- Identify the factor that keeps repeating. - Count how many copies of that factor are multiplied. - In a power, the large number and the small raised number have different roles.

1. The factor \(4\) appears three times, so the power is \(4^3\). 2. The base is \(4\), and the exponent is \(3\). The exponent counts the repeated factors.

\(4^3\). Base: \(4\). Exponent: \(3\) repeated factors.

- Expand each power as repeated multiplication if needed. - Count all the factors with the same base. - Think about which operation joins the two lists of factors.

1. The product is three factors of \(2\) followed by two more factors of \(2\). 2. There are five factors altogether, so \(2^3\cdot2^2=2^{3+2}=2^5\).

\(2^5\), because the product contains five factors of \(2\).

- Write out repeated factors if the powers feel unfamiliar. - Each canceled pair has value one. - Count the uncanceled factors.

1. Expand numerator and denominator into six and four factors of \(5\), respectively. 2. Cancel four matching factors from top and bottom. 3. Two factors remain in the numerator, so the result is \(5^2\).

Cancel four matching factors of \(5\) in the numerator and denominator. Two factors remain above, so the answer is \(5^2\).

- Compare consecutive values in the pattern. - Identify the operation that takes \(27\) to \(9\), then \(9\) to \(3\). - Continue that same operation one more time.

1. Each step divides the previous value by \(3\). 2. After \(3^1=3\), divide by \(3\) again to get \(3^0=1\).

\(3^0=1\). Each step divides the value by \(3\).

- Compare the values of the two powers already given. - Keep the same operation when moving to a smaller exponent. - Do the values change sign as you continue the pattern? Check rather than guess from the exponent's sign.

1. When the exponent decreases by one, divide the value by the base \(2\). 2. Thus, \(2^{-1}=1\div2=\frac{1}{2}\). 3. Divide once more: \(2^{-2}=\frac{1}{2}\div2=\frac{1}{4}\).

\(2^{-1}=\frac{1}{2}\), \(2^{-2}=\frac{1}{4}\); each step divides by \(2\).

- Identify the base in each expression. - Write each expression as a multiplication before evaluating it. - If the minus sign is not part of the base, when is it applied?

1. In \((-3)^2\), the base is \(-3\), so \((-3)(-3)=9\). 2. In \(-3^2\), only \(3\) is squared, and the negative sign remains outside: \(-(3^2)=-9\). 3. Therefore \(-3^2\) is negative.

\(-3^2=-9\) is negative; \((-3)^2=9\) is positive. The parentheses make \(-3\) the base of the first power. Without them, only \(3\) is squared and the minus sign stays outside.

- Work with the coefficients and variables separately. - Add exponents when multiplying powers with the same base. - A variable with no written exponent has exponent \(1\).

1. Multiply the coefficients: \(2 \cdot 3 = 6\). 2. Add exponents for the same base: \(x^2 \cdot x^3 = x^{2+3} = x^5\). 3. Since each factor contains \(y^1\), \(y \cdot y = y^{1+1} = y^2\). 4. The product is \(6x^5y^2\).

d) \(6x^5y^2\)

- Determine whether the negative sign is included in the base being squared. - A quotient of two negative numbers is positive. - The product of two negative values is positive.

1. a) The parentheses make the negative part of the base: \((-3)^2 = 9\), so \(\frac{9}{-9} = -1\). 2. b) The exponent applies only to \(3\): \(-3^2 = -(3^2) = -9\), so \(\frac{-9}{-9} = 1\). 3. c) Inside the fraction, \(-3^2 = -9\). Therefore, \(-\frac{-9}{9} = -(-1) = 1\).

a) \(-1\) b) \(1\) c) \(1\)

- First determine the sign of the power in the numerator. Does an odd or even exponent matter? - Consider what the negative sign in front of the fraction does. - Recall where positive and negative numbers lie relative to \(0\).

1. Since \(101\) is odd, \((-1)^{101} = -1\). 2. Substitute: \(x = -\frac{-1}{5} = \frac{1}{5}\). 3. Because \(\frac{1}{5} > 0\), the point is to the right of \(0\).

The value is to the right of \(0\) because \(x = \frac{1}{5}\), which is positive.

- What sign results when two negative numbers are multiplied? - How does an even or odd number of negative factors affect the sign? - You can write each row as a power.

1. Row 8 contains eight negative factors. Because eight is even, the product is positive. 2. Row 4 is \((-3)^4\). Pairing the factors gives \(9 \cdot 9 = 81\).

a) Positive, because the product has an even number of negative factors. b) \(81\)

- Determine exactly what the exponent applies to. Is the negative sign inside parentheses? - Consider how multiplying an even or odd number of negative factors affects the sign. - Follow the order of operations when a negative sign is outside a power.

1. a) \((-3)^4 = (-3) \cdot (-3) \cdot (-3) \cdot (-3) = 81\). 2. b) \(-3^4 = -(3 \cdot 3 \cdot 3 \cdot 3) = -81\). 3. c) \((-10)^3 = (-10) \cdot (-10) \cdot (-10) = -1000\). 4. d) \(-10^3 = -(10 \cdot 10 \cdot 10) = -1000\). 5. e) Since \(100\) is even, \((-1)^{100} = 1\). 6. In a), the parentheses make \(-3\) the base. In b), the exponent applies only to \(3\), and the negative sign is applied afterward.

a) \(81\) b) \(-81\) c) \(-1000\) d) \(-1000\) e) \(1\) In a), the base is \(-3\). In b), the base is \(3\), and the negative sign is outside the power.

- Rewrite each quantity as a power with base \(10\). - When multiplying or dividing powers with the same base, what happens to the exponents? - Count the zeros in \(100{,}000\).

1. Rewrite each quantity as a power of \(10\): (1) \(10^4\), (2) \(10^3\), (3) \(10^{2+3} = 10^5\), (4) \(10^5\), and (5) \(10^{7-1} = 10^6\). 2. Compare the exponents: \(3 < 4 < 5 < 6\). 3. Therefore, \(1000 < 10^4 < 10^2 \cdot 10^3 = 100{,}000 < 10^7 \div 10\).

\(1000 < 10^4 < 10^2 \cdot 10^3 = 100{,}000 < 10^7 \div 10\)

- Check whether the exponent is even or odd. - Notice whether a negative sign is inside or outside parentheses. - A negative exponent creates a reciprocal; it does not make the value negative.

1. \((-1)^{10}=1\) because the exponent is even, while \((-1)^{11}=-1\). Therefore, \(1>-1\). 2. The expression \(-4^2\) means \(-(4^2)=-16\), while \((-4)^2=16\). Therefore, \(-16<16\). 3. \(3^{-2}=\frac{1}{9}\), and \((-3)^{-2}=\frac{1}{(-3)^2}=\frac{1}{9}\). Therefore, the expressions are equal.

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

- Add exponents when multiplying powers with the same base. - Subtract the denominator exponent when dividing. - Recall that a nonzero number to the zero power equals \(1\).

1. For a), \(a^{-4}a^4=a^0=1\), so the statement is false. 2. For b), \(\frac{b^2}{b^{-3}}=b^{2-(-3)}=b^5\), so the statement is true. 3. For c), \((2^{-1})^3=2^{-3}=\frac{1}{8}\), so the statement is true.

a) False. The correct equation is \(a^{-4}\cdot a^4=1\). b) True. c) True.

- Rewrite decimal powers of ten using exponents. - Add exponents when multiplying and subtract them when dividing. - Multiply exponents when raising a power to a power.

1. For a), \(100{,}000=10^5\), so \(\frac{1}{100000}=10^{-5}\). 2. For b), \(0.001=10^{-3}\), so \(10^{-3}\cdot10^{-4}=10^{-7}\). 3. For c), \(10^{-2}\div10^5=10^{-2-5}=10^{-7}\). 4. For d), \((10^{-3})^2\cdot10^4=10^{-6+4}=10^{-2}\).

a) \(10^{-5}\) b) \(10^{-7}\) c) \(10^{-7}\) d) \(10^{-2}\)

- Add exponents when multiplying powers with the same base. - Square the negative base before combining powers. - Subtracting a negative exponent is addition.

1. For a), \(a^{5-8+4}=a\). 2. For b), \((-x)^2=x^2\), so \(x^2x^{-5}=x^{-3}=\frac{1}{x^3}\). 3. For c), \(b^{2-(-3)}=b^5\).

a) \(a\) b) \(\frac{1}{x^3}\) c) \(b^5\)

- An exponent outside parentheses applies to every factor inside. - Add exponents when multiplying powers with the same base. - Distinguish a negative sign inside the base from one outside a power.

1. a) True: \((5z)^2 = 5^2z^2 = 25z^2\). 2. b) False: \((-x)^2 = (-x)(-x) = x^2\). 3. c) False: powers with the same base are multiplied by adding exponents, so \(a^4 \cdot a^2 = a^6\). 4. d) True: \((2k)^3 = 2^3k^3 = 8k^3\).

a) True b) False; \((-x)^2 = x^2\) c) False; \(a^4 \cdot a^2 = a^6\) d) True

- Rewrite every negative exponent as a reciprocal. - Apply the exponent to the coefficient inside parentheses. - Put all four expressions in the same form before comparing.

1. \(A=\frac{4}{x^2}\). 2. \(B=4x^{-2}=\frac{4}{x^2}\). 3. \(C=(2x)^{-2}\cdot16=\frac{16}{(2x)^2}=\frac{4}{x^2}\). 4. \(D=4\cdot\frac{1}{x^{-2}}=4x^2\), so it is not equivalent to the others for all nonzero \(x\).

D is not equivalent. A, B, and C simplify to \(\frac{4}{x^2}\), while D simplifies to \(4x^2\).

- Identify whether the negative sign is part of the base. - Rewrite each negative exponent as a reciprocal. - Compare the resulting decimal values.

1. \(10^{-2}=\frac{1}{100}=0.01\). 2. \((-10)^{-2}=\frac{1}{(-10)^2}=0.01\). 3. \(-10^{-2}=-\frac{1}{100}=-0.01\). 4. Therefore, \(-0.01<0.01=0.01\), so III \(<\) I \(=\) II.

a) I: \(0.01\); II: \(0.01\); III: \(-0.01\) b) III \(<\) I \(=\) II

- Work with coefficients and powers separately when multiplying. - In part b), simplify the expression in parentheses first.

1. a) Multiply the coefficients and add the exponents: \(5 \cdot 3 = 15\) and \(a^3 \cdot a^2 = a^5\). The result is \(15a^5\). 2. b) First simplify the divisor: \(3b^2 \cdot 2b = 6b^3\). Then \(18b^6 \div 6b^3 = 3b^{6-3} = 3b^3\).

a) \(15a^5\) b) \(3b^3\)

- Multiply exponents when raising a power to a power. - Add exponents when multiplying powers with the same base.

