To decide how to round, test the midpoint: \((2.55)^5\approx107.82\). Since \(100<107.82\), \(\sqrt[5]{100}<2.55\), so the value rounds to \(2.5\).

\(\sqrt[5]{100}\approx2.5\)

- Square each input before taking the fourth root. - Rewrite a radical as a rational exponent and reduce the exponent. - Recall what an absolute value does to a negative number.

a) For \(x=16\), \(A=\sqrt[4]{16^2}=4\) and \(B=\sqrt{|16|}=4\). For \(x=-16\), \(A=\sqrt[4]{(-16)^2}=4\) and \(B=\sqrt{|-16|}=4\). b) When \(x>0\), \(A=(x^2)^{\frac{1}{4}}=x^{\frac{2}{4}}=x^{\frac{1}{2}}=\sqrt{x}\). Also, \(|x|=x\), so \(B=\sqrt{x}\). Thus \(A=B\). c) In \(A\), the radicand \(x^2\) is nonnegative for every real \(x\). In \(B\), the absolute value makes the radicand nonnegative; without it, \(\sqrt{x}\) would not be real for negative \(x\).

a) Both expressions equal \(4\) for \(x=16\) and for \(x=-16\). b) \((x^2)^{\frac{1}{4}}=x^{\frac{1}{2}}=\sqrt{x}=\sqrt{|x|}\) for \(x>0\). c) The square in \(A\) and the absolute value in \(B\) each make the radicand nonnegative.

- Reduce the radical index and exponent by their common factor. - An even-index root may introduce an absolute value. - Use each condition to determine the sign inside the absolute value.

1) \(\sqrt[6]{(b-3)^2}=\sqrt[3]{|b-3|}\). Since \(b<3\), \(b-3<0\), so \(|b-3|=3-b\). The result is \(\sqrt[3]{3-b}\). 2) \(\sqrt[10]{(x-1)^{10}}=|x-1|\). Since \(x<1\), \(x-1<0\), so \(|x-1|=1-x\).

1) \(\sqrt[3]{3-b}\) 2) \(1-x\)

- Rewrite each radical as a power with a fractional exponent. - Add exponents when multiplying and subtract when dividing like bases. - Use a common denominator for the exponent fractions. - Convert the simplified rational exponent back to a radical.

1) \(\sqrt[3]{7}\cdot\sqrt{7}=7^{\frac{1}{3}+\frac{1}{2}}=7^{\frac{5}{6}}=\sqrt[6]{7^5}\). 2) \(\frac{\sqrt[4]{x^3}}{\sqrt[8]{x}}=x^{\frac{3}{4}-\frac{1}{8}}=x^{\frac{5}{8}}=\sqrt[8]{x^5}\). 3) \(\sqrt[5]{a^2}\cdot\sqrt[10]{a}=a^{\frac{2}{5}+\frac{1}{10}}=a^{\frac{1}{2}}=\sqrt{a}\).

1) \(\sqrt[6]{7^5}\) 2) \(\sqrt[8]{x^5}\) 3) \(\sqrt{a}\)

- The radical index becomes the denominator of a rational exponent. - The exponent on the radicand becomes the numerator. - A negative exponent produces a reciprocal. - Treat a parenthesized sum as one base.

1) \(\sqrt[5]{x^3}=x^{\frac{3}{5}}\). 2) \(a^{-\frac{2}{3}}=\frac{1}{a^{\frac{2}{3}}}=\frac{1}{\sqrt[3]{a^2}}\). 3) \(\frac{1}{\sqrt[4]{y}}=y^{-\frac{1}{4}}\). 4) \((m+n)^{\frac{3}{4}}=\sqrt[4]{(m+n)^3}\).

1) \(x^{\frac{3}{5}}\) 2) \(\frac{1}{\sqrt[3]{a^2}}\) 3) \(y^{-\frac{1}{4}}\) 4) \(\sqrt[4]{(m+n)^3}\)

- Match the radical index with the denominator of the exponent. - Rewrite the decimal exponent as a fraction. - For part b), taking the cube root first makes the calculation simple. - A negative exponent produces a reciprocal.

a) \(\sqrt[5]{y^2}=y^{\frac{2}{5}}\). b) \(27^{\frac{2}{3}}=(\sqrt[3]{27})^2=3^2=9\). c) Since \(-0.75=-\frac{3}{4}\), \(b^{-0.75}=b^{-\frac{3}{4}}=\frac{1}{\sqrt[4]{b^3}}\).

a) \(y^{\frac{2}{5}}\) b) \(9\) c) \(\frac{1}{\sqrt[4]{b^3}}\)

- Rewrite each radical as a rational exponent. - Multiply exponents when raising a power to another power. - Compare the radical index with the outside exponent.

1. True: \((\sqrt[5]{15})^5=(15^{\frac{1}{5}})^5=15\). 2. True: \((\sqrt{2})^6=(2^{\frac{1}{2}})^6=2^3=8\). 3. False: \((\sqrt[3]{3})^6=(3^{\frac{1}{3}})^6=3^2=9\). The corrected result is \(9\). 4. True: \(\sqrt[4]{5^8}=5^{\frac{8}{4}}=5^2=25\).

1. True 2. True 3. False; the correct value is \(9\). 4. True

- Rewrite each radical as a rational exponent. - Apply an outside exponent to every factor in a product. - Track the sign of a negative factor raised to an even or odd power.

a) \((\sqrt[6]{x^4})^3=(x^4)^{\frac{3}{6}}=x^2\). b) \((-2\sqrt[4]{y^3})^4=(-2)^4(\sqrt[4]{y^3})^4=16y^3\). c) \((0.5\sqrt[3]{2a^2})^3=(0.5)^3(\sqrt[3]{2a^2})^3=0.125(2a^2)=0.25a^2\).

a) \(x^2\) b) \(16y^3\) c) \(0.25a^2\)

- An even outer exponent removes a negative sign; an odd outer exponent preserves it. - Rewrite radicals as rational exponents when useful. - Apply the outer power to every factor in a product.

a) \((-\sqrt[4]{10})^4=(\sqrt[4]{10})^4=10\) because the outer exponent is even. b) \((-\sqrt[5]{3})^5=-(\sqrt[5]{3})^5=-3\) because the outer exponent is odd. c) \((-2\sqrt{x})^2=(-2)^2(\sqrt{x})^2=4x\). d) \((\sqrt[3]{y^2})^6=(y^{\frac{2}{3}})^6=y^4\).

a) \(10\) b) \(-3\) c) \(4x\) d) \(y^4\)

- Apply the outside exponent to every factor. - Rewrite a radical as a rational exponent while simplifying. - Add exponents when multiplying powers with the same base. - Convert any remaining fractional exponent back to a radical.

1) \((5a^2\sqrt{a})^2=5^2(a^2)^2(\sqrt{a})^2=25a^5\). 2) \((b^3\sqrt[3]{b})^2=b^6(\sqrt[3]{b})^2=b^6\sqrt[3]{b^2}\).

1) \(25a^5\) 2) \(b^6\sqrt[3]{b^2}\)

- A radical followed by the matching power returns its nonnegative radicand. - Simplify the expression inside each radical. - Check that the variable condition makes the radical defined.

a) \((\sqrt{3.5\cdot4+11})^2=3.5\cdot4+11=14+11=25\). b) \((\sqrt[3]{5^2+2})^3=5^2+2=25+2=27\). c) Because \(x\geq0\), the radicand is defined and nonnegative. Therefore \((\sqrt{x+\sqrt{x}})^2=x+\sqrt{x}\).

a) \(25\) b) \(27\) c) \(x+\sqrt{x}\)

- Apply an outer power to every factor in a product. - Multiply exponents when raising a power to a power. - Simplify the numerator and denominator separately before dividing.

a) \((3\sqrt{y+2})^2=3^2(\sqrt{y+2})^2=9(y+2)=9y+18\). b) \((k^2\sqrt[3]{k})^3=(k^2)^3(\sqrt[3]{k})^3=k^6\cdot k=k^7\). c) \(\frac{(\sqrt{50})^2-(\sqrt{2})^2}{(\sqrt[3]{4})^3}=\frac{50-2}{4}=12\).

a) \(9y+18\) b) \(k^7\) c) \(12\)

- Work from the inner radical outward. - Add exponents when multiplying powers with the same base. - Multiply exponents when applying the outer radical. - Rewrite \(8\) as a power of \(2\).

a) \(\sqrt[3]{a^2\sqrt[4]{a}}=\left(a^2a^{\frac{1}{4}}\right)^{\frac{1}{3}}=a^{\left(2+\frac{1}{4}\right)\frac{1}{3}}=a^{\frac{3}{4}}\). b) Since \(8=2^3\), \(\sqrt{8\sqrt[3]{2}}=\left(2^3\cdot2^{\frac{1}{3}}\right)^{\frac{1}{2}}=2^{\left(3+\frac{1}{3}\right)\frac{1}{2}}=2^{\frac{5}{3}}\).

a) \(a^{\frac{3}{4}}\) b) \(2^{\frac{5}{3}}\)

- Rewrite radicals as rational exponents. - Multiply exponents in a power raised to another power. - A negative exponent takes the reciprocal. - Simplify the expression inside parentheses before applying the outside exponent.

a) \((\sqrt{7})^{-2}=(7^{\frac{1}{2}})^{-2}=7^{-1}=\frac{1}{7}\). b) \((\sqrt[3]{a^6})^{\frac{1}{2}}=(a^2)^{\frac{1}{2}}=a\) because \(a\geq0\). c) \(\sqrt[3]{\frac{1}{64}}=\frac{1}{4}\), so \(\left(\frac{1}{4}\right)^{-2}=4^2=16\).

a) \(\frac{1}{7}\) b) \(a\) c) \(16\)

- Rewrite each radical as a rational exponent. - Add exponents when multiplying powers with the same base. - Subtract the denominator exponent. - Write a variable with no visible exponent as a first power.

a) \(\frac{x^3}{\sqrt[5]{x^2}}=x^{3-\frac{2}{5}}=x^{\frac{13}{5}}\). b) \(\frac{\sqrt{a}\sqrt[3]{a}}{a}=a^{\frac{1}{2}+\frac{1}{3}-1}=a^{-\frac{1}{6}}\).

a) \(x^{\frac{13}{5}}\) b) \(a^{-\frac{1}{6}}\)

- Compare the base with \(1\). - Consider whether the exponent is positive, negative, or zero. - Recall the meaning of a negative exponent. - Recall the zero-exponent rule.

1. In part a), the base is greater than \(1\) and the exponent is positive, so the value is greater than \(1\). 2. In part b), the base is between \(0\) and \(1\) and the exponent is positive, so the value is less than \(1\). 3. In part c), a negative exponent takes the reciprocal: \(\left(\frac{1}{2}\right)^{-3}=2^3\), so the value is greater than \(1\). 4. In part d), \(\left(\frac{5}{4}\right)^{-0.2}=\frac{1}{\left(\frac{5}{4}\right)^{0.2}}\). The denominator is greater than \(1\), so the value is less than \(1\). 5. In part e), any nonzero base raised to the zero power equals \(1\). Since \(\sqrt{2}-1\ne0\), the value equals \(1\).

a) Greater than \(1\): the base is greater than \(1\) and the exponent is positive. b) Less than \(1\): the base is between \(0\) and \(1\) and the exponent is positive. c) Greater than \(1\): \(\left(\frac{1}{2}\right)^{-3}=2^3\). d) Less than \(1\): the negative exponent makes the value the reciprocal of a number greater than \(1\). e) Equal to \(1\): the nonzero base is raised to the zero power.

- Use a positive base other than \(1\) so equality of powers determines equality of exponents. - The exponent law turns the square of \(a^p\) into a power whose exponent involves \(p\). - After finding one value of \(p\), check why that same exponent also works when the base is \(1\).