1. Raise the power to a power: \((y^2)^3=y^6\). 2. Multiply powers with the same base: \(y^{-5}y^6=y^{-5+6}=y\).

\(y\)

- Rewrite \(9\) and \(0.1\) as powers with matching bases. - Add exponents when multiplying and subtract when dividing. - Follow multiplication and division from left to right.

1. For a), write \(9=3^2\): \(3^{6-7+2}=3\). 2. For b), \(10^{-2}\div10^{-5}=10^3\), and \(0.1^2=(10^{-1})^2=10^{-2}\). 3. Therefore, \(10^3\cdot10^{-2}=10\).

a) \(3\) b) \(10\)

- Combine like bases by adding or subtracting exponents. - Apply the outside exponent before simplifying a quotient.

1. \(\frac{x^8\cdot x^{-5}}{x^2}=x^{8-5-2}=x\). 2. The numerator is \(a^6b^{-4}\). Dividing by \(a^4b^{-4}\) gives \(a^2\).

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

- An exponent tells how many times the base is used as a factor. - Distinguish repeated addition from repeated multiplication. - Group equal factors before writing powers.

1. Six equal factors of \(8\) give \(8^6\). 2. Four equal factors of \(z\) give \(z^4\). 3. Group equal factors: \(3^3a^2b^4\). 4. \(x^4\) represents repeated multiplication, so \(x \cdot x \cdot x \cdot x = x^4\). The sum \(x + x + x + x\) equals \(4x\).

1) \(8^6\) 2) \(z^4\) 3) \(3^3a^2b^4\) 4) \(x \cdot x \cdot x \cdot x\), because exponents represent repeated multiplication.

- Rearrange factors so equal variables are next to each other. - A plus sign separates terms; do not combine factors across it. - Use the definition of a power with \(n\) equal factors.

1. Nine factors of \(c\) give \(c^9\). 2. Group equal factors in each term: \(4^2r^3 + s^5\). 3. There are three factors of \(x\) and four factors of \(y\), so the product is \(x^3y^4\). 4. A product of \(n\) factors of \(p\) is \(p^n\).

1) \(c^9\) 2) \(4^2r^3 + s^5\) 3) \(x^3y^4\) 4) \(p^n\)

- An exponent tells how many equal factors are multiplied. - Identify the complete base of each power. - Keep coefficients separate from exponents on variables.

1. \(6^3 = 6 \cdot 6 \cdot 6\). 2. Four factors of \(b\) give \(b^4\). 3. Two factors of \(x\) and three factors of \(y\) give \(3x^2y^3\). 4. The entire binomial is the base: \((a + b)^2 = (a + b)(a + b)\). 5. Two factors of \(m\) give \(7m^2\).

1) \(6 \cdot 6 \cdot 6\) 2) \(b^4\) 3) \(3x^2y^3\) 4) \((a + b)(a + b)\) 5) \(7m^2\)

- First rewrite one repeated product using exponents. - Then count how many times that identical term is added. - Keep an entire binomial together when it is the repeated factor.

1. Each repeated product is \(x^3\), and it appears twice, so the expression is \(2x^3\). 2. Each repeated product is \(a^2b^3\), and it appears three times, so the expression is \(3a^2b^3\). 3. Each repeated product is \((m + n)^2\), and it appears four times, so the expression is \(4(m + n)^2\).

1) \(2x^3\) 2) \(3a^2b^3\) 3) \(4(m + n)^2\)

- Determine whether the negative sign is part of the base. - Compare \(-a^n\) with \((-a)^n\). - Evaluate a power before applying a negative sign outside it.

1. \(-4^2 = -(4^2) = -16\). 2. \((-4)^2 = 16\). 3. Since \(50\) is even, \((-1)^{50} = 1\). 4. \(\left(-\frac{1}{2}\right)^4 = \frac{1}{16}\). 5. First, \((-2)^3 = -8\). Then \(-(-8) = 8\).

1) \(-16\) 2) \(16\) 3) \(1\) 4) \(\frac{1}{16}\) 5) \(8\)

- Add exponents when multiplying powers with the same base. - A variable without a written exponent has exponent \(1\). - Apply the sign rules separately from the exponent rule.

1. Add exponents when multiplying powers with the same base. 2. a) \(x^{4+5} = x^9\). 3. b) \(-a^{3+2} = -a^5\). 4. c) \(y^{1+4+2} = y^7\). 5. d) The negative signs multiply to a positive sign, giving \(m^{2+6} = m^8\). 6. e) \(b^{n+3}\).

a) \(x^9\) b) \(-a^5\) c) \(y^7\) d) \(m^8\) e) \(b^{n+3}\)

- Multiply the coefficients separately. - Add exponents when multiplying powers with the same base. - Apply the sign rules for multiplication.

1. a) \((6 \cdot 4)x^{2+5} = 24x^7\). 2. b) \((-3 \cdot 8)a^{4+3} = -24a^7\). 3. c) \((0.4 \cdot -5)y^{1+4} = -2y^5\). 4. d) \((-7 \cdot -2)b^{2+2} = 14b^4\). 5. e) \(\left(\frac{2}{5} \cdot 10\right)z^{6+2} = 4z^8\).

a) \(24x^7\) b) \(-24a^7\) c) \(-2y^5\) d) \(14b^4\) e) \(4z^8\)

- Multiply the coefficients separately. - Add exponents for each common variable base. - Apply the sign rules for multiplication.

1. \((-5 \cdot 4)a^{3+2}b^{1+4} = -20a^5b^5\). 2. \((0.8 \cdot -0.5)x^{2+1}y^{3+1} = -0.4x^3y^4\). 3. \((-6 \cdot -3)m^{4+2}n^{2+5} = 18m^6n^7\).

1) \(-20a^5b^5\) 2) \(-0.4x^3y^4\) 3) \(18m^6n^7\)

- Raise every factor inside the parentheses to the outside power. - Multiply exponents when raising a power to a power. - Use even and odd exponents to determine the sign.

1. \((4a^5)^2 = 4^2a^{5 \cdot 2} = 16a^{10}\). 2. \((-3b^2)^3 = (-3)^3b^{2 \cdot 3} = -27b^6\). 3. \(\left(\frac{1}{2}x^3\right)^4 = \frac{1}{16}x^{12}\). 4. \((-y^4)^2 = (-1)^2y^{4 \cdot 2} = y^8\).

1) \(16a^{10}\) 2) \(-27b^6\) 3) \(\frac{1}{16}x^{12}\) 4) \(y^8\)

- Add exponents when multiplying powers with the same base. - Subtract exponents when dividing powers with the same base.

1. a) Add exponents: \(10^{4+3} = 10^7\). 2. b) Subtract exponents: \(10^{7-2} = 10^5\). 3. c) Add all exponents: \(10^{2+2+2} = 10^6\).

a) \(10^7\) b) \(10^5\) c) \(10^6\)

- What do you get when you evaluate the expression inside the parentheses first? - Which property simplifies a power raised to another power? - How can a power of a product be applied to each factor?

1. For a), evaluate inside the parentheses first: \(4 \cdot 5 = 20\), so \(20^3 = 8000\). 2. Using the power of a product property: \((4 \cdot 5)^3 = 4^3 \cdot 5^3 = 64 \cdot 125 = 8000\). 3. For b), evaluate the inner power first: \(2^3 = 8\), so \(8^2 = 64\). 4. Using the power of a power property: \((2^3)^2 = 2^{3 \cdot 2} = 2^6 = 64\).

a) \(8000\) b) \(64\)

- Pay attention to whether the exponent on a negative base is even or odd. - Which property applies to a power raised to another power? - Does changing the order of the two exponent factors change their product?

1. Evaluate \(A\): \((-2)^2 = 4\), and then \(4^3 = 64\). 2. Evaluate \(B\): \((-2)^3 = -8\), and then \((-8)^2 = 64\). 3. Apply the power of a power property, \((a^m)^n = a^{mn}\). 4. \(A = (-2)^{2 \cdot 3} = (-2)^6\). 5. \(B = (-2)^{3 \cdot 2} = (-2)^6\). 6. Both expressions simplify to the same power, so they are equal.

a) \(A = 64\); \(B = 64\) b) \(A = (-2)^{2 \cdot 3} = (-2)^6\) and \(B = (-2)^{3 \cdot 2} = (-2)^6\). Therefore, \(A = B\).

- A reciprocal base reverses the sign of an exponent. - Add exponents when multiplying like bases. - Rewrite \(0.1\) as a power of \(10\).

1. \(2^4\cdot2^{-2}=2^{4-2}=2^2=4\). 2. \(\left(\frac{1}{3}\right)^{-3}=3^3\), so \(3^3\cdot3^{-2}=3\). 3. Since \(0.1=10^{-1}\), \(10^{-2}\cdot0.1^{-3}=10^{-2}\cdot(10^{-1})^{-3}=10^{-2}\cdot10^3=10\).

1) \(4\) 2) \(3\) 3) \(10\)

- A nonzero base raised to the zero power equals \(1\). - Follow parentheses, exponents, multiplication and division, then addition and subtraction. - Rewrite all quantities with base \(5\) when possible.

1. Since \(\left(\frac{17}{5}\right)^0=1\), the bracket equals \(6-2=4\). Thus \(4^{-2}=\frac{1}{16}=0.0625\). 2. Write all factors with base \(5\): \(25^2=5^4\) and \(0.2^{-2}=\left(\frac{1}{5}\right)^{-2}=5^2\). Therefore, \(\frac{5^{-1}\cdot5^4}{5^2}=5\).

1) \(\frac{1}{16}\), or \(0.0625\) 2) \(5\)

- A factor moved from the denominator to the numerator receives a negative exponent. - Rewrite numerical denominators as powers. - Keep coefficients and numerator factors unchanged.

1. Since \(10{,}000=10^4\), \(\frac{1}{10000}=10^{-4}\). 2. Since \(32=2^5\), \(\frac{1}{32}=2^{-5}\). 3. \(\frac{4}{x^3}=4x^{-3}\). 4. \(\frac{y^5}{z^2}=y^5z^{-2}\).

a) \(10^{-4}\) b) \(2^{-5}\) c) \(4x^{-3}\) d) \(y^5z^{-2}\)

- Treat a parenthetical expression as a single base. - Change the sign of an exponent when moving a factor across the fraction bar. - Use parentheses around a numerator with more than one term.

1. \(\frac{12}{x^5}=12x^{-5}\). 2. \(\frac{1}{(a+b)^3}=(a+b)^{-3}\). 3. \(\frac{1}{m^2}+\frac{1}{n^2}=m^{-2}+n^{-2}\). 4. \(\frac{x-y}{z^2}=(x-y)z^{-2}\).

1) \(12x^{-5}\) 2) \((a+b)^{-3}\) 3) \(m^{-2}+n^{-2}\) 4) \((x-y)z^{-2}\)

- Change an exponent’s sign when moving a factor across the fraction bar. - Convert \(\frac{5}{2}\) to a decimal for part 1. - In part 2, move all negative powers to the opposite side of the fraction bar.