1. For any positive \(a\ne1\), the required square is \(a\), so \(a^{2p}=a^1\). Equal powers of the same positive base other than \(1\) have equal exponents, so \(2p=1\) and \(p=\frac{1}{2}\). 2. This value of \(p\) is independent of the chosen base. Thus \((a^{\frac{1}{2}})^2=a\) for every \(a>0\), and \(a^{\frac{1}{2}}\) is positive. By the definition of the principal square root, \(a^{\frac{1}{2}}=\sqrt{a}\). For \(a=1\), both sides equal \(1\).

a) \(p=\frac{1}{2}\) b) For any \(a\ne1\), \(a^{2p}=a^1\) forces \(2p=1\). The same exponent \(p=\frac{1}{2}\) therefore works for every positive base, and the positive number \(a^{\frac{1}{2}}\) whose square is \(a\) is \(\sqrt{a}\). When \(a=1\), both sides are \(1\).

- Start with a positive base other than \(1\) so the exponent can be determined uniquely. - Translate “whose \(n\)th power is \(a\)” into an equation involving \((a^p)^n\). - After finding \(p\), connect the positive solution of the power equation to the principal \(n\)th root.

1. For any positive \(a\ne1\), the requirement is \((a^p)^n=a\). By the power-of-a-power law, \((a^p)^n=a^{np}\), so \(a^{np}=a^1\). 2. Because the base is positive and not \(1\), equal powers have equal exponents. Thus \(np=1\), so \(p=\frac{1}{n}\). 3. Therefore \((a^{\frac{1}{n}})^n=a\) for every positive \(a\). Since \(a^{\frac{1}{n}}\) is positive, it is the principal \(n\)th root of \(a\), so \(a^{\frac{1}{n}}=\sqrt[n]{a}\). When \(a=1\), both sides equal \(1\).

a) \(p=\frac{1}{n}\) b) \(a^{\frac{1}{n}}=\sqrt[n]{a}\). For any \(a\ne1\), the exponent law forces \(np=1\); the same exponent then works for all positive \(a\), including \(a=1\).

- Think about which part of \(\frac{m}{n}\) came from the relationship \(a^{1/n}=\sqrt[n]{a}\). - Separate the root operation from the outside power before evaluating. - Check whether the proposed radical form has rational exponent \(\frac23\) when converted back.

1. In a rational exponent \(\frac{m}{n}\), the denominator \(n\) gives the root index and the numerator \(m\) gives the power. 2. Therefore \(64^{2/3}\) can be written as \((\sqrt[3]{64})^2\), not \((\sqrt{64})^3\). 3. Since \(\sqrt[3]{64}=4\), the value is \(4^2=16\).

Devin reversed the roles of the numerator and denominator. The denominator \(3\) gives the cube root and the numerator \(2\) gives the square, so \(64^{\frac{2}{3}}=(\sqrt[3]{64})^2=16\).

- Start from the equation by undoing the fourth power while respecting that \(y\) is positive. - Convert the resulting root operation into an exponent before combining powers. - The two parts of a rational exponent record different operations rather than two unrelated numbers.

1. Because \(y>0\), taking the principal fourth root gives \(y=(7^3)^{1/4}\). 2. Using the power-of-a-power law, \((7^3)^{1/4}=7^{3/4}\). 3. In radical notation this is \(\sqrt[4]{7^3}\). 4. Using \(7^{1/4}=\sqrt[4]{7}\), the same value is \((\sqrt[4]{7})^3\). 5. In \(\frac34\), the denominator \(4\) records the fourth root and the numerator \(3\) records the third power.

a) \(y=7^{\frac{3}{4}}\) b) \(y=\sqrt[4]{7^3}\) c) \(y=(\sqrt[4]{7})^3\) The denominator \(4\) gives the root index, and the numerator \(3\) gives the power.

- Compare a fractional exponent with the way a root is written. - How could you check a proposed value by raising it to a whole-number power?

1. The denominator \(4\) means take the fourth root of \(16\). 2. The numerator \(3\) means cube that root. 3. Since \(\sqrt[4]{16}=2\), the value is \(2^3=8\).

The \(4\) means fourth root and the \(3\) means third power; the value is \(8\).

- Look at the denominator of the exponent before the numerator. - Decide whether the required root is odd or even.

1. Take the cube root first: \(\sqrt[3]{-8}=-2\). 2. Square it: \((-2)^2=4\). 3. An odd root can take a negative base in the real numbers.

\(4\); the cube root of a negative number is real.

- First find where the square root is real. - Then ask whether the root may appear in a denominator.

1. The square root \(x^{\frac{1}{2}}\) is real when \(x\ge0\). 2. The second expression equals \(\frac{1}{\sqrt{x}}\), so it also requires a nonzero denominator. 3. Thus it is defined when \(x>0\).

a) \(x\ge0\) b) \(x>0\) The reciprocal in b) excludes zero.

- Try a negative value of \(x\). - Remember that the principal square root cannot be negative.

1. Set \(x=-3\): \(\sqrt{(-3)^2}=\sqrt9=3\), not \(-3\). 2. The principal square root is nonnegative, so \(\sqrt{x^2}=|x|\).

For example, \(x=-3\) disproves the claim. The correct rule is \(\sqrt{x^2}=|x|\).

- What familiar number has a third power of \(27\)? - Check a small integer value for \(m\) before deciding if it works.

1. The cube root of \(27\) is \(3\). 2. We need \(3^m=9\), so \(m=2\).

\(m=2\), since \(27^{\frac{1}{3}}=3\) and \(3^2=9\).

- Evaluate both powers at the same positive input. - The denominator tells you what root to take first.

1. At \(x=4\), \(x^{\frac{1}{2}}=2\) while \(x^{\frac{3}{2}}=8\). 2. Curve p has the smaller value at that input, so p is \(y=x^{\frac{1}{2}}\). 3. Curve q is \(y=x^{\frac{3}{2}}\).

p: \(y=x^{\frac{1}{2}}\), which has value \(2\) at \(x=4\). q: \(y=x^{\frac{3}{2}}\), which has value \(8\) at \(x=4\).

- Decide which operation accepts negative inputs in the real numbers. - Compare how far each curve extends on either side of the vertical axis.

1. A square root is not real for negative inputs, so its graph has no points to the left of the y-axis. 2. Curve s begins at the origin and extends right only, so it shows \(y=\sqrt{x}\). 3. In rational-exponent form, the rule is \(y=x^{\frac{1}{2}}\).

Curve s; \(y=x^{\frac{1}{2}}\).

- Rewrite a square root as an exponent of \(\frac{1}{2}\). - Multiply exponents for a power of a power. - Use the positivity condition to avoid sign ambiguity.

Rewrite the square root using a rational exponent: \(\sqrt{a^{4m}}=(a^{4m})^{\frac{1}{2}}=a^{2m}\). a) \(\sqrt{3^4}=3^2=9\). b) \(\sqrt{2^{12}}=2^6=64\). c) \(\sqrt{x^{16}}=x^8\) for \(x>0\).

In general, \(\sqrt{a^{4m}}=a^{2m}\). a) \(9\) b) \(64\) c) \(x^8\)

- Test a positive number, a negative number, and \(0\) for \(x\). - Recall that the principal square root is never negative. - When is a square root defined in the real number system?

a) The principal square root is always nonnegative, so \(\sqrt{x^2} = |x|\). Therefore, \(\sqrt{x^2} = x\) only when \(x \geq 0\). b) The expression \(\sqrt{x}\) is real only when \(x \geq 0\). For every such value, \((\sqrt{x})^2 = x\). Therefore, the statement is true only when \(x \geq 0\). c) For \(\sqrt{-x}\) to be real, the radicand must satisfy \(-x \geq 0\). This is equivalent to \(x \leq 0\).

a) True only for \(x \geq 0\). b) True only for \(x \geq 0\). c) True only for \(x \leq 0\).

- What does taking a root do to a power when the base is positive? - Which quantities on the two sides must agree if an equation holds for every positive \(x\)? - In d), consider how multiplying powers of the same base changes their exponents.

1. In a), \(x^{\frac{n}{2}}=x^7\), so \(n=14\). 2. In b), since \(81=3^4\), \(3^{\frac{n}{2}}=3^4\), so \(n=8\). 3. In c), \(\sqrt[3]{x^{12}}=x^4\), so \(n=4\). 4. In d), \(\sqrt{x^{4+n}}=x^{\frac{4+n}{2}}=x^5\), so \(4+n=10\) and \(n=6\).

a) \(n=14\) b) \(n=8\) c) \(n=4\) d) \(n=6\)

- Rewrite a square root using exponent \(\frac{1}{2}\). - Use the power-of-a-power property. - Rewrite the decimal as a power of ten before taking its square root. - Taking a square root divides an exponent by \(2\).

1. A square root can be written using exponent \(\frac{1}{2}\). Therefore, \(\sqrt{10^{2n}}=(10^{2n})^{\frac{1}{2}}=10^{2n\cdot\frac{1}{2}}=10^n\). 2. a) \(\sqrt{10^6}=10^3=1000\). b) \(\sqrt{10^{-4}}=10^{-2}=0.01\). c) \(0.000001=10^{-6}\), so \(\sqrt{0.000001}=10^{-3}=0.001\).

1. \(\sqrt{10^{2n}}=(10^{2n})^{\frac{1}{2}}=10^n\) 2. a) \(1000\) b) \(0.01\) c) \(0.001\)

- A principal square root is always nonnegative. - Use \(\sqrt{u^2}=|u|\) when a square is involved. - Track the sign of a negative base raised to an even or odd exponent.

1. For \(x=4\) and \(k=2\), \(\sqrt{4^2}=4=4^{\frac{2}{2}}\), so the claim works in this case. 2. For \(x=-3\) and \(k=2\), \(\sqrt{(-3)^2}=3\), but \((-3)^{\frac{2}{2}}=-3\). The claim is false. 3. If \(x\ge0\), then \(\sqrt{x^k}=x^{\frac{k}{2}}\). 4. If \(x<0\) and \(k\) is even, then \(\sqrt{x^k}=|x|^{\frac{k}{2}}\). 5. If \(x<0\) and \(k\) is odd, then \(x^k<0\), so \(\sqrt{x^k}\) is not a real number.

a) The claim works: \(\sqrt{4^2}=4\). b) The claim fails: \(\sqrt{(-3)^2}=3\ne-3\). c) For \(x\ge0\), \(\sqrt{x^k}=x^{\frac{k}{2}}\). For \(x<0\) and even \(k\), \(\sqrt{x^k}=|x|^{\frac{k}{2}}\). For \(x<0\) and odd \(k\), the expression is not real.

- Undo a radical or power by applying the inverse exponent. - Rewrite \(243\) as a power of \(3\). - Take the cube root of the numerator and denominator separately in part d.

a) Raise both sides to the fourth power: \(x=3^4=81\). b) Raise both sides to the power \(\frac{2}{3}\): \(x=64^{\frac{2}{3}}=16\). c) Rewrite the equation as \(3^x=243\). Since \(243=3^5\), \(x=5\). d) Take the cube root: \(x=\sqrt[3]{\frac{27}{1000}}=\frac{3}{10}=0.3\).

a) \(x=81\) b) \(x=16\) c) \(x=5\) d) \(x=0.3\)

- Rewrite radicals as rational exponents. - Add or subtract exponents for products and quotients with the same base. - Use \(a>0\) when simplifying \(\sqrt{a^2}\).

1. \(\sqrt[3]{a^6}=a^2\), so expression 1 is equivalent. 2. \(\sqrt a\cdot a^{1.5}=a^{0.5+1.5}=a^2\), so expression 2 is equivalent. 3. \((a^4)^{\frac14}=a\), so expression 3 is not equivalent. 4. Since \(a>0\), \(\sqrt{a^2}=a\). Thus \(\frac{a^3}{a}=a^2\), so expression 4 is equivalent.

Expressions 1, 2, and 4 are equivalent to \(a^2\). Expression 3 equals \(a\).