1. Move the denominator factors to the numerator: \(T=2.5a^{-n}b^2c^3d^{-1}\). 2. Move factors with negative exponents to the denominator: \(T=\frac{5b^2c^3}{2a^nd}\).

1) \(2.5a^{-n}b^2c^3d^{-1}\) 2) \(\frac{5b^2c^3}{2a^nd}\)

- Treat every denominator factor as a factor with a negative exponent. - Constant denominators can also be written using negative exponents. - Keep parentheses around a multi-term base.

1. \(\frac{12}{a^5}=12a^{-5}\). 2. \(\frac{3x^2}{4y^3}=3\cdot4^{-1}x^2y^{-3}\). 3. \(\frac{2}{3(m+n)^4}=2\cdot3^{-1}(m+n)^{-4}\). 4. \(\frac{ab}{c}=abc^{-1}\).

1) \(12a^{-5}\) 2) \(3\cdot4^{-1}x^2y^{-3}\) 3) \(2\cdot3^{-1}(m+n)^{-4}\) 4) \(abc^{-1}\)

- Apply the exponent to both factors in \(xy\). - Subtract the denominator exponent for each variable. - Substitute only after simplifying the expression.

1. Rewrite the denominator: \((xy)^{-2}=x^{-2}y^{-2}\). 2. Divide powers with the same bases: \(A=x^{-1-(-2)}y^{-1-(-2)}=xy\). 3. For \(x=2\) and \(y=3\), \(A=2\cdot3=6\).

The simplified expression is \(xy\). For \(x=2\) and \(y=3\), the value is \(6\).

- Rewrite \(a^{-n}\) as a reciprocal. - Combine the two factors into one quotient. - Substitute only after the expression has no negative exponents.

1. Rewrite the negative power as a reciprocal: \(a^{-n}=\frac{1}{a^n}\). 2. Therefore, \(T=\frac{b^n}{a^n}=\left(\frac{b}{a}\right)^n\). 3. For \(a=2\), \(b=6\), and \(n=2\), \(T=\left(\frac62\right)^2=3^2=9\).

\(T=\frac{b^n}{a^n}=\left(\frac{b}{a}\right)^n\). The numerical value is \(9\).

- Compare each value with the one immediately to its left. - Decreasing an exponent by \(1\) divides the value by the base. - Continue the same pattern through exponent \(0\) and into the negative exponents.

1. Each time the exponent decreases by \(1\), divide the preceding value by \(5\). 2. \(5\div5=1\), so \(5^0=1\). 3. \(1\div5=\frac{1}{5}\), so \(5^{-1}=\frac{1}{5}\). 4. \(\frac{1}{5}\div5=\frac{1}{25}\), so \(5^{-2}=\frac{1}{25}\).

The missing values are \(1\), \(\frac{1}{5}\), and \(\frac{1}{25}\). Thus, \(5^0=1\), \(5^{-1}=\frac{1}{5}\), and \(5^{-2}=\frac{1}{25}\).

- What multiplication does the outside square ask you to do? - Expand only far enough to count the equal-base factors. - Compare the exponent you counted with each offered choice.

1. The square means \((2^3)^2=2^3\cdot2^3\). 2. Each copy has three factors of \(2\), giving six altogether. 3. Therefore \((2^3)^2=2^{3\cdot2}=2^6\), not \(2^5\).

\(2^6\), because the two copies contain six factors of \(2\).

- What is the exponent in the formula when there have been no folds? - Make a short list of layer counts for the first few folds. - Which exponent gives the number of layers in the question?

1. The number of layers must satisfy \(2^n=32\). 2. Since \(2^5=32\), \(n=5\).

\(5\) folds, because \(2^5=32\).

- Evaluate each expression separately. - In \(B\), determine whether the exponent applies to the negative sign. - Compare the resulting rational numbers on a number line or by using common denominators.

1. Evaluate \(A\): \((-1)^3 = -1\), so \(A = -\frac{1}{2}\). 2. Evaluate \(B\): the exponent applies before the leading negative sign, so \(-1^4 = -(1^4) = -1\). Thus, \(B = \frac{-1}{-4} = \frac{1}{4}\). 3. Evaluate \(C\): \((-1)^2 = 1\), so \(C = -\frac{1}{8}\). 4. Compare the values: \(-\frac{1}{2} < -\frac{1}{8} < \frac{1}{4}\). Therefore, \(A < C < B\).

\(A < C < B\)

- Recall how an even or odd exponent affects the sign of a power with a negative base. - On a number line, a negative number with greater absolute value is farther left. - Absolute value is distance from \(0\).

1. Expression A is \((-10)^3 = -1000\). 2. Expression B is \((-1)^4 \cdot 10^4 = 1 \cdot 10^4 = 10{,}000\). 3. Expression C is \((-10)^5 = -100{,}000\). 4. Since \(-100{,}000 < -1000 < 10{,}000\), Expression C has the least value. 5. Only Expression B is positive. 6. The absolute value of Expression B is \(|10{,}000| = 10{,}000\).

a) Expression C, with a value of \(-100{,}000\) b) Expression B only c) \(10{,}000\)

- Look for a pattern in the signs of consecutive terms. - Recall how even and odd exponents affect a negative base. - In part c), evaluate the power before applying the outside negative sign.

1. The first five terms are \((-2)^1 = -2\), \((-2)^2 = 4\), \((-2)^3 = -8\), \((-2)^4 = 16\), and \((-2)^5 = -32\). 2. Term 15 is \((-2)^{15}\). Since \(15\) is odd, the term is negative. Term 20 is \((-2)^{20}\). Since \(20\) is even, the term is positive. 3. First evaluate the power: \((-2)^4 = 16\). Then apply the outside negative sign: \(-16\). The expression is negative.

a) \(-2, 4, -8, 16, -32\) b) Term 15 is negative, and Term 20 is positive. c) Negative; \(-(-2)^4 = -16\)

- Every positive number is greater than every negative number. - Test both a positive number between \(0\) and \(1\) and a positive number greater than \(1\). - Consider what happens when a positive value is multiplied by a number between \(0\) and \(1\).

1. For \(x = -3\), \(x^2 = (-3)^2 = 9\) and \(x^3 = (-3)^3 = -27\). 2. For any negative \(x\), \(x^2\) is positive and \(x^3\) is negative. Therefore, \(x^2 > x^3\). 3. For \(x = 0.5\), \(x^2 = 0.25\) and \(x^3 = 0.125\), so \(x^2 > x^3\). 4. For \(x = 2\), \(x^2 = 4\) and \(x^3 = 8\), so \(x^3 > x^2\). Therefore, the conclusion does not hold for every positive \(x\).

a) \(x^2 = 9\) and \(x^3 = -27\) b) \(x^2\), because it is positive while \(x^3\) is negative for every negative \(x\) c) No. For \(x = 0.5\), \(0.25 > 0.125\), but for \(x = 2\), \(8 > 4\).

- Apply the properties for multiplying, dividing, and raising powers to powers. - In Expression D, distinguish addition from multiplication. - After substitution, compare the powers of \(10\).

1. Apply the exponent properties: \(A = x^5\), \(B = x^8\), \(C = x^7\), and \(D = 2x^5\). 2. At \(x = 10\), the values are \(A = 10^5 = 100{,}000\), \(B = 10^8 = 100{,}000{,}000\), \(C = 10^7 = 10{,}000{,}000\), and \(D = 2 \cdot 10^5 = 200{,}000\). 3. Expression B has the greatest value.

\(A = x^5\) \(B = x^8\) \(C = x^7\) \(D = 2x^5\) When \(x = 10\), Expression B has the greatest value, \(100{,}000{,}000\).

- Write \(4\), \(8\), \(16\), and \(32\) as powers of \(2\). - Use \((a^m)^n = a^{mn}\). - Once the bases match, compare the exponents.

1. Rewrite each base as a power of \(2\): (1) \((2^4)^3 = 2^{12}\), (2) \((2^2)^7 = 2^{14}\), (3) \(2^{11}\), (4) \((2^3)^4 = 2^{12}\), and (5) \((2^5)^2 = 2^{10}\). 2. Compare the exponents: \(10 < 11 < 12 < 14\). 3. Therefore, \(32^2\) has the least value, and \(4^7\) has the greatest value.

(1) \(2^{12}\) (2) \(2^{14}\) (3) \(2^{11}\) (4) \(2^{12}\) (5) \(2^{10}\) Least: \(32^2\) Greatest: \(4^7\)

- Multiply like bases by adding exponents and divide them by subtracting exponents. - For a power of a power, multiply the exponents. - A nonzero base raised to the zero power equals \(1\). - Apply an exponent to every factor inside parentheses.

1. \((x^3)^4=x^{12}\), so \(x^{12}\cdot x^{-5}=x^7\). 2. Divide the coefficients and subtract exponents: \(\frac{24y^6}{3y^2}=8y^4\). 3. \((-3a)^2=9a^2\), so \(9a^2\cdot a^3=9a^5\). 4. Since \(z^0=1\), \(\frac{z^0\cdot z^7}{z^3}=z^{7-3}=z^4\).

a) \(x^7\) b) \(8y^4\) c) \(9a^5\) d) \(z^4\)

- Apply the exponent rules to each base separately. - When dividing powers with the same base, subtract exponents. - A power outside parentheses applies to each factor inside. - An exponent containing a letter follows the same algebraic rules.

1. Subtract exponents for each base: \(a^{4-(-1)}b^{-2-3}=a^5b^{-5}=\frac{a^5}{b^5}\). 2. Evaluate the coefficient and exponents: \((0.2x^2)^3\cdot125x^{-6}=0.008x^6\cdot125x^{-6}=1\). 3. Subtract exponent expressions: \(\frac{x^{n+3}}{x^{n-1}}=x^{(n+3)-(n-1)}=x^4\).

a) \(\frac{a^5}{b^5}\) b) \(1\) c) \(x^4\)

- Simplify each expression until it is written in the form \(ax^n\). - Remember that adding like terms does not change the exponent. - An exponent outside parentheses applies to every factor inside the parentheses.

1. A: \((2x^2)^4 = 2^4(x^2)^4 = 16x^8\), so A is equivalent. 2. B: \(8x^4 + 8x^4 = 16x^4\), so B is not equivalent. 3. C: \(32x^9 \div (2x) = 16x^{9-1} = 16x^8\), so C is equivalent. 4. D: \((4x^4)^2 = 4^2(x^4)^2 = 16x^8\), so D is equivalent. 5. E: \(2 \cdot (2x^2)^3 = 2 \cdot 8x^6 = 16x^6\), so E is not equivalent.

A, C, and D are equivalent to \(16x^8\).

- Add exponents when multiplying powers with the same base. - Subtract exponents when dividing powers with the same base. - A negative exponent indicates a reciprocal.