- Isolate the power of \(x\) first. - Undo the power by taking the corresponding root. - An even power can lead to two real solutions, while an odd power has one real solution. - Round only after finding the exact value.

a) Add \(25\) and divide by \(3\): \(x^4=81\). Because the exponent is even, \(x=\pm\sqrt[4]{81}=\pm3\), so \(x\approx\pm3.00\). b) Subtract \(10\) and multiply by \(-2\): \(x^3=-80\). Thus \(x=\sqrt[3]{-80}=-\sqrt[3]{80}\approx-4.31\). c) Subtract \(7\) and divide by \(0.2\): \(x^5=-10\). Thus \(x=\sqrt[5]{-10}=-\sqrt[5]{10}\approx-1.58\).

a) \(x=\pm3\), so \(x\approx\pm3.00\) b) \(x=-\sqrt[3]{80}\approx-4.31\) c) \(x=-\sqrt[5]{10}\approx-1.58\)

- Isolate the power in each equation. - In part b, determine what the right side must be after isolating \(x^6\) so that \(x=0\) is the only solution. - In part c, divide by the factor outside the parentheses before taking a root.

a) Add \(20\) and divide by \(5\): \(x^6=24\). Because the exponent is even, \(x=\pm\sqrt[6]{24}\approx\pm1.70\). b) The only way an even power equals zero is when \(x=0\). Set \(x^6=0\): from \(5x^6-20=c\), this gives \(c=-20\). c) Divide by \(2\) and add \(5\): \(x^3=25\). Therefore, \(x=\sqrt[3]{25}\).

a) \(x\approx\pm1.70\) b) \(c=-20\) c) \(x=\sqrt[3]{25}\)

- Evaluate perfect roots when possible. - For two positive quantities, squaring can make a comparison easier. - Pay close attention to decimal place value.

1. \(\sqrt{0.49}=0.7\), and \(0.7>0.07\). 2. \(\sqrt[3]{27}=3\) and \(\sqrt{9}=3\), so the values are equal. 3. \(\sqrt[4]{\frac{1}{16}}=\frac{1}{2}\), and \(\frac{1}{2}>\frac{1}{4}\). 4. Both values are positive. Since \((4.5)^2=20.25>20\), \(\sqrt{20}<4.5\).

a) \(>\) b) \(=\) c) \(>\) d) \(<\)

- Isolate the power of \(x\). - Then take the corresponding root. - Check the sign of the isolated value and whether the exponent is even or odd.

a) Add \(32\) and divide by \(2\): \(x^4=16\). Therefore, \(x=\pm2\). b) Subtract \(10\) and multiply by \(3\): \(x^3=-27\). Therefore, \(x=-3\). c) Subtract \(12\) and divide by \(5\): \(x^6=-1\). An even power cannot be negative for real \(x\), so there is no real solution.

a) \(x=\pm2\) b) \(x=-3\) c) No real solution

- Rewrite familiar decimals or fractions as powers when useful. - Include both signs when an even power equals a positive number. - For part d, isolate \(x^5\) before taking the fifth root.

a) Since \(\left(\frac{1}{3}\right)^4=\frac{1}{81}\) and the exponent is even, \(x=\pm\frac{1}{3}\). b) Since \(3^6=729\) and the exponent is even, \(x=\pm3\). c) Since \((-0.5)^3=-0.125\), \(x=-0.5\). d) Subtract \(64\) and divide by \(2\): \(x^5=-32\). Since \((-2)^5=-32\), \(x=-2\).

a) \(\left\{-\frac{1}{3},\frac{1}{3}\right\}\) b) \(\{-3,3\}\) c) \(\{-0.5\}\) d) \(\{-2\}\)

- Write the positive solution as \(100^{\frac{1}{n}}\). - Think about how a base greater than \(1\) behaves when its exponent increases. - For part c, test consecutive positive even exponents and use the decreasing pattern to know when to stop.

a) When \(n=2\), \(x=\pm\sqrt{100}=\pm10\). When \(n=4\), \(x=\pm\sqrt[4]{100}=\pm\sqrt{10}\approx\pm3.16\). b) The positive solution is \(100^{\frac{1}{n}}\). As \(n\) increases, a number greater than \(1\) needs a smaller base to produce the same value \(100\). Therefore, the positive solution decreases and approaches \(1\). c) Test positive even exponents: \(100^{\frac{1}{6}}\approx2.15\), \(100^{\frac{1}{8}}\approx1.78\), \(100^{\frac{1}{10}}\approx1.58\), and \(100^{\frac{1}{12}}\approx1.47\). Since the positive solution decreases as \(n\) increases, only \(n=8\) and \(n=10\) give values between \(1.5\) and \(2.0\).

a) For \(n=2\), \(x=\pm10\). For \(n=4\), \(x=\pm\sqrt{10}\approx\pm3.16\). b) The positive solution decreases toward \(1\). c) \(n=8\) and \(n=10\)

- Relate the number of real solutions to whether \(n\) is even or odd. - For the second part, substitute both \(2\) and \(-2\) into \(x^n\). - In the last part, isolate the power term before taking a root.

1. To make \(x=-3\) the only real solution, \(n\) must be odd and \(c=(-3)^n\). Two possible pairs are \((n, c)=(1, -3)\) and \((n, c)=(3, -27)\). 2. To have the symmetric solutions \(x=\pm2\), \(n\) must be a positive even integer and \(c=2^n\). For example, \(n=2\) and \(c=4\). 3. Subtract \(10\): \(0.5x^3=500\). Divide by \(0.5\): \(x^3=1000\). Take the cube root: \(x=10\).

1. Sample pairs: \((n, c)=(1, -3)\) and \((n, c)=(3, -27)\) 2. \(n\) must be a positive even integer and \(c=2^n\). For example, \(n=2\) and \(c=4\). 3. \(x=10\)

- Isolate the power term first. - Use inverse operations in reverse order. - When the exponent is even and the isolated value is positive, include both signs.

a) Divide by \(3\): \(x^4=16\). Therefore, \(x=\pm2\). b) Subtract \(32\) and divide by \(4\): \(x^3=-8\). Therefore, \(x=-2\). c) Add \(12\) and multiply by \(2\): \(x^5=32\). Therefore, \(x=2\). d) Subtract \(100\) and divide by \(-1\): \(x^2=81\). Therefore, \(x=\pm9\).

a) \(x=\pm2\) b) \(x=-2\) c) \(x=2\) d) \(x=\pm9\)

- Add exponents when multiplying powers with the same base. - Combine cube roots in a quotient. - Rewrite \(81\) as a power of \(3\).

a) \(x^{\frac{1}{2}}x^{\frac{1}{4}}x^{\frac{1}{4}}=x^1=x\). b) \(\frac{\sqrt[3]{y^5}}{\sqrt[3]{y^2}}=\sqrt[3]{y^3}=y\). c) Since \(81=3^4\), \(\sqrt[4]{81^3}=\sqrt[4]{3^{12}}=3^3\). Dividing by \(3^2\) gives \(3\).

a) \(x\) b) \(y\) c) \(3\)

- Combine radicals with the same index. - Interpret \(a^{\frac{m}{n}}\) as an \(n\)th root raised to the \(m\)th power. - Evaluate the root before the remaining power when that keeps the numbers small.

a) \(A=\sqrt[3]{2\cdot4}=\sqrt[3]{8}=2\), while \(B=\sqrt{2\cdot8}=\sqrt{16}=4\). Therefore, \(B>A\). b) \(C=(\sqrt[3]{27})^2=3^2=9\), while \(D=(\sqrt[4]{16})^3=2^3=8\). Therefore, \(C>D\).

a) \(B>A\), because \(B=4\) and \(A=2\). b) \(C>D\), because \(C=9\) and \(D=8\).

- What power reverses the third power in the mass formula? - Make sure the mass unit matches the formula before calculating. - When a mass changes, can you compare the two height formulas without finding either height separately?

1. Rearrange \(m=0.05h^3\) to \(h^3=\frac{m}{0.05}\). Thus \(h=\left(\frac{m}{0.05}\right)^{\frac{1}{3}}\) for positive masses. 2. Convert \(3.2\,\text{kg}=3200\,\text{g}\); then \(h=\left(\frac{3200}{0.05}\right)^{\frac{1}{3}}=\left(64{,}000\right)^{\frac{1}{3}}=40\,\text{cm}\). 3. If \(m\) doubles, the new-to-old height ratio is \(\left(\frac{2m}{m}\right)^{\frac{1}{3}}=2^{\frac{1}{3}}\approx1.26\).

a) \(h=\left(\frac{m}{0.05}\right)^{\frac{1}{3}}\), with \(m\) in grams and \(h\) in centimeters. b) \(40\,\text{cm}\). c) \(2^{\frac{1}{3}}\approx1.26\) times the original height.

- In part a, isolate \(v^3\) before taking a cube root. - In part b, replace \(v\) by \(0.5v\) and compare the resulting expression with the original formula.

a) Substitute \(P=500\): \(500=0.2v^3\). Divide by \(0.2\): \(v^3=2500\). Therefore, \(v=\sqrt[3]{2500}\approx13.572\), which rounds to \(13.6\,\text{m/s}\). b) Replace \(v\) with \(0.5v\): \(P_{\text{new}}=0.2(0.5v)^3=0.125(0.2v^3)=0.125P_{\text{original}}\). The new power is \(12.5\%\) of the original power, so the decrease is \(100\%-12.5\%=87.5\%\).

a) \(13.6\,\text{m/s}\) b) The power decreases by \(87.5\%\).

- Convert the mass to grams so its unit matches the density. - How is a cube's volume related to its edge length? - Which fractional power lets you recover that edge length from the volume? - For part b, determine the volume scale factor when each dimension doubles.

1. Convert the mass: \(20\,\text{kg}=20{,}000\,\text{g}\). 2. Use \(V=\frac{m}{\rho}\): \(V=\frac{20000}{7.87}\approx2541.30\,\text{cm}^3\). 3. For a cube, \(V=a^3\), so \(a=\left(\frac{m}{\rho}\right)^{\frac{1}{3}}=\left(\frac{20000}{7.87}\right)^{\frac{1}{3}}\approx13.65\,\text{cm}\). 4. Doubling every edge multiplies volume and mass by \(2^3=8\). The new mass is \(8\cdot20\,\text{kg}=160\,\text{kg}\).

a) \(a=\left(\frac{m}{\rho}\right)^{\frac{1}{3}}\approx13.65\,\text{cm}\) b) \(160\,\text{kg}\)

- Find the original volume from the edge length, then use density to calculate the original mass. - Add the extra mass, then use density to find the new volume. - Take a cube root, equivalently a \(\frac{1}{3}\) power, to find the new edge length. - Compare the change in edge length with the original edge length to find the percent increase.

1. The original volume is \(V_1=(5.0\,\text{cm})^3=125\,\text{cm}^3\). Using the stated approximate density, the original mass is \(m_1\approx2412.5\,\text{g}\). 2. The new mass is \(m_2\approx2912.5\,\text{g}\), so the new volume is \(V_2\approx\frac{2912.5}{19.3}\approx150.9067\,\text{cm}^3\). 3. The new edge length is \(a_2=V_2^{\frac{1}{3}}\approx5.32398\,\text{cm}\), which rounds to \(5.32\,\text{cm}\). 4. Using the unrounded edge length, the percent increase is \(\frac{a_2-5.0}{5.0}\cdot100\%\approx6.4795\%\), which rounds to \(6.5\%\).

a) \(\approx2412.5\,\text{g}\) b) \(\approx5.32\,\text{cm}\); \(\approx6.5\%\) increase

- How does the volume formula simplify when the height equals the radius? - Convert liters to cubic centimeters before solving. - How could you check a proposed formula for \(r\) by substituting it into \(V\)? - In part c), how many times does the radius occur as a factor in the volume formula?

1. The volume of a cylinder is \(V=\pi r^2h\). Since \(h=r\), \(V=\pi r^3\). 2. Convert volume: \(1000\,\text{L}=1{,}000{,}000\,\text{cm}^3\). 3. Solve \(1{,}000{,}000=\pi r^3\): \(r=\left(\frac{1000000}{\pi}\right)^{\frac{1}{3}}\approx68.28\,\text{cm}\). 4. Replacing \(r\) by \(3r\) gives \(V_{\text{new}}=\pi(3r)^3=27\pi r^3=27V\).

a) \(V=\pi r^3\) b) \(r=\left(\frac{V}{\pi}\right)^{\frac{1}{3}}\approx68.28\,\text{cm}\) for \(V=1{,}000{,}000\,\text{cm}^3\). c) The volume is multiplied by \(27\), because \(V=\pi r^3\) and replacing \(r\) with \(3r\) multiplies the volume by \(3^3\).

- Convert decimal exponents to fractions. - A negative exponent takes the reciprocal. - Evaluate each exact value before comparing.

1. \(A=16^{\frac{1}{4}}=2\). 2. \(B=9^{\frac{1}{2}}=3\). 3. \(C=(\sqrt[3]{27})^2=9\). 4. \(D=\frac{1}{\sqrt{25}}=\frac{1}{5}=0.2\). 5. Therefore, \(D<A<B<C\).

\(D<A<B<C\)

- Rewrite all radicals as rational exponents. - Use a common denominator for the exponent fractions. - Add exponents in the numerator and subtract the denominator exponent.

1. Rewrite each radical as a rational exponent: \(\frac{a^{\frac{2}{3}}a^{\frac{1}{4}}}{a^{\frac{5}{12}}}\). 2. Combine exponents: \(\frac{2}{3}+\frac{1}{4}-\frac{5}{12}=\frac{8}{12}+\frac{3}{12}-\frac{5}{12}=\frac{1}{2}\). 3. The expression is \(a^{\frac{1}{2}}=\sqrt a\).

Exponent form: \(a^{\frac{1}{2}}\) Radical form: \(\sqrt a\)

- Rewrite every radical as a rational exponent. - Reduce each exponent fraction. - With the same positive base, compare the exponents.

a) \(\sqrt[6]{x^9}=x^{\frac{9}{6}}=x^{\frac{3}{2}}\), and \(\sqrt{x^3}=x^{\frac{3}{2}}\). The statement is true. b) \(\sqrt[4]{x^2}=x^{\frac{1}{2}}\), while \(\sqrt[8]{x^6}=x^{\frac{3}{4}}\). The statement is false. c) \(\frac{1}{\sqrt[3]{x}}=x^{-\frac{1}{3}}\), and \(\sqrt[6]{x^{-2}}=x^{-\frac{2}{6}}=x^{-\frac{1}{3}}\). The statement is true.

a) True b) False c) True

- For the same radicand greater than \(1\), compare the radical indices. - For the same index, compare the radicands. - Use a common index to compare \(a\) and \(d\).

1. With the same base greater than \(1\), a larger radical index gives a smaller value. Thus \(b<a\) and \(d<c\). 2. With the same radical index, the larger radicand gives the larger value. Thus \(b<d\) and \(a<c\). 3. Compare \(a\) and \(d\) using index \(12\): \(a=\sqrt[12]{625}\) and \(d=\sqrt[12]{1000}\). Since \(625<1000\), \(a<d\). 4. Therefore, \(b<a<d<c\). Numerically, \(b\approx1.50\), \(a\approx1.71\), \(d\approx1.78\), and \(c\approx2.15\).

\(b<a<d<c\), with approximate values \(1.50<1.71<1.78<2.15\)

- Combine radicals with matching indices in a quotient or product. - A root of a root multiplies the radical indices. - Rewrite the decimal as a fraction before applying the negative exponent.

a) \(\frac{\sqrt[3]{2^7}}{\sqrt[3]{2}}=\sqrt[3]{2^6}=2^2=4\). b) \(\sqrt{\sqrt[3]{729}}=\sqrt[6]{729}=3\), because \(3^6=729\). c) \((8\cdot27)^{\frac{1}{3}}=8^{\frac{1}{3}}27^{\frac{1}{3}}=2\cdot3=6\). d) \(0.5^{-2}=\left(\frac{1}{2}\right)^{-2}=4\), and \(\sqrt[4]{16}=2\). The product is \(8\).

a) \(4\) b) \(3\) c) \(6\) d) \(8\)

- Use \(2\) as the common base. - Rewrite radicals and reciprocals as exponents. - For a base greater than \(1\), larger exponents give larger values.

1. \(T_1=\sqrt[4]{2^2}=2^{\frac{1}{2}}\). 2. \(T_2=\left(\frac{1}{2}\right)^{-1}=2^1\). 3. \(T_3=2^{-\frac{1}{2}}\). 4. \(T_4=(2^2)^{\frac{1}{3}}=2^{\frac{2}{3}}\). 5. Since \(-\frac{1}{2}<\frac{1}{2}<\frac{2}{3}<1\), the order is \(T_3<T_1<T_4<T_2\).

\(T_3<T_1<T_4<T_2\)

- Rewrite every factor as a power of \(x\). - Add numerator exponents and subtract the denominator exponent. - Use a common denominator for the exponent fractions.

1. Rewrite the radicals: \(\sqrt[3]{x^2}=x^{\frac{2}{3}}\) and \(\sqrt x=x^{\frac{1}{2}}\). 2. Combine exponents: \(x^{\frac{2}{3}+\frac{1}{2}-1}=x^{\frac{4}{6}+\frac{3}{6}-\frac{6}{6}}=x^{\frac{1}{6}}\). 3. For \(x=64\), the original expression is \(\frac{16\cdot8}{64}=2\), and the simplified expression is \(64^{\frac{1}{6}}=2\).

The simplified expression is \(x^{\frac{1}{6}}\). For \(x=64\), both forms equal \(2\).

- Rewrite both sides of each equation with the same base. - A reciprocal can be written using a negative exponent. - When equal powers have the same positive base other than \(1\), their exponents are equal.

a) Rewrite both sides with base \(2\): \(2^{\frac{15}{x}}=2^3\). Thus \(\frac{15}{x}=3\), so \(x=5\). b) \(\frac{7^{\frac{5}{3}}}{7^k}=7^{\frac{5}{3}-k}=7^{\frac{2}{3}}\). Therefore, \(\frac{5}{3}-k=\frac{2}{3}\), so \(k=1\). c) \(\frac{1}{\sqrt[4]{100}}=\frac{1}{(10^2)^{\frac{1}{4}}}=10^{-\frac{1}{2}}\). Therefore, \(k=-\frac{1}{2}\).

a) \(x=5\) b) \(k=1\) c) \(k=-\frac{1}{2}\)

- Rewrite negative exponents as reciprocals, or reciprocals as negative exponents. - Combine factors with the same base before simplifying. - Treat a parenthesized expression as one base. - Reduce exponents that contain the variable \(n\).

a) Since \(\frac{1}{x^{-6}}=x^6\), \(\sqrt[3]{\frac{27}{x^{-6}}}=\sqrt[3]{3^3x^6}=3x^2\). b) \(\sqrt[n]{a^{3n}b^{2n}}=a^3b^2\), so \(\frac{1}{\sqrt[n]{a^{3n}b^{2n}}}=a^{-3}b^{-2}\). c) \(\frac{\sqrt[3]{x^2}\sqrt[3]{x^7}}{x^2}=\frac{\sqrt[3]{x^9}}{x^2}=\frac{x^3}{x^2}=x\). d) Because \(a-b>0\), \(\sqrt[4]{(a-b)^{12}}=(a-b)^{\frac{12}{4}}=(a-b)^3\).

a) \(3x^2\) b) \(a^{-3}b^{-2}\) c) \(x\) d) \((a-b)^3\)

- Multiply the exponents in a power raised to another power. - Rewrite every radical as a rational exponent. - Recall the value of a nonzero number raised to the zero power. - Identify the cube root of \(27\).

a) \(\left(z^{\frac{2}{3}}\right)^{-\frac{3}{2}}=z^{\frac{2}{3}\cdot\left(-\frac{3}{2}\right)}=z^{-1}=\frac{1}{z}\). b) \(\frac{\sqrt{x}\sqrt[3]{x}}{\sqrt[6]{x^5}}=x^{\frac{1}{2}+\frac{1}{3}-\frac{5}{6}}=x^{\frac{3}{6}+\frac{2}{6}-\frac{5}{6}}=x^0=1\). c) \((27a^6)^{\frac{1}{3}}=27^{\frac{1}{3}}a^{\frac{6}{3}}=3a^2\).

a) \(\frac{1}{z}\) b) \(1\) c) \(3a^2\)

- For nested radicals, simplify from the inside out. - Rewrite \(8\) as a power that works well with an exponent of \(\frac{2}{3}\). - A variable in the denominator has an exponent of \(-1\) when moved to the numerator. - Express every factor with the same base before combining exponents.

a) \(\sqrt{a\sqrt[3]{a}}=\left(a^1a^{\frac{1}{3}}\right)^{\frac{1}{2}}=\left(a^{\frac{4}{3}}\right)^{\frac{1}{2}}=a^{\frac{2}{3}}\). b) \((8x^3)^{\frac{2}{3}}=8^{\frac{2}{3}}x^2=4x^2\), so \(\frac{(8x^3)^{\frac{2}{3}}}{4x}=\frac{4x^2}{4x}=x\). c) \(\frac{z^{-\frac{1}{2}}\sqrt[4]{z^6}}{z}=z^{-\frac{1}{2}+\frac{6}{4}-1}=z^{-\frac{2}{4}+\frac{6}{4}-\frac{4}{4}}=z^0=1\).

a) \(a^{\frac{2}{3}}\) b) \(x\) c) \(1\)

- Rewrite nested radicals as powers and multiply the exponents. - Expand the square of a binomial carefully. - Opposite exponents add to zero when their powers are multiplied. - Express all factors in part c) as powers of \(z\).

a) \(\sqrt{x\sqrt{x}}=\left(x\cdot x^{\frac{1}{2}}\right)^{\frac{1}{2}}=\left(x^{\frac{3}{2}}\right)^{\frac{1}{2}}=x^{\frac{3}{4}}\). b) \(\left(y^{\frac{1}{3}}+y^{-\frac{1}{3}}\right)^2=y^{\frac{2}{3}}+2y^{\frac{1}{3}}y^{-\frac{1}{3}}+y^{-\frac{2}{3}}=y^{\frac{2}{3}}+2+y^{-\frac{2}{3}}\). c) \(\frac{\sqrt[3]{z^4}z^{-1}}{\sqrt[6]{z}}=z^{\frac{4}{3}-1-\frac{1}{6}}=z^{\frac{8}{6}-\frac{6}{6}-\frac{1}{6}}=z^{\frac{1}{6}}\).

a) \(x^{\frac{3}{4}}\) b) \(y^{\frac{2}{3}}+2+y^{-\frac{2}{3}}\) c) \(z^{\frac{1}{6}}\)

- When squaring a binomial, include the middle product and square each coefficient. - Rewrite radicals as rational exponents. - Use a common denominator when adding and subtracting exponent fractions.

a) Expand the binomial: \((2\sqrt a+3\sqrt b)^2=(2\sqrt a)^2+2(2\sqrt a)(3\sqrt b)+(3\sqrt b)^2=4a+12\sqrt{ab}+9b\). b) \(\sqrt[4]{x^3}\sqrt[8]{x^2}=x^{\frac{3}{4}}x^{\frac{2}{8}}=x^{\frac{3}{4}+\frac{1}{4}}=x\). c) \(\frac{y^2\sqrt y}{\sqrt[3]{y^2}}=y^{2+\frac{1}{2}-\frac{2}{3}}=y^{\frac{12}{6}+\frac{3}{6}-\frac{4}{6}}=y^{\frac{11}{6}}\).

a) \(4a+12\sqrt{ab}+9b\) b) \(x\) c) \(y^{\frac{11}{6}}\)

- Work from the inner radical outward. - Write \(a\) as \(a^1\). - Multiply exponents when raising a power to another power.

1. Rewrite the inner radical: \(\sqrt[3]{a^2}=a^{\frac{2}{3}}\). 2. Combine the factors inside the outer radical: \(a\cdot a^{\frac{2}{3}}=a^{\frac{5}{3}}\). 3. Apply the outer cube root: \(\left(a^{\frac{5}{3}}\right)^{\frac{1}{3}}=a^{\frac{5}{9}}\).

\(a^{\frac{5}{9}}\)

- Rewrite both radicals using rational exponents. - Subtract denominator exponents from numerator exponents. - Be careful when subtracting a negative exponent.

1. Rewrite the radicals: \(\sqrt[3]{x^2y}=x^{\frac{2}{3}}y^{\frac{1}{3}}\) and \(\sqrt{xy^{-1}}=x^{\frac{1}{2}}y^{-\frac{1}{2}}\). 2. Subtract exponents for each base: \(x^{\frac{2}{3}-\frac{1}{2}}y^{\frac{1}{3}-(-\frac{1}{2})}\). 3. Simplify the exponents: \(\frac{2}{3}-\frac{1}{2}=\frac{1}{6}\) and \(\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\). 4. The expression is \(x^{\frac{1}{6}}y^{\frac{5}{6}}\).