1. For a), add exponents: \(4^{5+(-7)}=4^{-2}=\frac{1}{16}\). 2. For b), subtract exponents: \(3^{-2-2}=3^{-4}=\frac{1}{81}\). 3. For c), multiply exponents: \(2^{(-3)\cdot(-2)}=2^6=64\). 4. For d), add exponents: \(10^{-3+5-2}=10^0=1\). 5. For e), take the reciprocal and use a positive exponent: \(\left(\frac{5}{2}\right)^2=\frac{25}{4}=6.25\).

a) \(\frac{1}{16}\) b) \(\frac{1}{81}\) c) \(64\) d) \(1\) e) \(\frac{25}{4}\)

- Simplify every expression to a numerical value. - Distinguish a negative base from a negative exponent. - Apply the quotient rule to expression \(E\).

1. \(A=\left(\frac{1}{3}\right)^{-2}=3^2=9\). 2. \(B=3^{-2}=\frac{1}{9}\). 3. \(C=\frac{1}{9}\) and \(D=9\). 4. \(E=3^{5-7}=3^{-2}=\frac{1}{9}\). 5. \(F=(-3)^2=9\). 6. Therefore, \(A,D,F\) have value \(9\), and \(B,C,E\) have value \(\frac{1}{9}\).

Value \(9\): \(A,D,F\) Value \(\frac{1}{9}\): \(B,C,E\)

- Apply the outer exponent to both factors inside the parentheses. - Subtract exponents when dividing powers with the same base. - Move a factor with a negative exponent across the fraction bar.

1. Apply the power of a product and power of a power rules: \((x^2y^{-3})^{-2}=x^{-4}y^6\). 2. Divide powers with the same base: \(\frac{x^{-4}y^6}{x^{-1}y^4}=x^{-4-(-1)}y^{6-4}=x^{-3}y^2\). 3. Rewrite without negative exponents: \(x^{-3}y^2=\frac{y^2}{x^3}\).

\(\frac{y^2}{x^3}\)

- Write each power as repeated multiplication. - Cancel matching factors in the numerator and denominator. - Recall the definition of a negative exponent.

1. Expand and cancel: \(\frac{a\cdot a\cdot a}{a\cdot a\cdot a\cdot a\cdot a}=\frac{1}{a^2}\). 2. By the definition of a negative exponent, \(a^{-2}=\frac{1}{a^2}\), so Paul’s claim is true. 3. For \(a\ne0\), the quotient rule is \(\frac{a^m}{a^n}=a^{m-n}\). When \(n>m\), the exponent \(m-n\) is negative.

a) \(\frac{1}{a^2}\) b) \(\frac{1}{a^2}=a^{-2}\), so the claim is true. c) For \(a\ne0\), \(\frac{a^m}{a^n}=a^{m-n}\).

- Add exponents when multiplying powers with the same base. - Multiply exponents when raising a power to a power. - When dividing, subtract the denominator exponent.

1. For a), add exponents in the numerator and subtract the denominator exponent: \(x^{7+(-3)-2}=x^2\). 2. For b), multiply exponents, then add: \(y^{(-2)\cdot4+5}=y^{-3}\). 3. For c), subtract the exponent in the denominator: \(b^{4-(-2)}=b^6\).

a) \(x^2\) b) \(y^{-3}\) c) \(b^6\)

- Apply the product, quotient, and power-of-a-power rules. - Distinguish addition of like terms from multiplication of powers. - Recall that \(x^0=1\) for \(x\ne0\).

1. \(A=x^{3+4}=x^7\). 2. \(B=x^{10-3}=x^7\). 3. \(C=x^{(-1)\cdot(-7)}=x^7\). 4. \(D=2x^7\), so it is not equivalent. 5. \(E=x^7\). 6. \(F=x^{14-2}=x^{12}\), so it is not equivalent. 7. Since \(x^0=1\), \(G=x^7\).

A, B, C, E, and G

- Simplify the exponent in each expression. - Pay attention to whether a negative sign is part of the base or outside the power. - An even exponent removes the sign of a nonzero base.

1. Expression 1 is \(a^{-2+(-4)}=a^{-6}\). 2. Expression 2 is \(a^{3\cdot(-2)}=a^{-6}\). 3. Expression 3 is \(a^{4-10}=a^{-6}\). 4. Expression 4 equals \(a^{-6}\) because the exponent is even. 5. Expression 5 equals \(a^{-6}\) by definition. 6. Expression 6 is the opposite of \(a^{-6}\), so it is not equivalent. 7. Expression 7 is \(a^{(-3)\cdot(-2)}=a^6\), so it is not equivalent.

Expressions 1, 2, 3, 4, and 5 are equivalent to \(a^{-6}\). Expressions 6 and 7 are not.

- Combine all powers of \(b\) before deciding which exponent is needed. - In part b), which part of the original expression is being raised to the negative power? - Can you rewrite that power as a fraction before checking any numbers?

1. Combine the left-hand exponents: \(\frac{b^{n-2}}{b^5}=b^{n-7}\). 2. For the identity to hold for every nonzero \(b\), \(n-7=3\), giving \(n=10\). 3. The exponent \(-2\) belongs to \(x\) only, so \(4x^{-2}=4\cdot\frac{1}{x^2}=\frac{4}{x^2}\) for all \(x\ne0\). 4. At \(x=2\), both \(4\cdot2^{-2}\) and \(\frac{4}{2^2}\) equal \(1\). This check is not the proof of equivalence; the exponent rule is.

a) \(n=10\) b) \(4x^{-2}=\frac{4}{x^2}\), since the negative exponent applies to \(x\), not \(4\). Both forms equal \(1\) at \(x=2\).

- Add exponents when multiplying powers with the same base. - Subtract exponents when dividing. - Rewrite negative exponents as reciprocals at the end.

1. For a), \(a^{-4+7}=a^3\). 2. For b), \(b^{2-6}=b^{-4}=\frac{1}{b^4}\). 3. For c), \(c^{(-2)\cdot(-4)}=c^8\). 4. For d), \(\frac{15}{5}x^{3-8}=3x^{-5}=\frac{3}{x^5}\). 5. For e), \((y^{-3})^{-2}=y^6\).

a) \(a^3\) b) \(\frac{1}{b^4}\) c) \(c^8\) d) \(\frac{3}{x^5}\) e) \(y^6\)

- Rewrite each side as a single power of its letter. - Turn a reciprocal into an equivalent negative power when useful. - For the last equation, what exponent gives a value of \(1\) for any nonzero base?

1. For a), \(n+5=-3\), so \(n=-8\). 2. For b), \(a^{4n}=a^{-8}\), so \(4n=-8\) and \(n=-2\). 3. For c), \(n-(-2)=5\), so \(n=3\). 4. For d), \(1=c^0\). Thus \(-3+n=0\) and \(n=3\).

a) \(n=-8\) b) \(n=-2\) c) \(n=3\) d) \(n=3\)

- Apply the outer exponent to every factor inside the parentheses. - Combine powers with the same base. - Rewrite any remaining negative exponent as a reciprocal.

1. Apply the exponent to each factor: \((3a^{-2}b)^2=9a^{-4}b^2\). 2. Multiply by \(a^5\): \(9a^{-4}b^2a^5=9ab^2\). 3. Divide by \(9b^3\): \(\frac{9ab^2}{9b^3}=ab^{-1}\). 4. Rewrite without a negative exponent: \(ab^{-1}=\frac{a}{b}\).

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

- Use the definition of a negative exponent. - Multiply exponents when raising a power to a power. - Subtract exponents when dividing powers with the same base.

1. By the definition of negative exponents, \(k^{-4}=\frac{1}{k^4}\). Therefore, both students are correct. 2. For b), \((a^2)^{-3}=a^{-6}\). 3. Divide powers with the same base: \(a^{-6}\div a^{-4}=a^{-6-(-4)}=a^{-2}\). 4. Rewrite without a negative exponent: \(a^{-2}=\frac{1}{a^2}\).

a) Both are correct because \(k^{-4}=\frac{1}{k^4}\). b) \(\frac{1}{a^2}\)

- Square both the coefficient and the variable. - Simplify the numerator before dividing. - Rewrite the final negative exponent as a reciprocal.

1. Expand the power: \((3a)^2=9a^2\). 2. Combine the factors in the numerator: \(9a^2a^{-5}=9a^{-3}\). 3. Divide by \(3a^{-1}\): \(\frac{9a^{-3}}{3a^{-1}}=3a^{-3-(-1)}=3a^{-2}\). 4. Rewrite without a negative exponent: \(3a^{-2}=\frac{3}{a^2}\).

\(\frac{3}{a^2}\)

- Combine the numerator before dividing in part a). - Multiply exponents when raising a power to a power. - Keep careful track of subtracting a negative exponent.

1. For a), combine the numerator and then divide: \(x^{3+(-7)-(-2)}=x^{-2}\). 2. For b), raise the power to a power, then multiply: \(y^{(-2)\cdot(-3)+(-10)}=y^{6-10}=y^{-4}\).

a) \(x^{-2}\) b) \(y^{-4}\)

- Add exponents when multiplying powers with the same base. - Subtract exponents when dividing. - A negative exponent on a fraction reverses the fraction.

1. For a), \(3^{5+(-7)}=3^{-2}=\frac{1}{9}\). 2. For b), \(\left(\frac{1}{2}\right)^{-3}=2^3\), so \(2^3\cdot2^{-2}=2\). 3. For c), \(10^{-4-(-6)}=10^2=100\). 4. For d), \((-5)^{3+(-3)}=(-5)^0=1\).

a) \(\frac{1}{9}\) b) \(2\) c) \(100\) d) \(1\)

- Rewrite numbers using a common base when possible. - Convert a terminating decimal to a fraction before applying exponent rules. - A negative exponent on a fraction gives the reciprocal.

1. For a), write \(8=2^3\): \(2^3\cdot2^{-5}=2^{-2}=\frac{1}{4}\). 2. For b), \(\left(\frac{4}{3}\right)^{-1}=\frac{3}{4}\), so the product is \(\left(\frac{3}{4}\right)^3=\frac{27}{64}\). 3. For c), write \(0.25=\frac{1}{4}\): \(\left(\frac{1}{4}\right)^{-1}4^{-2}=4^1\cdot4^{-2}=\frac{1}{4}\). 4. For d), \(7^{2-5-(-3)}=7^0=1\).

a) \(\frac{1}{4}\) b) \(\frac{27}{64}\) c) \(\frac{1}{4}\) d) \(1\)

- Evaluate each power exactly before ordering. - Rewrite \(25\) as a power of \(5\). - Combine bases when the exponents are equal.

1. The values are \(2^{-2}=\frac{1}{4}\), \((-2)^2=4\), \(2^0=1\), and \((-2)^{-2}=\frac{1}{4}\). Therefore, \(2^{-2}=(-2)^{-2}<2^0<(-2)^2\). 2. For b), write \(25=5^2\). Then \(x+2=-1\), so \(x=-3\). 3. For c), use the common exponent: \(2^9\cdot0.5^9=(2\cdot0.5)^9=1\).

a) \(2^{-2}=(-2)^{-2}<2^0<(-2)^2\) b) \(x=-3\) c) \(1\)

- Add exponents when multiplying and subtract them when dividing. - Multiply exponents when raising a power to a power. - Rewrite negative exponents as reciprocals.

1. For a), \(a^{5+(-9)}=a^{-4}=\frac{1}{a^4}\). 2. For b), \(x^{-3-(-5)}=x^2\). 3. For c), \(y^{(-2)\cdot4}=y^{-8}=\frac{1}{y^8}\). 4. For d), \(b^{2+(-2)}=b^0=1\).

a) \(\frac{1}{a^4}\) b) \(x^2\) c) \(\frac{1}{y^8}\) d) \(1\)

- Combine powers with the same base. - A negative exponent on a fraction reverses the fraction. - Apply an exponent to every factor inside parentheses.

1. For a), \(z^{-3+7-2}=z^2\). 2. For b), take the reciprocal and cube: \(\left(\frac{y}{x^2}\right)^3=\frac{y^3}{x^6}\). 3. For c), \((3a^{-2})^2a^5=9a^{-4}a^5=9a\).

a) \(z^2\) b) \(\frac{y^3}{x^6}\) c) \(9a\)

- Divide coefficients and subtract exponents in part a). - Apply each outer exponent before combining powers with the same base. - In part c), decide whether the odd exponent preserves or changes the negative sign.

1. For a), divide the coefficients and subtract exponents: \(\frac{4x^{-2}}{2x^{-4}}=2x^{-2-(-4)}=2x^2\). 2. For b), \((x^{-2}y^3)^2(x^3y^{-4})=x^{-4}y^6x^3y^{-4}=x^{-1}y^2=\frac{y^2}{x}\). 3. For c), \((-a^2)^{-3}=-a^{-6}\) because the exponent is odd. Then \(-a^{-6}a^7=-a\).

a) \(2x^2\) b) \(\frac{y^2}{x}\) c) \(-a\)

- A negative exponent creates a reciprocal. - Moving a factor across a fraction bar changes the exponent’s sign. - An exponent applies only to the base immediately before it unless parentheses show otherwise.

1. In a), a negative exponent means a reciprocal, not a negative value: \(3^{-2}=\frac{1}{3^2}=\frac{1}{9}\). 2. In b), moving a power across the fraction bar changes the sign of the exponent: \(\frac{1}{x^5}=x^{-5}\). 3. In c), the exponent applies only to \(a\), not to the coefficient \(2\): \(2a^{-3}=\frac{2}{a^3}\).

a) A negative exponent means a reciprocal, not a negative value: \(3^{-2}=\frac{1}{9}\). b) Moving a power across the fraction bar changes the exponent sign: \(\frac{1}{x^5}=x^{-5}\). c) The exponent applies only to \(a\), not to \(2\): \(2a^{-3}=\frac{2}{a^3}\).

- Add exponents when multiplying powers with the same base. - Subtract the denominator exponent when dividing. - Be careful when subtracting a negative number.

1. Simplify the numerator: \(y^{-3}y^5=y^{-3+5}=y^2\). 2. Divide by \(y^{-2}\): \(y^{2-(-2)}=y^4\). 3. Sarah is correct. Lucas incorrectly treated subtraction of \(-2\) as subtraction of \(2\).

Sarah is correct; the expression simplifies to \(y^4\).

- Apply the product, quotient, and power-of-a-power rules. - A nonzero base raised to the zero power equals \(1\). - Distribute a subtraction sign across grouped terms in an exponent.

1. For a), \(a^n a^{-n}=a^{n-n}=a^0=1\), so the statement is true. 2. For b), \((a^{-1})^n=a^{-n}=\frac{1}{a^n}\), so the statement is true. 3. For c), \(\frac{a^n}{a^{n+2}}=a^{n-(n+2)}=a^{-2}=\frac{1}{a^2}\), so the statement is false.

a) True. b) True. c) False; the left side equals \(\frac{1}{a^2}\).

- A negative exponent on a fraction reverses the fraction. - Apply an exponent to every factor in parentheses. - Subtract exponents carefully when dividing.

1. For a), \((3y^2)^{-2}=\frac{1}{9y^4}\), so \(\frac{27y^5}{9y^4}=3y\). 2. For b), \(\left(\frac{k}{2}\right)^{-3}=\left(\frac{2}{k}\right)^3=\frac{8}{k^3}\), so the product is \(\frac{8}{k}\). 3. For c), \((m^{-2})^3=m^{-6}\), so \(m^{-4-(-6)}=m^2\).

a) \(3y\) b) \(\frac{8}{k}\) c) \(m^2\)

- Simplify the numerator and denominator separately. - Recall that a nonzero number to the zero power equals \(1\). - Rewrite negative powers before combining factors.

1. For a), \((u^3v^{-2})^{-1}=u^{-3}v^2\). Dividing by \(u^{-4}v\) gives \(u^{-3-(-4)}v^{2-1}=uv\). 2. For b), \(\left(\frac{2}{x^2}\right)^{-2}=\frac{x^4}{4}\), and \((xy^0)^{-3}=x^{-3}\) because \(y^0=1\). 3. Multiply: \(\frac{x^4}{4}x^{-3}=\frac{x}{4}\).

a) \(uv\) b) \(\frac{x}{4}\)

- Apply powers to both numerator and denominator. - Rewrite division as multiplication by the reciprocal. - A negative exponent on a fraction reverses the fraction.

1. For a), \(\frac{x^4}{y^4}\cdot\frac{y^2}{x^2}=\frac{x^2}{y^2}\), where \(x\ne0\) and \(y\ne0\). 2. For b), \(\frac{a^2}{b^2}\cdot\frac{b}{a^3}=\frac{1}{ab}\), where \(a\ne0\) and \(b\ne0\). 3. For c), \(\left(\frac{n}{m}\right)^2\cdot\frac{m^2}{n^2}=1\), where \(m\ne0\) and \(n\ne0\).

a) \(\frac{x^2}{y^2}\), where \(x\ne0\) and \(y\ne0\) b) \(\frac{1}{ab}\), where \(a\ne0\) and \(b\ne0\) c) \(1\), where \(m\ne0\) and \(n\ne0\)

- Apply each exponent to the coefficient and every variable factor. - Multiply exponents when raising a power to a power. - Combine powers with the same fractional base.

1. For a), \(\left(\frac{2a}{b^2}\right)^3=\frac{8a^3}{b^6}\) and \(\left(\frac{b}{a}\right)^4=\frac{b^4}{a^4}\). 2. Multiply and simplify: \(\frac{8a^3b^4}{b^6a^4}=\frac{8}{ab^2}\). 3. For b), \(\left(\left(\frac{x}{y}\right)^{-2}\right)^3=\left(\frac{x}{y}\right)^{-6}=\left(\frac{y}{x}\right)^6\). 4. Divide powers with the same base: \(\left(\frac{y}{x}\right)^{6-4}=\frac{y^2}{x^2}\).

a) \(\frac{8}{ab^2}\) b) \(\frac{y^2}{x^2}\)

- Compare each value with the value immediately above it. - Ask what operation matches a decrease of \(1\) in the exponent. - Express the values below \(3^0\) as reciprocals of positive powers.

1. Each time the exponent decreases by \(1\), divide the previous value by \(3\). 2. Therefore, \(3^0=3\div3=1\), \(3^{-1}=1\div3=\frac13\), and \(3^{-2}=\frac13\div3=\frac19\). 3. Since \(3^2=9\), the value \(3^{-2}=\frac19\) is \(\frac{1}{3^2}\). 4. The same pattern gives \(a^{-n}=\frac{1}{a^n}\) for every nonzero \(a\) and positive integer \(n\).

a) \(3^0=1\), \(3^{-1}=\frac13\), and \(3^{-2}=\frac19\) b) Decreasing the exponent by \(1\) divides the value by \(3\), so \(3^0=1\) and \(3^{-2}=\frac19=\frac{1}{3^2}\). c) \(a^{-n}=\frac{1}{a^n}\)

- Treat a repeated binomial as a single base. - Add exponents when multiplying and subtract them when dividing. - Rewrite a negative exponent as a reciprocal.

1. For a), \(x^{3-5}=x^{-2}=\frac{1}{x^2}\). 2. For b), \(a^{-2-(-4)}=a^2\). 3. For c), \((y-1)^{2-3}=(y-1)^{-1}=\frac{1}{y-1}\). 4. For d), \(2z^{-3+1}=2z^{-2}=\frac{2}{z^2}\).

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

- Apply a negative exponent to every factor inside parentheses. - Subtract exponents when dividing powers with the same base. - Recall that a nonzero base to the zero power equals \(1\).

1. For a), \((2a)^{-3}=\frac{1}{8}a^{-3}\), so the product is \(a\). 2. For b), \((xy)^{-1}=x^{-1}y^{-1}\). Dividing gives \(x^{-2-(-1)}y^{2-(-1)}=x^{-1}y^3=\frac{y^3}{x}\). 3. For c), \(\left(\frac{b^2}{3}\right)^{-2}=9b^{-4}\), so multiplying by \(b^4\) gives \(9\).

a) \(a\) b) \(\frac{y^3}{x}\) c) \(9\)

- Apply the outside exponent to each factor in the product. - Compare the exponents after simplifying part b). - Check which number is the base of the power in each expression in part c).

1. The numerator in a) is \(u^{-3}v^6\). Dividing gives \(u^{-3-(-4)}v^{6-5}=uv\). 2. For b), \(\frac{x^n}{x^{-3}}=x^{n+3}\). An identity for all nonzero \(x\) requires \(n+3=2\), so \(n=-1\). 3. In \((-3)^{-2}\), the negative sign is part of the base and the even power gives a positive result. In \(-3^{-2}\), the negative sign is outside the power.

a) \(uv\) b) \(n=-1\) c) \((-3)^{-2}\) is positive because the negative base is squared. In \(-3^{-2}\), the minus sign stays outside the power, so the result is negative.

- Determine whether the negative sign is part of the base. - Add exponents when multiplying like bases and subtract them when dividing. - Multiply exponents for a power of a power. - Rewrite a negative exponent as a reciprocal.

1. \((-5)^2=25\), while \(-5^2=-(5^2)=-25\). 2. The denominator is \(x^2\cdot x^3=x^5\), so \(\frac{x^6}{x^5}=x\). 3. \((2a^3)^4=2^4(a^3)^4=16a^{12}\). 4. \(10^3\cdot10^{-5}=10^{-2}=\frac{1}{100}=0.01\).

a) \(25\) and \(-25\) b) \(x\) c) \(16a^{12}\) d) \(0.01\)

- A negative exponent produces a reciprocal. - Rewrite \(32\) as a power of \(2\). - Use the quotient property when two powers have the same exponent.