\(x^{\frac{1}{6}}y^{\frac{5}{6}}\)

- Solve the surface-area formula for the common variable \(a\). - Substitute that expression into the volume formula. - Rewrite the square root as a rational exponent. - Multiply exponents when raising a power to another power.

1. Solve the surface-area formula for \(a\): \(S=6a^2\), so \(a=\sqrt{\frac{S}{6}}\). 2. Substitute into the volume formula: \(V=\left(\sqrt{\frac{S}{6}}\right)^3\). 3. Rewrite the square root as a rational exponent: \(V=\left(\left(\frac{S}{6}\right)^{\frac{1}{2}}\right)^3=\left(\frac{S}{6}\right)^{\frac{3}{2}}\). 4. Therefore, \(k=6\) and \(n=\frac{3}{2}\).

\(V=\left(\frac{S}{6}\right)^{\frac{3}{2}}\), so \(k=6\) and \(n=\frac{3}{2}\).

- Replace \(V\) in the formula with eight times the original volume. - Distribute the exponent over the product. - Interpret \(8^{\frac{2}{3}}\) using a cube root and a square. - Compare the new expression with the original surface-area expression.

1. Let the original volume be \(V_1\), so \(S_1=6V_1^{\frac{2}{3}}\). 2. The new volume is \(V_2=8V_1\). Therefore, \(S_2=6(8V_1)^{\frac{2}{3}}\). 3. Apply the power-of-a-product property: \(S_2=6\cdot8^{\frac{2}{3}}V_1^{\frac{2}{3}}\). 4. Since \(8^{\frac{2}{3}}=(\sqrt[3]{8})^2=2^2=4\), \(S_2=4(6V_1^{\frac{2}{3}})=4S_1\).

The surface area, and therefore the material needed, increases by a factor of \(4\).

- Test each claim with simple numerical examples. - Consider a base between \(0\) and \(1\). - Interpret an exponent such as \(\frac{1}{2}\) as a root. - A single valid counterexample disproves a universal statement.

a) The statement is false. For example, with \(a=3\) and \(n=1\), \(3^1=3\), but doubling the exponent gives \(3^2=9\), not \(6\). b) The statement is false. It is true when \(a>1\), but not when \(0<a<1\). For example, \(0.25^{\frac{1}{2}}=0.5\), and \(0.5>0.25\).

a) False. For example, \(3^1=3\), but \(3^2=9\), not \(6\). b) False. For example, \(0.25^{\frac{1}{2}}=0.5>0.25\).

- Which operation reverses raising a length to the third power? - A cube root can also be written using an exponent of \(\frac{1}{3}\). - Represent the new edge length as \(ka\). - Translate each volume change into an equation involving \((ka)^3\).

a) Let the new edge length be \(ka\). Then \((ka)^3=\frac{1}{8}a^3\), so \(k^3=\frac{1}{8}\) and \(k=\frac{1}{2}\). b) \((ka)^3=2a^3\), so \(k^3=2\) and \(k=\sqrt[3]{2}=2^{\frac{1}{3}}\). c) A \(50\%\) reduction leaves one-half of the volume. Thus, \((ka)^3=\frac{1}{2}a^3\), so \(k=\sqrt[3]{\frac{1}{2}}=\left(\frac{1}{2}\right)^{\frac{1}{3}}=2^{-\frac{1}{3}}\).

a) \(\frac{1}{2}\) b) \(\sqrt[3]{2}=2^{\frac{1}{3}}\) c) \(\sqrt[3]{\frac{1}{2}}\), equivalently \(2^{-\frac{1}{3}}\)

- Use the relationship between a square's area and its side length. - A square root can also be written using an exponent of \(\frac{1}{2}\). - What decimal factor remains after a \(20\%\) decrease? - When a product is squared, each factor is squared.

a) Let the original area be \(A_1=s_1^2\) and the new area be \(A_2=2A_1\). Since \(A_2=s_2^2\), \(s_2^2=2s_1^2\). Taking the positive square root gives \(s_2=\sqrt{2}s_1=2^{\frac{1}{2}}s_1\). The side length must be multiplied by \(\sqrt{2}\), or equivalently \(2^{\frac{1}{2}}\). b) A \(20\%\) reduction in side length gives \(s_{\text{new}}=0.8s_{\text{old}}\). Then \(A_{\text{new}}=(0.8s_{\text{old}})^2=0.64A_{\text{old}}\). The area lost is \(100\%-64\%=36\%\).

a) Multiply the side length by \(\sqrt{2}=2^{\frac{1}{2}}\). b) \(36\%\) of the original area is lost.

- Distribute the factor outside each set of parentheses. - Add exponents when multiplying powers with the same base. - Recall that a nonzero number raised to the zero power equals \(1\). - Rewrite a negative exponent as a reciprocal when useful.

a) Distribute and add exponents on like bases: \(x^{-3}(x^5-2x^3+4x^2)=x^2-2x^0+4x^{-1}=x^2-2+\frac{4}{x}\). b) \(z^{\frac{1}{2}}(z^{\frac{3}{2}}-5z^{-\frac{1}{2}})=z^2-5z^0=z^2-5\). c) \(2a^2b^{-2}(ab^2-3b^3)=2a^3b^0-6a^2b=2a^3-6a^2b\).

a) \(x^2-2+\frac{4}{x}\) b) \(z^2-5\) c) \(2a^3-6a^2b\)

- Apply the power-of-a-product property before distributing in part a). - Rewrite radicals as rational exponents. - Add exponents on like bases after multiplying. - Multiply numerical coefficients separately from variable factors.

a) \((xy^2)^3=x^3y^6\). Distribute: \(x^3y^6(x^{-2}y^{-4}+3x^{-3}y^{-6})=xy^2+3\). b) Rewrite the radicals as powers and distribute: \(a^{\frac{1}{2}}(2a^{\frac{3}{2}}-4a^{-\frac{1}{2}}+a^{-\frac{1}{2}})=2a^2-4+1=2a^2-3\). c) Distribute the coefficient and variables: \(\frac{2}{3}u^2v\left(6u^{-2}-9v^{-1}+\frac{3}{2}uv^2\right)=4v-6u^2+u^3v^3\).

a) \(xy^2+3\) b) \(2a^2-3\) c) \(4v-6u^2+u^3v^3\)

- Rewrite decimal exponents as fractions. - Add or subtract exponents when the bases are the same. - A negative exponent represents a reciprocal. - Rewrite radicals as rational exponents.

a) Since \(1.25=\frac{5}{4}\), \(x^{1.25}=x^{\frac{5}{4}}=\sqrt[4]{x^5}\). b) \(\sqrt[3]{a^2}\cdot a^{\frac{1}{3}}=a^{\frac{2}{3}}a^{\frac{1}{3}}=a\). c) Since \(-0.25=-\frac{1}{4}\), \(16^{-0.25}=16^{-\frac{1}{4}}=\frac{1}{\sqrt[4]{16}}=\frac{1}{2}\). d) \(\frac{\sqrt{z^3}}{\sqrt[4]{z^2}}=z^{\frac{3}{2}-\frac{2}{4}}=z^{\frac{3}{2}-\frac{1}{2}}=z\).

a) \(\sqrt[4]{x^5}\) b) \(a\) c) \(\frac{1}{2}\) d) \(z\)

- Work from the inner radical outward and rewrite roots as rational exponents. - In part b), simplify the expression inside the sixth root before taking that root. - In part c), rewrite both sides as powers of the same base. - Relate each radical index to the denominator of a rational exponent.

1. In part a), \(\sqrt[4]{x^2\sqrt[3]{x}}=\left(x^2x^{\frac{1}{3}}\right)^{\frac{1}{4}}=\left(x^{\frac{7}{3}}\right)^{\frac{1}{4}}=x^{\frac{7}{12}}\). 2. In part b), \(\sqrt[3]{y^6}=y^2\), so \(\sqrt[6]{y^4\sqrt[3]{y^6}}=\sqrt[6]{y^6}=y\). 3. In part c), rewrite both sides with base \(5\): \(5^{\frac{k}{3}}=5^2\). Thus, \(\frac{k}{3}=2\), so \(k=6\).

a) \(x^{\frac{7}{12}}\) b) \(y\) c) \(k=6\)

- Interpret an exponent of \(\frac{2}{3}\) using a cube root and a square. - Isolate the power of \(V\) before undoing its exponent. - Use the reciprocal exponent to solve for the base. - Check part c) against the result in part a).

a) Substitute \(V=125\): \(m=0.4\cdot125^{\frac{2}{3}}\). Since \(125^{\frac{2}{3}}=(\sqrt[3]{125})^2=25\), \(m=0.4\cdot25=10\). Thus the mass is \(10\,\text{kg}\). b) Solve for \(V\): \(\frac{m}{0.4}=V^{\frac{2}{3}}\), so \(2.5m=V^{\frac{2}{3}}\). Raise both sides to the \(\frac{3}{2}\) power: \(V=(2.5m)^{\frac{3}{2}}\). c) For \(m=10\), \(V=(2.5\cdot10)^{\frac{3}{2}}=25^{\frac{3}{2}}=125\). Thus the volume is \(125\,\text{L}\).

a) \(10\,\text{kg}\) b) \(V=(2.5m)^{\frac{3}{2}}\) c) \(125\,\text{L}\)

- Rewrite the decimal exponent as a fraction. - To solve for the base, raise both sides to the reciprocal exponent. - Distribute a power over a product. - Evaluate \(81^{\frac{3}{4}}\) using a fourth root and a cube.

a) Since \(0.75=\frac{3}{4}\), when \(x=16\), \(y=3\cdot16^{\frac{3}{4}}=3\cdot8=24\). b) Divide by \(3\): \(\frac{y}{3}=x^{\frac{3}{4}}\). Raise both sides to the \(\frac{4}{3}\) power: \(x=\left(\frac{y}{3}\right)^{\frac{4}{3}}\). c) If \(x\) is replaced by \(81x\), then \(y_{\text{new}}=3(81x)^{\frac{3}{4}}=3\cdot81^{\frac{3}{4}}x^{\frac{3}{4}}\). Since \(81^{\frac{3}{4}}=(\sqrt[4]{81})^3=27\), \(y_{\text{new}}=27y\).

a) \(y=24\) b) \(x=\left(\frac{y}{3}\right)^{\frac{4}{3}}\) c) \(y\) is multiplied by \(27\).

- Rewrite each radical as a rational exponent. - Add exponents when multiplying like bases and subtract when dividing. - Multiply exponents when raising a power to another power.

a) \(\sqrt{x}\cdot x^2=x^{\frac{1}{2}+2}=x^{\frac{5}{2}}\). b) \(\frac{\sqrt[3]{y^2}}{y}=y^{\frac{2}{3}-1}=y^{-\frac{1}{3}}\). c) \(\left(a^{0.5}a^{\frac{1}{4}}\right)^2=\left(a^{\frac{1}{2}+\frac{1}{4}}\right)^2=\left(a^{\frac{3}{4}}\right)^2=a^{\frac{3}{2}}\). d) \(\frac{z^{1.5}}{\sqrt{z^3}}=\frac{z^{\frac{3}{2}}}{z^{\frac{3}{2}}}=z^0=1\).

a) \(x^{\frac{5}{2}}\) b) \(y^{-\frac{1}{3}}\) c) \(a^{\frac{3}{2}}\) d) \(z^0=1\)

- Compare even and odd radical indices. - Check a proposed root by raising it to the radical index. - A principal even root is nonnegative.

1. True, because \((-5)^3=-125\). 2. False, because a principal even root is never negative. 3. False, because odd roots of negative numbers are defined. In fact, \(\sqrt[5]{-1}=-1\).

1. True, because \((-5)^3=-125\). 2. False, because a principal fourth root cannot be negative. 3. False, because odd roots of negative numbers are real; \(\sqrt[5]{-1}=-1\).