1. \(2^{-3}=\frac{1}{8}=0.125\), while \(-2^3=-8\). Therefore, \(2^{-3}>-2^3\). 2. \((x^2)^3=x^6\), and \(x^4\cdot x^{-1}=x^3\). Thus \(\frac{x^6}{x^3}=x^3\). 3. \(\frac{15^3}{5^3}=\left(\frac{15}{5}\right)^3=3^3=27\). 4. Since \(\frac{1}{32}=2^{-5}\), matching exponents gives \(x=-5\).

a) \(2^{-3}=0.125>-8=-2^3\) b) \(x^3\) c) \(27\) d) \(x=-5\)

- Expand powers of products before combining like bases. - Rewrite \(125\) as a power of \(5\). - In part c, rewrite every power using base \(2\).

1. Expand the denominator: \(3(ab)^2=3a^2b^2\). Then \(\frac{18a^5b^{-2}}{3a^2b^2}=6a^3b^{-4}=\frac{6a^3}{b^4}\). 2. Since \(125=5^3\), \(5^{8-n}=5^3\). Therefore, \(8-n=3\) and \(n=5\). 3. Rewrite \(4^k=(2^2)^k=2^{2k}\). The numerator is \(2^{2k+1}\), so the quotient is \(2^{2k+1-2k}=2\).

a) \(\frac{6a^3}{b^4}\) b) \(n=5\) c) \(2\)

- Determine every factor to which the exponent applies. - Check whether the exponent applies to the entire quantity in parentheses. - Treat the numerical coefficient and the power of \(x\) consistently.

1. The exponent applies to both factors inside the parentheses. Karim squared \(x\) but did not square \(3\). 2. Correctly expand the power: \((3x)^2 = 3^2x^2 = 9x^2\). 3. Multiply: \(4x^2 \cdot 9x^2 = 36x^4\).

Karim failed to square the coefficient \(3\). The correct result is \(36x^4\).

- Substitute \(a = 3\) and follow the order of operations. - Parentheses identify the complete base of a power. - Exponents are evaluated before multiplication when there are no grouping symbols.

1. Substitute \(a = 3\): \(T_1 = (2 \cdot 3)^3 = 6^3 = 216\). 2. For the second expression, \(T_2 = 2 \cdot 3^3 = 2 \cdot 27 = 54\). 3. In \(T_1\), the parentheses make \(2a\) the entire base, so the product is cubed. In \(T_2\), the exponent applies only to \(a\), and multiplication by \(2\) occurs afterward.

a) \(T_1 = 216\) and \(T_2 = 54\) b) In \(T_1\), the entire product \(2a\) is cubed. In \(T_2\), only \(a\) is cubed before multiplying by \(2\).

- Multiply exponents when raising a power to a power. - Add exponents in the numerator. - Subtract the denominator exponent carefully.

1. Raise the power to a power: \((a^2)^{-2}=a^{-4}\). 2. Combine the numerator: \(a^{-3}a^{-4}=a^{-7}\). 3. Divide: \(a^{-7}\div a^{-5}=a^{-7-(-5)}=a^{-2}\).

\(a^{-2}\)

- Determine whether a negative sign is inside or outside the base. - A variable with no written exponent has exponent \(1\). - Add exponents when multiplying powers with the same base. - Subtract exponents when dividing powers with the same nonzero base.

1. a) \((-4)^2 = 16\), \(-4^2 = -(4^2) = -16\), and \((-2)^3 = -8\). 2. b) Since \(x = x^1\), add the exponents: \(x^{2+5+1} = x^8\). 3. c) First divide: \(y^8 \div y^3 = y^{8-3} = y^5\). Then multiply: \(y^5 \cdot y^2 = y^7\).

a) \(16, -16, -8\) b) \(x^8\) c) \(y^7\)

- Rewrite the decimal as a power of \(10\). - Add exponents when multiplying powers with the same base. - Equal powers with the same base have equal exponents.

1. Write \(0.01\) as a power of \(10\): \(0.01=10^{-2}\). 2. Combine the powers on the left: \(10^4\cdot10^x=10^{4+x}\). 3. Set the exponents equal: \(4+x=-2\). 4. Solve to get \(x=-6\).

\(x=-6\)

- Rewrite \(8\) and \(32\) as powers of \(2\). - Apply the power-of-a-power rule. - Compare the exponents after the bases match.

1. Write every term as a power of \(2\): \(8^{-1}=(2^3)^{-1}=2^{-3}\) and \(\frac{1}{32}=2^{-5}\). 2. Combine the left side: \(2^n\cdot2^{-3}=2^{n-3}\). 3. Set exponents equal: \(n-3=-5\). 4. Solve to get \(n=-2\).

\(n=-2\)

- Rewrite the negative exponent as a reciprocal. - Isolate \(a^2\). - Remember both square roots of a positive number.

1. Rewrite the negative exponent: \(25a^{-2}=\frac{25}{a^2}\). 2. Solve \(\frac{25}{a^2}=4\): \(25=4a^2\), so \(a^2=\frac{25}{4}\). 3. Taking both square roots gives \(a=\pm\frac{5}{2}\).

\(a=-\frac{5}{2}\) or \(a=\frac{5}{2}\)

- Simplify the numerator and denominator separately. - Multiply exponents in the denominator. - Subtract the denominator exponent when dividing.

1. Simplify the numerator: \(x^4x^{-7}=x^{-3}\). 2. Simplify the denominator: \((x^{-2})^2=x^{-4}\). 3. Divide: \(x^{-3}\div x^{-4}=x^{-3-(-4)}=x\).

\(x\)

- A negative exponent on a fraction reverses the fraction. - Apply the exponent to every factor in a product. - Cancel common powers at the end.

1. Reverse the first fraction and square: \(\left(\frac{a^2}{b}\right)^{-2}=\frac{b^2}{a^4}\). 2. Rewrite the second factor: \((ab^3)^{-1}=\frac{1}{ab^3}\). 3. Multiply: \(\frac{b^2}{a^4}\cdot\frac{1}{ab^3}=\frac{1}{a^5b}\).

\(\frac{1}{a^5b}\)

- Rewrite the decimal as a fraction or a power with a useful base. - Combine powers with the same exponent when possible. - Pay attention to whether a negative sign is part of the base.

1. For a), \(0.25=4^{-1}\), so \(0.25^{-3}=4^3\), and \(2^4=4^2\). Thus \(4^{3-5+2}=4^0=1\). 2. For b), \((-6)^3\div3^3=(-2)^3\). Then \((-2)^3\cdot(-2)^{-1}=(-2)^2=4\).

a) \(1\) b) \(4\)

- Factor \(12\) into powers of \(2\) and \(3\). - Simplify the numerator first. - Subtract the denominator exponents when dividing.

1. Factor \(12=3\cdot2^2\): \(12^3=3^3\cdot2^6\). 2. The numerator becomes \(3^3\cdot2^{6-4}=3^3\cdot2^2\). 3. Divide by the denominator: \(3^{3-2}\cdot2^{2-(-1)}=3\cdot2^3\). 4. Evaluate: \(3\cdot8=24\).

\(24\)

- Rewrite the negative exponent as a reciprocal before substituting. - Compare the denominator when the input doubles. - What can you say about the square of a nonzero number?

1. Since \(x^{-2}=\frac{1}{x^2}\), \(E(x)=\frac{3}{x^2}\). 2. Substitute: \(E(1)=3\), \(E(2)=\frac{3}{4}\), and \(E(4)=\frac{3}{16}\). 3. Replacing \(x\) by \(2x\) multiplies its square by \(4\), so \(E(x)\) is divided by \(4\). 4. The numerator \(3\) is positive and \(x^2>0\) whenever \(x\ne0\), so \(E(x)>0\).

a) \(E(x)=\frac{3}{x^2}\) b) \(E(1)=3\), \(E(2)=\frac{3}{4}\), \(E(4)=\frac{3}{16}\) c) The value is divided by \(4\) each time \(x\) is doubled because \((2x)^2=4x^2\), making the denominator four times as large. d) Both \(3\) and \(x^2\) are positive for nonzero \(x\).

- Add exponents when multiplying like bases and subtract them when dividing. - Rewrite a negative exponent as a reciprocal. - Simplify inside parentheses before applying an outside exponent.

1. \(x^4\cdot x^{-6}=x^{-2}=\frac{1}{x^2}\). 2. \(\frac{y^{-2}}{y^3}=y^{-5}=\frac{1}{y^5}\). 3. \((-2z)^{-3}=\frac{1}{(-2z)^3}=-\frac{1}{8z^3}\). 4. \((a^2\cdot a^{-3})^2=(a^{-1})^2=a^{-2}=\frac{1}{a^2}\).

a) \(\frac{1}{x^2}\) b) \(\frac{1}{y^5}\) c) \(-\frac{1}{8z^3}\) d) \(\frac{1}{a^2}\)

- Apply an outside exponent to every factor in parentheses. - A negative exponent reverses a fraction. - Combine like bases only after expanding the powers.

1. Divide coefficients and combine exponents: \(5a^{4-7-(-2)}=5a^{-1}=\frac{5}{a}\). 2. \((2b^{-2})^3=8b^{-6}\) and \((b^2)^2=b^4\). Their product is \(8b^{-2}=\frac{8}{b^2}\). 3. \(\left(\frac{x}{y^2}\right)^{-2}=\frac{y^4}{x^2}\). Multiplying by \(x^{-1}\) gives \(\frac{y^4}{x^3}\).

a) \(\frac{5}{a}\) b) \(\frac{8}{b^2}\) c) \(\frac{y^4}{x^3}\)

- What part of the volume formula stays the same when the edge length changes? - Compare the new and old volumes using a ratio before finding a percent increase. - How does replacing \(a\) with \(3a\) affect the power in the formula?

1. The fixed factor \(\frac{\sqrt{2}}{12}\) cancels in the ratio of volumes: \(\frac{V_{12}}{V_{10}}=\frac{12^3}{10^3}=\left(\frac{12}{10}\right)^3=1.728\). 2. The percent increase is \((1.728-1)\cdot100\%=72.8\%\). 3. If the edge length triples, \(V_{\text{new}}=\frac{\sqrt{2}}{12}(3a)^3=27\cdot\frac{\sqrt{2}}{12}a^3=27V_{\text{old}}\).

a) \(72.8\%\) increase, since \(\left(\frac{12}{10}\right)^3=1.728\). b) The volume is multiplied by \(27\), because \(\frac{\sqrt{2}}{12}(3a)^3=27\cdot\frac{\sqrt{2}}{12}a^3\).

- Combine variable expressions in exponents just as you combine numerical exponents. - A zero exponent gives \(1\) for a nonzero base. - Determine the sign before combining powers.