- How do the area of a square and the side length relate? - Express the cube's volume using its side length before substituting your expression for \(s\). - What happens to an exponent when a power is raised to another power?

1. The side length is \(s=A^{\frac{1}{2}}=(2^8\cdot5^2)^{\frac{1}{2}}=2^4\cdot5=80\,\text{m}\). 2. The cube's volume is \(V=s^3=(A^{\frac{1}{2}})^3=A^{\frac{3}{2}}=80^3=512{,}000\,\text{m}^3\). 3. Since \(80=2^4\cdot5\), \(V=(2^4\cdot5)^3=2^{12}\cdot5^3\,\text{m}^3\).

a) \(s=A^{\frac{1}{2}}=80\,\text{m}\) b) \(V=A^{\frac{3}{2}}=512{,}000\,\text{m}^3\) c) \(V=2^{12}\cdot5^3\,\text{m}^3\)

- Evaluate each radical before combining the terms. - Even roots of negative numbers are not real. - Odd roots preserve the sign of the radicand.

a) \(\sqrt[3]{-125}+\sqrt[4]{16}=-5+2=-3\). b) \(\sqrt[6]{-64}\) has no real value, so the expression is undefined over the real numbers. c) \(\sqrt[5]{-1}-\sqrt[3]{-1}=-1-(-1)=0\). d) \(\sqrt[4]{0.0001}=0.1\) and \(\sqrt[3]{-0.001}=-0.1\), so the quotient is \(\frac{0.1}{-0.1}=-1\).

a) \(-3\) b) Undefined c) \(0\) d) \(-1\)

- An even-index radical of an even power may produce an absolute value. - Use the given inequality to determine the sign of the expression inside the absolute value. - Compare \((a-b)^2\) and \((b-a)^2\).

a) \(\sqrt[4]{(x-y)^4}=|x-y|\). Since \(x<y\), \(x-y<0\), so \(|x-y|=y-x\). b) \(\sqrt[6]{(a-b)^{12}}=(a-b)^2\). For \(a=3\) and \(b=5\), the value is \((3-5)^2=4\). After switching the values, the result is \((5-3)^2=4\), so it does not change. c) \(\sqrt{(z-10)^6}=|(z-10)^3|\). Since \(z<10\), \(z-10<0\), so \((z-10)^3<0\). Therefore the expression simplifies to \(-(z-10)^3=(10-z)^3\).

a) \(y-x\) b) \(4\); the value does not change. c) \((10-z)^3\)

- For an even index, \(\sqrt[n]{u^n}=|u|\). - Determine the sign of each expression inside an absolute value. - Compare radicals by comparing their nonnegative radicands.

1) \(\sqrt{(-13)^2}=|-13|=13\). 2) \(\sqrt[4]{(-3)^4}=|-3|=3\). 3) \(\sqrt{(4-\sqrt{17})^2}=|4-\sqrt{17}|\). Since \(4=\sqrt{16}<\sqrt{17}\), the expression inside the absolute value is negative. The result is \(\sqrt{17}-4\). 4) \(\sqrt[6]{(2-\sqrt{3})^6}=|2-\sqrt{3}|\). Since \(2=\sqrt{4}>\sqrt{3}\), the expression inside the absolute value is positive. The result is \(2-\sqrt{3}\).

1) \(13\) 2) \(3\) 3) \(\sqrt{17}-4\) 4) \(2-\sqrt{3}\)

- For an odd index, \(\sqrt[n]{u^n}=u\); for an even index, the result is \(|u|\). - Determine whether \(\sqrt{5}-3\) and \(3-\pi\) are positive or negative. - Ask when a number's absolute value equals its opposite.

a) \(\sqrt[3]{(-5)^3}=-5\) and \(\sqrt{(-5)^2}=|-5|=5\), so the sum is \(0\). b) \(\sqrt[4]{(\sqrt{5}-3)^4}=|\sqrt{5}-3|\). Since \(\sqrt{5}<3\), the result is \(3-\sqrt{5}\). c) \(\sqrt{(3-\pi)^2}=|3-\pi|\). Since \(\pi>3\), the result is \(\pi-3\). d) The equation is equivalent to \(|x|=-x\). This is true exactly when \(x\leq0\).

a) \(0\) b) \(3-\sqrt{5}\) c) \(\pi-3\) d) \(x\leq0\)

- Rewrite each radical as a power with a fractional exponent. - Reduce each exponent fraction. - Compare the reduced exponents.

1. Rewrite each radical as a rational exponent. 2. \(A=y^{\frac{3}{6}}=y^{\frac{1}{2}}\). 3. \(B=y^{\frac{2}{4}}=y^{\frac{1}{2}}\). 4. \(C=y^{\frac{5}{10}}=y^{\frac{1}{2}}\). 5. All three exponent fractions reduce to \(\frac{1}{2}\), so the expressions are equivalent and equal \(\sqrt{y}\).

\(A=B=C=y^{\frac{1}{2}}=\sqrt{y}\).

- First compare each radicand with nearby integer powers. - Test tenths between \(3\) and \(4\). - Look for one benchmark value that lies between the two numbers.

1. Since \(3^3=27\) and \(4^3=64\), \(a\) is between \(3\) and \(4\). More precisely, \((3.1)^3=29.791\) and \((3.2)^3=32.768\), so \(3.1<a<3.2\). 2. Since \(3^4=81\) and \(4^4=256\), \(b\) is between \(3\) and \(4\). More precisely, \((3.0)^4=81\) and \((3.1)^4=92.3521\), so \(3.0<b<3.1\). 3. Because \(a>3.1\) and \(b<3.1\), it follows that \(a>b\).

\(3.1<a<3.2\) and \(3.0<b<3.1\), so \(a>b\).

- Reducing an even radical index and an even exponent may introduce an absolute value. - Compare the terms in each difference before removing the absolute value. - You can compare square roots by comparing their nonnegative radicands.

a) \(\sqrt[4]{(3-\sqrt{11})^2}=\sqrt{|3-\sqrt{11}|}\). Since \(3=\sqrt{9}<\sqrt{11}\), the difference is negative, so the result is \(\sqrt{\sqrt{11}-3}\). b) \(\sqrt[6]{(\sqrt{2}-\sqrt{7})^2}=\sqrt[3]{|\sqrt{2}-\sqrt{7}|}\). Since \(\sqrt{2}<\sqrt{7}\), the difference is negative, so the result is \(\sqrt[3]{\sqrt{7}-\sqrt{2}}\). c) \(\sqrt[10]{(4-\sqrt{17})^2}=\sqrt[5]{|4-\sqrt{17}|}\). Since \(4=\sqrt{16}<\sqrt{17}\), the difference is negative, so the result is \(\sqrt[5]{\sqrt{17}-4}\).

a) \(\sqrt{\sqrt{11}-3}\) b) \(\sqrt[3]{\sqrt{7}-\sqrt{2}}\) c) \(\sqrt[5]{\sqrt{17}-4}\)

- Evaluate the original expression before judging the proposed simplification. - Compare the real domains of the original and proposed expressions. - For \(x<5\), determine the sign of \(x-5\) before removing an absolute value.

1. \(T(1)=\sqrt[4]{(1-5)^2}=\sqrt[4]{16}=2\). 2. The proposed expression gives \(\sqrt{1-5}=\sqrt{-4}\), which is not a real number, while the original expression equals \(2\). Therefore, the proposed simplification is not equivalent to the original expression. 3. In general, \(\sqrt[4]{(x-5)^2}=\sqrt{|x-5|}\). 4. When \(x<5\), \(x-5<0\), so \(|x-5|=5-x\). Thus \(T(x)=\sqrt{5-x}\).

a) \(2\) b) At \(x=1\), the original expression equals \(2\), but \(\sqrt{x-5}=\sqrt{-4}\) is not real. c) \(\sqrt{5-x}\)

- Compute the power first in each example. - Compare the behavior of negative bases under even and odd powers. - Remember that the principal value of an even-index root is never negative.

1. \(\sqrt[4]{(-3)^4}=\sqrt[4]{81}=3\), not \(-3\), so the claim is false for this even index and negative input. 2. \(\sqrt[3]{(-2)^3}=\sqrt[3]{-8}=-2\), so the claim is true in this odd-index example. 3. A principal even root is nonnegative. Therefore, for even \(n\), the correct formula is \(\sqrt[n]{a^n}=|a|\).

a) False: \(\sqrt[4]{(-3)^4}=3\). b) True: \(\sqrt[3]{(-2)^3}=-2\). c) For even \(n\), \(\sqrt[n]{a^n}=|a|\).

- Rewrite radicals as rational exponents. - Work from the inner radical outward in part a). - Reduce the exponent fraction in part b). - Add exponents when multiplying powers with the same base.

a) \(\sqrt{b\sqrt[3]{b}}=\left(b\cdot b^{\frac{1}{3}}\right)^{\frac{1}{2}}=b^{\frac{2}{3}}=\sqrt[3]{b^2}\). b) \(\sqrt[4]{49p^2}=\sqrt[4]{(7p)^2}=(7p)^{\frac{2}{4}}=\sqrt{7p}\). c) \(\sqrt[3]{2}\cdot\sqrt[6]{2}=2^{\frac{1}{3}}2^{\frac{1}{6}}=2^{\frac{1}{3}+\frac{1}{6}}=2^{\frac{1}{2}}=\sqrt{2}\).

a) \(\sqrt[3]{b^2}\) b) \(\sqrt{7p}\) c) The exponents add to \(\frac{1}{2}\), so the equation is true.

- Rewrite every radical as a rational exponent. - Add numerator exponents and subtract denominator exponents. - Reduce the final exponent fraction. - For an improper fractional exponent, separate the whole-number part.

1) \(y^{\frac{2}{3}+\frac{5}{6}-\frac{1}{2}}=y^{\frac{4}{6}+\frac{5}{6}-\frac{3}{6}}=y\). 2) \(k^{\frac{2}{3}+\frac{3}{4}-\frac{1}{12}}=k^{\frac{8}{12}+\frac{9}{12}-\frac{1}{12}}=k^{\frac{4}{3}}=k\sqrt[3]{k}\). 3) \(z^{\frac{4}{5}-\frac{2}{15}}=z^{\frac{12}{15}-\frac{2}{15}}=z^{\frac{2}{3}}=\sqrt[3]{z^2}\).

1) \(y\) 2) \(k\sqrt[3]{k}\) 3) \(\sqrt[3]{z^2}\)

- Rewrite each radical as a rational exponent. - Add exponents when multiplying like bases. - Multiply exponents for nested powers. - Interpret a negative exponent as a reciprocal.

1. In part a), \(\sqrt[3]{x^2}\cdot\sqrt[3]{x}=x^{\frac{2}{3}}x^{\frac{1}{3}}=x\), so the statement is true. 2. In part b), \(\sqrt{\sqrt[3]{a}}=\left(a^{\frac{1}{3}}\right)^{\frac{1}{2}}=a^{\frac{1}{6}}\), not \(a^{\frac{1}{5}}\) in general, so the statement is false. 3. In part c), \(16^{-\frac{1}{4}}=\frac{1}{16^{\frac{1}{4}}}=\frac{1}{2}=0.5\), so the statement is true.

a) True: \(x^{\frac{2}{3}}x^{\frac{1}{3}}=x^{1}=x\). b) False: \(\left(a^{\frac{1}{3}}\right)^{\frac{1}{2}}=a^{\frac{1}{6}}\), not \(a^{\frac{1}{5}}\) in general. c) True: \(16^{-\frac{1}{4}}=\frac{1}{16^{\frac{1}{4}}}=\frac{1}{2}=0.5\).

- A radical in the denominator can be written using a negative exponent. - Evaluate the fourth root of \(16\) before applying the cube. - Write \(x\) as \(x^1\) and add exponents.

a) Since \(\sqrt[4]{x^3}=x^{\frac{3}{4}}\), \(T=x^{-\frac{3}{4}}\). b) When \(x=16\), \(T=16^{-\frac{3}{4}}=\frac{1}{(\sqrt[4]{16})^3}=\frac{1}{2^3}=\frac{1}{8}\). c) \(T\cdot x=x^{-\frac{3}{4}}x^1=x^{\frac{1}{4}}=\sqrt[4]{x}\).

a) \(T=x^{-\frac{3}{4}}\) b) \(\frac{1}{8}\) c) \(T\cdot x=x^{\frac{1}{4}}=\sqrt[4]{x}\)

- Apply an outside exponent to every factor in a product. - Rewrite radicals as rational exponents. - Multiply exponents for a power raised to another power. - Subtract exponents when dividing like bases.

1. \((3\sqrt[3]{2})^3=3^3(\sqrt[3]{2})^3=27\cdot2=54\). 2. \(\frac{(\sqrt[6]{5})^{12}}{5}=\frac{5^{\frac{12}{6}}}{5}=5\). 3. \((\sqrt[4]{4})^2=(4^{\frac{1}{4}})^2=4^{\frac{1}{2}}=2\). 4. \(\frac{(\sqrt[3]{y})^9}{y^2}=y^{\frac{9}{3}-2}=y^{3-2}=y\).

1. \(54\) 2. \(5\) 3. \(2\) 4. \(y\)

- Rewrite the fourth root as an exponent of \(\frac{1}{4}\). - Distribute the exponent over each product. - Subtract denominator exponents from numerator exponents. - Convert the remaining fractional exponent back to a radical.

1. Rewrite the denominator as a rational exponent and distribute the exponents: \(\frac{x^{\frac{2}{4}}y^{\frac{3}{4}}}{x^{\frac{1}{4}}y^{-\frac{1}{4}}}\). 2. Subtract exponents for each base: \(x^{\frac{2}{4}-\frac{1}{4}}y^{\frac{3}{4}-(-\frac{1}{4})}=x^{\frac{1}{4}}y\). 3. Convert the remaining fractional exponent to a radical: \(y\sqrt[4]{x}\).

\(y\sqrt[4]{x}\)

- Rewrite nested radicals from the inside outward. - Distribute each rational exponent over its product. - Subtract exponents when dividing like bases. - Express \(64\) as a power of \(2\).

a) Rewrite the radicals: \(T=\frac{(a^3b^{\frac{2}{3}})^{\frac{1}{2}}}{(ab)^{\frac{1}{3}}}=\frac{a^{\frac{3}{2}}b^{\frac{1}{3}}}{a^{\frac{1}{3}}b^{\frac{1}{3}}}=a^{\frac{3}{2}-\frac{1}{3}}=a^{\frac{7}{6}}\). b) For \(a=64=2^6\), \(T=64^{\frac{7}{6}}=(2^6)^{\frac{7}{6}}=2^7=128\).

a) \(a^{\frac{7}{6}}\) b) \(128\)

- Rewrite each pair with a common radical index. - Use the least common multiple of the two indices. - Once the indices match, compare the radicands.

1. In part a), use a common index of \(6\): \(\sqrt[3]{5}=\sqrt[6]{25}\) and \(\sqrt{3}=\sqrt[6]{27}\). Since \(25<27\), \(\sqrt[3]{5}<\sqrt{3}\). 2. In part b), use a common index of \(12\): \(\sqrt[4]{10}=\sqrt[12]{1000}\) and \(\sqrt[3]{4}=\sqrt[12]{256}\). Since \(1000>256\), \(\sqrt[4]{10}>\sqrt[3]{4}\). 3. In part c), use a common index of \(6\): \(\sqrt[3]{3}=\sqrt[6]{9}\). Since \(10>9\), \(\sqrt[6]{10}>\sqrt[3]{3}\).

a) \(\sqrt[3]{5}<\sqrt{3}\) because \(\sqrt[3]{5}=\sqrt[6]{25}\) and \(\sqrt{3}=\sqrt[6]{27}\), with \(25<27\). b) \(\sqrt[4]{10}>\sqrt[3]{4}\) because the common-index forms are \(\sqrt[12]{1000}\) and \(\sqrt[12]{256}\). c) \(\sqrt[6]{10}>\sqrt[3]{3}\) because \(\sqrt[3]{3}=\sqrt[6]{9}\) and \(10>9\).

- Apply the cube to every factor in the product. - Rewrite the fifth root as a rational exponent. - Separate an improper fractional exponent into a whole-number and fractional part. - Analyze the sign of each factor for \(x>0\).

1) Apply the exponent to each factor: \(T=(-2)^3x^3(\sqrt[5]{x^2})^3=-8x^3x^{\frac{6}{5}}=-8x^{3+\frac{6}{5}}=-8x^{\frac{21}{5}}=-8x^4\sqrt[5]{x}\). 2) For \(x>0\), both \(x^4\) and \(\sqrt[5]{x}\) are positive. Multiplying them by \(-8\) makes \(T\) negative.

1) \(-8x^4\sqrt[5]{x}\) 2) Yes. The variable factors are positive for \(x>0\), and the coefficient is negative.

- Apply the outside cube to both factors inside the parentheses. - A cube and a cube root undo each other. - Subtract exponents when dividing powers with the same base.

1. Apply the cube to each factor: \(\left(\frac{a^2}{b}\right)^3\left(\sqrt[3]{\frac{b^2}{a^3}}\right)^3\). 2. This gives \(\frac{a^6}{b^3}\cdot\frac{b^2}{a^3}\). 3. Subtract exponents on like bases: \(a^{6-3}b^{2-3}=a^3b^{-1}=\frac{a^3}{b}\).

\(\frac{a^3}{b}\)

- Rewrite all radicals as rational exponents. - Simplify the numerator and denominator separately. - Subtract exponents when dividing like bases. - Convert a negative exponent to a reciprocal.

1. Simplify the numerator: \(\sqrt{x\sqrt[3]{x^2}}=\left(x\cdot x^{\frac{2}{3}}\right)^{\frac{1}{2}}=x^{\frac{5}{6}}\). 2. Simplify the denominator: \(x\sqrt[6]{x}=x^{1+\frac{1}{6}}=x^{\frac{7}{6}}\). 3. Divide the powers: \(x^{\frac{5}{6}-\frac{7}{6}}=x^{-\frac{1}{3}}=\frac{1}{\sqrt[3]{x}}\).

\(\frac{1}{\sqrt[3]{x}}\)

- Rewrite each radical as a rational exponent. - Simplify the expression inside the parentheses first. - Add or subtract exponents on like bases. - Multiply exponents when raising a power to another power.

1. Rewrite the radicals: \(\sqrt{a}=a^{\frac{1}{2}}\) and \(\sqrt[3]{a^2}=a^{\frac{2}{3}}\). 2. Simplify inside the parentheses: \(\frac{a^{2+\frac{1}{2}}}{a^{\frac{2}{3}}}=a^{\frac{5}{2}-\frac{2}{3}}=a^{\frac{11}{6}}\). 3. Raise the power to the sixth power: \(\left(a^{\frac{11}{6}}\right)^6=a^{11}\).

\(a^{11}\)

- Apply the outside exponent to each factor. - A cube root raised to the sixth power produces a square. - Multiply the resulting rational expressions. - Cancel like factors by subtracting exponents.

1. Apply the sixth power to both factors: \(\left(\frac{x-y}{x}\right)^6\left(\sqrt[3]{\frac{x^2}{x-y}}\right)^6\). 2. Simplify each factor: \(\frac{(x-y)^6}{x^6}\left(\frac{x^2}{x-y}\right)^2\). 3. Multiply and subtract exponents on like factors: \(\frac{(x-y)^6x^4}{x^6(x-y)^2}=\frac{(x-y)^4}{x^2}\).

\(\frac{(x-y)^4}{x^2}\)

- Distribute the outside exponent over the product. - Multiply exponents in the numerator. - Subtract denominator exponents from numerator exponents. - Recall that a nonzero number raised to the zero power equals \(1\).

1. Apply the fourth power to each factor in the numerator: \((x^{\frac{1}{4}}y^{-\frac{1}{2}})^4=xy^{-2}\). 2. Divide powers with the same bases: \(\frac{xy^{-2}}{xy^{-3}}=x^{1-1}y^{-2-(-3)}\). 3. This simplifies to \(x^0y=y\).

\(y\)

- In the exponent method, represent each square root with exponent \(\frac{1}{2}\). - In the nested-radical method, preserve equality when moving a positive factor under a square root. - Check that both methods produce the same overall exponent on \(p\).

1. Using rational exponents, \(\sqrt{p^5\sqrt{p}}=\left(p^5p^{\frac{1}{2}}\right)^{\frac{1}{2}}=p^{\left(5+\frac{1}{2}\right)\frac{1}{2}}=p^{\frac{11}{4}}=\sqrt[4]{p^{11}}\). 2. Since \(p>0\), \(p^5=\sqrt{p^{10}}\). Thus \(\sqrt{p^5\sqrt{p}}=\sqrt{\sqrt{p^{10}}\sqrt{p}}=\sqrt{\sqrt{p^{11}}}=\sqrt[4]{p^{11}}\). 3. Both methods confirm Mikkel's claim.

The claim is true. Both methods give \(\sqrt[4]{p^{11}}\).

- Rewrite both radicals as rational exponents. - Distribute exponents over products and quotients. - Combine powers with the same base. - Give the final exponents a common denominator to form one radical.

1. Rewrite the expression with rational exponents: \(Q=\left(a^2b^{-1}(ba^{-1})^{\frac{1}{2}}\right)^{\frac{1}{3}}\). 2. Distribute the square-root exponent and combine like bases inside the cube root: \(Q=\left(a^{2-\frac{1}{2}}b^{-1+\frac{1}{2}}\right)^{\frac{1}{3}}=\left(a^{\frac{3}{2}}b^{-\frac{1}{2}}\right)^{\frac{1}{3}}\). 3. Multiply exponents: \(Q=a^{\frac{1}{2}}b^{-\frac{1}{6}}=a^{\frac{3}{6}}b^{-\frac{1}{6}}=\left(\frac{a^3}{b}\right)^{\frac{1}{6}}\). 4. Therefore \(Q=\sqrt[6]{\frac{a^3}{b}}\). The work uses the power-of-a-product, product-of-powers, and power-of-a-power properties.

\(Q=\sqrt[6]{\frac{a^3}{b}}\)

- Rewrite every quantity as a power of \(2\). - Add exponents on factors with the same base. - A reciprocal has a negative exponent. - Compare the final exponents.

a) The left side is \(\left(2^1\cdot2^{\frac{2}{3}}\right)^{\frac{1}{2}}=2^{\frac{5}{6}}\). The right side is \(\left(2^2\cdot2^{\frac{1}{2}}\right)^{\frac{1}{3}}=2^{\frac{5}{6}}\). Therefore the equation is true. b) Since \(\frac{1}{2}=2^{-1}\), \(\sqrt[3]{\frac{1}{2}\sqrt{\frac{1}{2}}}=\left(2^{-1}2^{-\frac{1}{2}}\right)^{\frac{1}{3}}=2^{-\frac{1}{2}}\). Comparing exponents gives \(k=-\frac{1}{2}\).

a) Both sides equal \(2^{\frac{5}{6}}\). b) \(k=-\frac{1}{2}\)

- Rewrite both sides using rational exponents. - Simplify the product inside the square root before applying the outer exponent. - Subtract exponents when dividing like bases. - Multiply exponents for nested powers.

1. The left side is \(\left(a\cdot a^{\frac{3}{4}}\right)^{\frac{1}{2}}=a^{\frac{7}{8}}\). 2. The right side is \(\sqrt[8]{a^7}=a^{\frac{7}{8}}\), so the equation is true for every \(a>0\). 3. For the second expression, \(\sqrt[4]{\frac{z}{\sqrt{z}}}=\left(z^{1-\frac{1}{2}}\right)^{\frac{1}{4}}=z^{\frac{1}{8}}=\sqrt[8]{z}\).

The equation is true because both sides equal \(a^{\frac{7}{8}}\). The second expression simplifies to \(\sqrt[8]{z}\).

- Rewrite all radicals as rational exponents. - Add exponents on like bases before applying the outside exponent. - Multiply exponents in a power raised to another power. - Simplify each expression exactly before comparing.

a) \((\sqrt{3}\cdot\sqrt[6]{3})^{-3}=\left(3^{\frac{1}{2}}3^{\frac{1}{6}}\right)^{-3}=\left(3^{\frac{2}{3}}\right)^{-3}=3^{-2}=\frac{1}{9}\). b) \(A=(\sqrt[4]{16})^3=2^3=8\). Also, \(B=(16^{\frac{1}{3}})^{\frac{3}{2}}=16^{\frac{1}{2}}=4\). Since \(8>4\), \(A>B\).

a) \(\frac{1}{9}\) b) \(A\) is greater because \(A=8\) and \(B=4\).