1. The numerator exponent in part a is \((2n+1)+(n-2)=3n-1\). Dividing by \(a^{3n-1}\) gives \(a^0=1\). 2. \((2b^3)^{-2}=\frac{1}{4}b^{-6}\). Multiplying by \(4b^6\) gives \(1\). 3. Since \((-x)^4=x^4\), the quotient is \(x^{4-2-5}=x^{-3}=\frac{1}{x^3}\).

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

- Expand each power as repeated multiplication. - Identify whether a coefficient is inside or outside the base. - Count how many times each variable appears as a factor.

1. \(5a^2 = 5 \cdot a \cdot a\), while \((5a)^2 = (5a)(5a) = 25a^2\). The expressions are not equivalent. 2. \(x^2 \cdot x^4\) contains six factors of \(x\), so it equals \(x^6\). The claim is false because exponents are added when powers with the same base are multiplied. 3. The product has three factors of \(z\) and three factors of \(w\), so it is \(z^3w^3\).

1) No; \(5a^2\) and \((5a)^2 = 25a^2\) are not equivalent. 2) False; \(x^2 \cdot x^4 = x^6\). 3) \(z^3w^3\)

- A power outside parentheses applies to every factor inside. - Follow the order of operations when substituting. - Count repeated factors separately in each term.

1. The repeated factor is \(3y\), so \((3y)(3y) = (3y)^2 = 9y^2\). 2. \(2x^3 = 2 \cdot 3^3 = 54\), while \((2x)^3 = (2 \cdot 3)^3 = 216\). 3. The products become \(m^3 + n^2\).

1) \((3y)^2 = 9y^2\) 2) \(54\) and \(216\) 3) \(m^3 + n^2\)

- A coefficient tells how many equal terms are added. - An exponent tells how many equal factors are multiplied. - Substitute a simple value such as \(x = 2\) as a check.

1. Expand \(3x^2\) as repeated addition: \(3x^2 = x \cdot x + x \cdot x + x \cdot x\). 2. Each product in the second expression is \(x^3\), and it appears twice, so \(x \cdot x \cdot x + x \cdot x \cdot x = 2x^3\). 3. Since \(3x^2\) and \(2x^3\) are not equivalent, Lina is incorrect. She switched the roles of the coefficient and exponent.

Lina is incorrect. \(3x^2 = x \cdot x + x \cdot x + x \cdot x\), while \(x \cdot x \cdot x + x \cdot x \cdot x = 2x^3\). She confused the coefficient with the exponent.

- Pay attention to which quantity the exponent applies to. - Equivalent expressions must have the same value for every allowable input. - The area of a square is the square of its side length.

1. For \(T_1\), add \(n\) and \(3\), then square the sum. For \(T_2\), square \(n\) and \(3\) separately, then add. 2. At \(n = 2\), \(T_1 = 25\) and \(T_2 = 13\). At \(n = 5\), \(T_1 = 64\) and \(T_2 = 34\). Since the values differ for the same inputs, the expressions are not equivalent. 3. The new side length is \(n + 3\), so the area is \((n + 3)^2\,\text{cm}^2\). Thus, \(T_1\) represents the area.

1) \(T_1\) is the square of a sum; \(T_2\) is a sum of squares. 2) For \(n = 2\): \(T_1 = 25\), \(T_2 = 13\). For \(n = 5\): \(T_1 = 64\), \(T_2 = 34\). They are not equivalent. 3) \(T_1\); the area is \((n + 3)^2\,\text{cm}^2\).

- A negative base raised to an even power gives a positive result. - A negative base raised to an odd power gives a negative result. - Convert a mixed number to an improper fraction before raising it to a power.

1. \((-8)^2 = 64\). 2. \((-3)^3 = -27\). 3. \(\left(-\frac{2}{5}\right)^2 = \frac{4}{25}\). 4. \((-0.1)^3 = -0.001\). 5. Convert the mixed number: \(-1\frac{1}{2} = -\frac{3}{2}\). Then \(\left(-\frac{3}{2}\right)^2 = \frac{9}{4} = 2\frac{1}{4}\).

1) \(64\) 2) \(-27\) 3) \(\frac{4}{25}\) 4) \(-0.001\) 5) \(2\frac{1}{4}\)

- Identify whether the negative sign is inside or outside the base. - Track the sign of a product with an even or odd number of negative factors. - Compare the order of operations in the two expressions.

1. For \(n = 2\), \(A = (-5)^2 = 25\) and \(B = -5^2 = -25\). They are not equal. 2. For \(n = 3\), \(A = (-5)^3 = -125\) and \(B = -5^3 = -125\). They are equal. 3. When \(n\) is a positive odd integer, a product of \(n\) negative factors is negative, so \((-x)^n = -x^n\) for every real \(x\).

a) \(A = 25\), \(B = -25\); they are not equal. b) \(A = -125\), \(B = -125\); they are equal. c) The equation is true for every positive odd integer exponent \(n\).

- First determine whether each power of a negative base is positive or negative. - Add exponents when multiplying powers with the same base. - Subtract exponents when dividing powers with the same nonzero base.

1. a) \((-a)^2 = a^2\), so \(a^2 \cdot a^4 = a^6\). 2. b) \((-a)^3 = -a^3\), so \(-a^3 \cdot a^3 = -a^6\). 3. c) \((-x)^4 = x^4\), so \(x^7 \div x^4 = x^3\). 4. d) Add the exponents on the common base: \((-y)^5 = -y^5\).

a) \(a^6\) b) \(-a^6\) c) \(x^3\) d) \(-y^5\)

- Add exponents for powers with the same base. - Write a simple equation for the missing exponent. - In part c), account for both the exponent and the negative sign.

1. Since \(7+\square=12\), the missing exponent is \(5\). 2. The exponent on the product is \(n+4\), so the missing exponent is \(4\). 3. The missing factor must supply a negative sign and five more factors of \(y\), so it is \(-y^5\). 4. Add the exponent expressions: \((k+2)+(k-1)=2k+1\).

a) \(5\) b) \(4\) c) \(-y^5\) d) \(2k+1\)

- Compare the coefficients and the exponent totals separately. - For powers with the same base, examine what repeated multiplication does to the exponent count. - Test the stated rule using the numerical example before generalizing it.

1. The missing coefficient is \(12\div3=4\), and the missing exponent is \(9-4=5\), so the factor is \(4x^5\). 2. The missing coefficient is \(20\div(-5)=-4\), and the missing exponent is \(3-2=1\), so the factor is \(-4a\). 3. \(y^k\cdot y^5=y^{k+5}\). 4. \(2^3\cdot2^2=8\cdot4=32=2^5\), not \(2^6\). The exponents are added, not multiplied.

a) \(4x^5\) b) \(-4a\) c) \(y^{k+5}\) d) Paul is incorrect; \(2^3 \cdot 2^2 = 2^{3+2} = 2^5 = 32\).

- Handle the numerical coefficient and each variable base separately. - For the missing factor, compare what is already present with the target product. - Keep letter-valued exponents symbolic while applying the same-base power rule.

1. The coefficient product is \(-\frac{4}{7} \cdot \frac{14}{8} = -1\). Add exponents to get \(-x^7y^{n+3}\). 2. Divide coefficients and subtract exponents: \(12 \div (-3) = -4\), \(5 - 2 = 3\), and \(7 - 4 = 3\). The factor is \(-4a^3b^3\). 3. Multiply coefficients and add exponents: \((-1) \cdot 1.2u^{3+k}v^{2+1} = -1.2u^{k+3}v^3\).

1) \(-x^7y^{n+3}\) 2) \(-4a^3b^3\) 3) \(-1.2u^{k+3}v^3\)

- Multiply the coefficients. - Add exponents on common bases. - Simplify exponent expressions after adding them.

1. a) \((7 \cdot 3)x^{n+4} = 21x^{n+4}\). 2. b) \((-5 \cdot -2)a^{(k+2)+3} = 10a^{k+5}\). 3. c) \(\left(\frac{1}{3} \cdot -9\right)y^{m+(m-1)} = -3y^{2m-1}\).

a) \(21x^{n+4}\) b) \(10a^{k+5}\) c) \(-3y^{2m-1}\)

- Compare the coefficient of the known factor with the coefficient of the product. - Compare the exponent totals on the same base. - Substitute the completed factor to check both parts at once.

1. Divide the target coefficient by the known coefficient: \(-10 \div 2 = -5\). 2. If the missing exponent is \(r\), then \(n + r = n + 3\), so \(r = 3\). 3. The missing factor is \(-5b^3\).

\(-5b^3\)

- A variable without a written exponent has exponent \(1\). - Add exponents when multiplying powers with the same base. - Check the coefficient and each variable separately.

1. The coefficients give \(12\), but the exponents must be added: \(a^{2+1}b^{1+3} = a^3b^4\). The correct result is \(12a^3b^4\). 2. The coefficients give \(10\), and the variable powers give \(x^4y^3\). This result is correct. 3. The coefficient is \(\frac{2}{3} \cdot 6 = 4\), and the variable powers give \(m^{2+3}n^{4+1} = m^5n^5\). The correct result is \(4m^5n^5\).

1) Incorrect; \(12a^3b^4\) 2) Correct 3) Incorrect; \(4m^5n^5\)

- Use the volume formula \(V = abc\). - Group coefficients and like variable bases. - Count how many factors of \(2\) enter the volume when edges are doubled.

1. a) \(V = abc = (2x^2y)(1.5xy^2)(4x^3)\). Multiply coefficients and add exponents: \(V = 12x^6y^3\). 2. b) Doubling one edge doubles the volume, so the new volume is \(24x^6y^3\). 3. c) Doubling all three edges multiplies the volume by \(2^3 = 8\), so the new volume is \(96x^6y^3\).

a) \(12x^6y^3\) b) \(24x^6y^3\) c) \(96x^6y^3\); the volume is multiplied by \(8\).

- Track every factor that lies inside an outside power. - For a power raised to another power, compare a repeated-product expansion with the proposed exponent. - When a fractional coefficient is squared, check both parts of the fraction.

1. a) Incorrect. The coefficient must be cubed: \((2z^4)^3 = 2^3z^{12} = 8z^{12}\). 2. b) Correct: \((-5k^2)^2 = 25k^4\). 3. c) Incorrect. Exponents are multiplied when raising a power to a power: \((a^3)^2 = a^6\). 4. d) Correct: \(\left(\frac{2}{3}m\right)^2 = \frac{4}{9}m^2\).

a) Incorrect; \(8z^{12}\). The coefficient was multiplied by the exponent instead of raised to the power. b) Correct c) Incorrect; \(a^6\). The exponents were added instead of multiplied. d) Correct

- Add exponents when multiplying powers with the same base. - Apply sign rules when multiplying coefficients. - Treat exponent expressions as grouped quantities before simplifying.