- Rewrite every radical as a rational exponent. - Add numerator exponents and subtract the denominator exponent. - Use a common denominator for the exponent fractions. - For the numerical value, take the sixth root before raising to the fifth power.

a) Rewrite the radicals: \(T=y^{\frac{1}{3}+2-\frac{3}{2}}\). Combine the exponents: \(\frac{1}{3}+2-\frac{3}{2}=\frac{2}{6}+\frac{12}{6}-\frac{9}{6}=\frac{5}{6}\). Thus \(T=y^{\frac{5}{6}}\), so \(k=\frac{5}{6}\). b) For \(y=64\), \(T=64^{\frac{5}{6}}=(\sqrt[6]{64})^5=2^5=32\).

a) \(T=y^{\frac{5}{6}}\), so \(k=\frac{5}{6}\). b) \(32\)

- Rewrite terminating decimals as fractions when helpful. - Multiply exponents in a power raised to another power. - A negative exponent represents a reciprocal. - Any nonzero number raised to the zero power equals \(1\).

1. Since \(0.125=\frac{1}{8}=2^{-3}\), \((0.125)^{-\frac{1}{3}}=(2^{-3})^{-\frac{1}{3}}=2\). 2. \((0.25)^{0.5}=\sqrt{\frac{1}{4}}=\frac{1}{2}\), so the product of the first two factors is \(1\). 3. \((3^2)^{1.5}=3^{2\cdot\frac{3}{2}}=3^3=27\). 4. \(\sqrt[4]{81^3}=(81^{\frac{1}{4}})^3=3^3=27\). 5. Since \(5.7\ne0\), \((5.7)^0=1\). 6. Therefore \(A=1+27-27+1=2\).

\(2\)

- Compare the graphs near \(x=0\) and for larger positive values of \(x\). - Evaluate each function at simple inputs such as \(1\), \(2\), and \(4\). - Decide whether each graph becomes steeper or less steep as \(x\) increases. - Which function is decreasing for \(x>0\)?

1. Graph \(c\) is the one that decreases for positive inputs; \(x^{-1}=\frac{1}{x}\), so it represents \(f\). 2. Since \(g(4)=\sqrt4=2\) and the curve grows more slowly for larger inputs, \(g\) matches graph \(a\). 3. The rule \(h(x)=x\sqrt{x}\) grows faster than \(x\) for \(x>1\) and matches graph \(b\). 4. The rule \(k(x)=\frac{1}{4}x^2\) gives \(k(2)=1\) and \(k(4)=4\), matching graph \(d\).

1) Graph \(c\): \(x^{-1}=1/x\) decreases as positive \(x\) increases. 2) Graph \(a\): \(g(4)=2\) and the curve gets less steep. 3) Graph \(b\): \(h(x)=x\sqrt{x}\) gets steeper for \(x>1\). 4) Graph \(d\): \(k(2)=1\) and \(k(4)=4\), matching the curve's values.

- Read the inputs of the marked points from the graph. - A point lies on a curve if its coordinates satisfy that rule. - Decide which function takes reciprocals for positive inputs. - Compare how the two increasing curves change as \(x\) grows.

1. Function A gives \(f(4)=1.5\cdot\sqrt4=3\), matching marked point P on \(u\). 2. Function B gives \(g(4)=0.5\cdot4^{\frac{3}{2}}=4\), matching Q on \(v\). 3. Function C gives \(h(2)=2\cdot\frac{1}{2}=1\), matching R on \(w\). 4. Curve \(w\) decreases for positive inputs, as a reciprocal power does. Curve \(u\) increases but becomes less steep; curve \(v\) increases and becomes steeper.

a) A: \(u\); B: \(v\); C: \(w\). b) Point checks: \(f(4)=3\), \(g(4)=4\), \(h(2)=1\). Curve \(u\) becomes less steep, \(v\) becomes steeper, and \(w\) decreases.

- Evaluate each rule at \(x=1\). - Identify which graph decreases as \(x\) increases. - Use a second input to distinguish graphs that intersect. - Compare the shapes of square-root, quadratic, cubic, and reciprocal functions.

1. Graph \(p\) passes through \((1, 3)\), matching \(f(1)=3\). Thus, graph \(p\) represents \(f\). 2. Graph \(r\) decreases for positive \(x\) and passes through \((1, 4)\), matching \(h(x)=\frac{4}{x}\). 3. Graphs \(q\) and \(s\) meet at \((2, 4)\). At \(x=1\), graph \(q\) has value \(1\), matching \(g(1)=1\), while graph \(s\) has value \(0.5\), matching \(k(1)=0.5\). 4. The remaining rule, \(m(x)=x^{-2}\), would have values \(m(1)=1\) and \(m(2)=0.25\), which do not match any displayed graph.

Graph \(p\): \(f(x)=3x^{0.5}\); graph \(q\): \(g(x)=x^2\); graph \(r\): \(h(x)=\frac{4}{x}\); graph \(s\): \(k(x)=0.5x^3\)

- A negative leading coefficient makes the quadratic outputs negative for positive inputs. - Functions with negative exponents decrease for positive inputs. - Evaluate the rules at \(x=1\) and \(x=4\). - Recall that \(x^{0.5}=\sqrt{x}\).

1. Graph \(z\) passes through the origin and has negative outputs for positive inputs, so it matches \(s(x)=-0.2x^2\). 2. Graph \(v\) decreases and passes through \((1, 1)\) and \((4, 0.5)\). Since \(4^{-0.5}=\frac{1}{\sqrt{4}}=0.5\), it matches \(q\). 3. Graph \(w\) passes through \((1, 1.5)\) and \((4, 3)\), matching \(r(x)=1.5\sqrt{x}\). 4. Graph \(u\) passes through \((1, 1)\) and has a value near \(1.6\) at \(x=4\). Since \(\sqrt[3]{4}\approx1.587\), it matches \(p\). 5. The rule \(t(x)=2x^{-1}\) would pass through \((1, 2)\), so it is the unused rule.

Graph \(u\): \(p(x)=\sqrt[3]{x}\); graph \(v\): \(q(x)=x^{-0.5}\); graph \(w\): \(r(x)=1.5\sqrt{x}\); graph \(z\): \(s(x)=-0.2x^2\)

- The point with \(x=1\) makes it easy to find \(a\). - Use the second point to form an equation for \(n\). - Rewrite \(\frac{1}{2}\) as a power of \(2\).

1. Read the marked points: \(P(1, 2)\) and \(Q(2, 1)\). 2. Substitute \(P\): \(2=a(1^n)\), so \(a=2\). 3. Substitute \(Q\): \(1=2\cdot2^n\), so \(2^n=\frac{1}{2}=2^{-1}\). 4. Therefore, \(n=-1\), and \(f(x)=2x^{-1}=\frac{2}{x}\).

\(f(x)=2x^{-1}\), or equivalently \(f(x)=\frac{2}{x}\)

- Substitute the point with \(x=1\) first to find \(a\). - Use the second point to form an equation for \(n\). - Rewrite both sides as powers with the same base.

1. Substitute \(P(1, 0.5)\): \(0.5=a(1^n)\), so \(a=0.5\). 2. Substitute \(Q(4, 4)\): \(4=0.5\cdot4^n\). 3. Multiply by \(2\): \(8=4^n\). 4. Write both sides with base \(2\): \(2^3=(2^2)^n=2^{2n}\). 5. Thus, \(3=2n\), so \(n=\frac{3}{2}\). Therefore, \(g(x)=0.5x^{\frac{3}{2}}\). 6. The rule gives \(g(1)=0.5\) and \(g(4)=4\) and has the same increasing, upward-curving shape shown for \(x\geq0\), so it matches the graph.

\(a=0.5\), \(n=\frac{3}{2}\), and \(g(x)=0.5x^{\frac{3}{2}}\)

- Rewrite the fraction \(\frac{m}{n}\) as a product involving \(\frac{1}{n}\). - For part b), raise the rational-exponent expression to the \(n\)th power and simplify its exponent. - In the numerical example, \(32\) has a convenient integer fifth root.

1. Since \(\frac{m}{n}=m\cdot\frac{1}{n}\), the power-of-a-power law gives \((a^{\frac{1}{n}})^m=a^{\frac{m}{n}}\). 2. Also, \((a^{\frac{m}{n}})^n=a^{\left(\frac{m}{n}\right)n}=a^m\). 3. Because \(a>0\), \(a^{\frac{m}{n}}\) is positive. It is therefore the principal \(n\)th root of \(a^m\), so \(a^{\frac{m}{n}}=\sqrt[n]{a^m}\). 4. For \(32^{\frac{3}{5}}\), first taking the fifth root gives \((\sqrt[5]{32})^3=2^3=8\). 5. First taking the third power gives \(\sqrt[5]{32^3}=\sqrt[5]{2^{15}}=2^3=8\). Both interpretations agree.

a) \(a^{\frac{m}{n}}=(a^{\frac{1}{n}})^m\) b) \((a^{\frac{m}{n}})^n=a^m\), so \(a^{\frac{m}{n}}=\sqrt[n]{a^m}\) c) Both interpretations give \(32^{\frac{3}{5}}=8\).

- For (1), test two simple nonzero perfect squares. - For (2), consider separately what happens when \(x\) is negative, zero, or positive. - For (3), determine when the negative of a square can be nonnegative.

(1) Statement (1) is false. For example, let \(a = 9\) and \(b = 16\). Then \(\sqrt{9} + \sqrt{16} = 3 + 4 = 7\), but \(\sqrt{9 + 16} = \sqrt{25} = 5\). (2) Since \(\sqrt{x^2} = |x|\), the equation is \(|x| = -x\). This is true for every \(x \leq 0\), so there are infinitely many solutions. Statement (2) is true. (3) The radicand must be nonnegative: \(-(x-5)^2 \geq 0\). Since \((x-5)^2 \geq 0\), its opposite is nonnegative only when \((x-5)^2 = 0\). Thus, \(x = 5\) is the only value in the domain. Statement (3) is true.

(1) False. For example, \(\sqrt{9} + \sqrt{16} = 7\), while \(\sqrt{25} = 5\). (2) True. Every \(x \leq 0\) is a solution. (3) True. The expression is defined only for \(x = 5\).

- Rewrite both radicals with the same index. - Use the least common multiple of \(3\) and \(4\). - Once the indices match, compare the radicands.

1. Use a common radical index. The least common multiple of \(3\) and \(4\) is \(12\). 2. \(u=\sqrt[12]{5^4}=\sqrt[12]{625}\), and \(v=\sqrt[12]{8^3}=\sqrt[12]{512}\). 3. Since \(625>512\), \(\sqrt[12]{625}>\sqrt[12]{512}\). Therefore, \(u>v\).

\(\sqrt[3]{5}=\sqrt[12]{625}\) and \(\sqrt[4]{8}=\sqrt[12]{512}\). Since \(625>512\), \(\sqrt[3]{5}>\sqrt[4]{8}\).

- Rewrite negative exponents as reciprocals. - Combine the two fractions in the denominator. - Simplify the complex fraction before applying the outside exponent. - Multiply exponents in a power raised to another power.

1. Rewrite the denominator: \(x^{-k}+y^{-k}=\frac{1}{x^k}+\frac{1}{y^k}=\frac{x^k+y^k}{x^ky^k}\). 2. The fraction inside the parentheses becomes \(\frac{x^k+y^k}{\frac{x^k+y^k}{x^ky^k}}=x^ky^k=(xy)^k\). 3. Apply the outside exponent: \(((xy)^k)^{-\frac{1}{k}}=(xy)^{-1}=\frac{1}{xy}\). 4. For \(x=0.25\) and \(y=8\), the value is \(\frac{1}{0.25\cdot8}=\frac{1}{2}\).

The simplified expression is \(\frac{1}{xy}\). Its value is \(\frac{1}{2}\).