1. Multiply coefficients: \(1.2 \cdot (-5) = -6\). Add exponents: \(a^{n+(n+1)} = a^{2n+1}\) and \(b^{k+3k} = b^{4k}\). The result is \(-6a^{2n+1}b^{4k}\). 2. The coefficient product is \(-\frac{2}{5} \cdot \frac{15}{4} = -\frac{3}{2}\). Add exponents: \(x^{(m-1)+3} = x^{m+2}\) and \(y^{2+(m+1)} = y^{m+3}\). The result is \(-\frac{3}{2}x^{m+2}y^{m+3}\).

1) \(-6a^{2n+1}b^{4k}\) 2) \(-\frac{3}{2}x^{m+2}y^{m+3}\)

- Treat the coefficient and each variable base as separate matching conditions. - Check which exponent condition is already satisfied before solving for the unknown one. - Verify the completed product against every factor on the right side.

1. Compare coefficients: \(4\cdot\square=-12\), so \(\square=-3\). 2. The powers of \(a\) already match because \(a^k\cdot a^{k+1}=a^{2k+1}\). 3. Compare the powers of \(b\): \(2+n=7\), so \(n=5\).

\(\square=-3\) and \(n=5\)

- Raise each factor inside parentheses to the outside power. - Multiply exponents when raising a power to a power. - Add exponents when multiplying powers with the same base.

1. a) \((5a^2)^3 = 125a^6\). Then \(125a^6 \cdot 2a^5 = 250a^{11}\). 2. b) \((-2x^3)^4 = 16x^{12}\). Then \(16x^{12} \cdot (-3x^2) = -48x^{14}\).

a) \(250a^{11}\) b) \(-48x^{14}\)

- Simplify each powered monomial separately. - An even power makes the negative coefficient positive. - Multiply coefficients and add exponents at the end.

1. \((2xy^3)^3 = 2^3x^3y^9 = 8x^3y^9\). 2. \((-0.5x^2y)^4 = (-0.5)^4x^8y^4 = \frac{1}{16}x^8y^4\). 3. Multiply: \(8 \cdot \frac{1}{16} = \frac{1}{2}\), \(x^{3+8} = x^{11}\), and \(y^{9+4} = y^{13}\). 4. The result is \(\frac{1}{2}x^{11}y^{13}\).

\(\frac{1}{2}x^{11}y^{13}\)

- Apply each outside exponent to both factors inside parentheses. - Think about how an even or odd number of negative factors affects a product. - Test a different small outside exponent in part c) and check the sign.

1. \(A=(-2)^4a^{12}=16a^{12}\), and \(B=(-4)^2a^{12}=16a^{12}\). They are equivalent. 2. The even outside exponent makes the negative coefficient positive. 3. An odd outside exponent makes it negative. For example, \((-2a^3)^3=-8a^9\).

a) \(A=B=16a^{12}\) b) The outside exponent is even, so the negative coefficient \(-2\) is raised to an even power and becomes positive. c) For example, \((-2a^3)^3=-8a^9\).

- Simplify each powered monomial first. - Determine the sign from the parity of the outside exponent. - Add exponents when multiplying powers with the same base.

1. \((-3a^kb^2)^3 = -27a^{3k}b^6\). 2. \((2a^2b^n)^2 = 4a^4b^{2n}\). 3. Multiply the results: \((-27) \cdot 4 = -108\), \(a^{3k}a^4 = a^{3k+4}\), and \(b^6b^{2n} = b^{2n+6}\). 4. The simplified expression is \(-108a^{3k+4}b^{2n+6}\).

\(-108a^{3k+4}b^{2n+6}\)

- Work from the innermost power outward. - Track the sign after an even or odd power. - Compare the final exponents and signs.

1. \(T_1 = (x^2)^3 = x^6\). 2. \(T_2 = (-x^3)^2 = x^6\). 3. Both expressions simplify to \(x^6\), so the claim is correct.

The claim is true because both expressions simplify to \(x^6\).

- Substitute the scaled variables and compare the new expression with the old one. - Use the power of a product to separate the scale factor. - For the second part, consider the numerator and denominator effects separately.

1. Doubling both variables changes the expression to \(\frac{(2x)^n}{2y}=2^{n-1}T\). 2. Since \(T\ne0\), a factor of \(4\) means \(2^{n-1}=2^2\), giving \(n=3\). 3. Doubling \(x\) now multiplies the numerator by \(2^3=8\). Multiplying \(y\) by \(4\) divides the ratio by \(4\). 4. The net factor is \(\frac{8}{4}=2\).

a) \(n=3\) b) The value of \(T\) is multiplied by \(2\).

- Determine whether the negative sign is inside the base. - Use even and odd exponents to determine signs. - Evaluate powers before addition or subtraction.

1. \((-4)^2 = 16\), while \(-4^2 = -16\). 2. Since \(101\) is odd, \((-1)^{101} = -1\). Since \(202\) is even, \((-1)^{202} = 1\). 3. \((-2)^5 + (-3)^3 = -32 - 27 = -59\). 4. Since \(2n\) is even, \((-1)^{2n} = 1\), so \(-(-1)^{2n} = -1\).

1) \(16\) and \(-16\) 2) \(-1\) and \(1\) 3) \(-59\) 4) \(-1\)

- Apply powers before multiplication, addition, or subtraction. - Raise both numerator and denominator of a fraction to the power. - Use parentheses when substituting negative values.

1. \(\left(-\frac{2}{3}\right)^3 = -\frac{8}{27}\). 2. \(0.2^2 - (-0.1)^3 = 0.04 - (-0.001) = 0.041\). 3. \((-1)^7 \cdot \left(\frac{1}{2}\right)^2 + 0.75 = -0.25 + 0.75 = 0.5\). 4. \((-2)^4 - (-2)^3 = 16 - (-8) = 24\).

1) \(-\frac{8}{27}\) 2) \(0.041\) 3) \(0.5\) 4) \(24\)

- What happens to the exponents when a power is raised to another power? - How does an even outer exponent affect a negative factor? - When dividing powers with the same base, subtract the exponents. - Apply the outer exponent to every factor inside the parentheses.

1. Apply the exponent to every factor: \((-3a^2b^k)^4 = (-3)^4(a^2)^4(b^k)^4 = 81a^8b^{4k}\). 2. Raise the numerator and denominator to the third power: \(\left(\frac{2x^n}{y^3}\right)^3 = \frac{2^3(x^n)^3}{(y^3)^3} = \frac{8x^{3n}}{y^9}\). 3. Simplify the numerator, then divide powers with the same base: \(\frac{(m^4q^2)^p}{m^{3p}} = \frac{m^{4p}q^{2p}}{m^{3p}} = m^{4p-3p}q^{2p} = m^pq^{2p}\).

1. \(81a^8b^{4k}\) 2. \(\frac{8x^{3n}}{y^9}\) 3. \(m^pq^{2p}\)

- Simplify the first quotient one variable at a time. - Rewrite factors in the second fraction with exponent notation before multiplying. - Check which variables have a total exponent of \(0\).

1. Simplify the first fraction by subtracting exponents: \(a^{3-(-1)}b^{-2-2}c^{4-1}=a^4b^{-4}c^3\). 2. Multiply by the second fraction: \(a^4b^{-4}c^3\cdot b^4a^{-2}c^{-3}\). 3. Combine exponents: \(a^{4-2}b^{-4+4}c^{3-3}=a^2b^0c^0=a^2\). 4. Since the exponents of \(b\) and \(c\) are \(0\), those factors equal \(1\), so the result depends only on \(a\).

\(a^2\). The factors involving \(b\) and \(c\) cancel.

- Simplify the powers in the numerator and denominator separately. - Apply the outer exponent to every factor in a product. - When dividing powers with the same base, subtract the exponents. - Treat the numerical coefficients and variable factors separately.

1. Expand the powers in the numerator: \((4x^3y)^2 = 16x^6y^2\) and \((3z^2)^3 = 27z^6\). 2. Multiply the numerator: \(16 \cdot 27x^6y^2z^6 = 432x^6y^2z^6\). 3. Expand the denominator: \(12x^4(yz)^2 = 12x^4y^2z^2\). 4. Divide coefficients and subtract exponents of like bases: \(\frac{432}{12}x^{6-4}y^{2-2}z^{6-2} = 36x^2z^4\).

\(36x^2z^4\)

- Simplify the numerator and denominator separately before dividing. - When dividing powers with the same base, subtract the exponents. - Use parentheses when substituting a negative number for \(a\).

1. Simplify inside the brackets in the numerator: \((-a)^2 = a^2\) and \((-2a)^2 = 4a^2\), so their product is \(4a^4\). 2. Square the product: \((4a^4)^2 = 16a^8\). 3. Simplify the denominator: \((-2a^2)^3 = (-2)^3(a^2)^3 = -8a^6\). 4. Divide: \(\frac{16a^8}{-8a^6} = -2a^{8-6} = -2a^2\). 5. Substitute \(a = -2\): \(-2(-2)^2 = -2 \cdot 4 = -8\).

a) \(-2a^2\) b) \(-8\)

- Apply an exponent outside parentheses to every factor inside. - Use the power of a power property. - When dividing powers with the same base, subtract the exponents. - Recall the value of a nonzero base raised to the zero power.

1. Apply the exponent to each factor in the numerator: \((2a^n b^{2m})^5 = 2^5(a^n)^5(b^{2m})^5 = 32a^{5n}b^{10m}\). 2. Simplify the power in the denominator: \((b^m)^{10} = b^{10m}\). 3. Divide: \(\frac{32a^{5n}b^{10m}}{8a^{3n}b^{10m}} = 4a^{5n-3n}b^{10m-10m}\). 4. Since \(b^0 = 1\), the expression simplifies to \(4a^{2n}\).

\(4a^{2n}\)

- Apply the outside exponent to the coefficient and both variable factors. - Treat \(x+y\) as one base. - Simplify the quotient before multiplying by the final reciprocal.

1. Expand the powered numerator: \([5(x+y)^{2k}z^n]^2=25(x+y)^{4k}z^{2n}\). 2. Divide by the denominator: \(\frac{25(x+y)^{4k}z^{2n}}{25(x+y)^{3k}z^{2n}}=(x+y)^k\). 3. Multiply by the last factor: \((x+y)^k\cdot\frac{1}{(x+y)^k}=1\). 4. Every variable factor cancels, so the final value is independent of \(x\), \(y\), and \(z\).

\(1\). All variable factors cancel.

- A negative exponent means take the reciprocal. - Notice whether a negative sign belongs to the base or stands outside the power. - Rewrite decimals as fractions when useful.

1. \(6^{-2}=\frac{1}{6^2}=\frac{1}{36}\). 2. \(\left(\frac{2}{3}\right)^{-3}=\left(\frac{3}{2}\right)^3=\frac{27}{8}\). 3. \((-0.2)^{-2}=\left(-\frac{1}{5}\right)^{-2}=(-5)^2=25\). 4. In \(-4^{-2}\), the negative sign is outside the power, so the value is \(-\frac{1}{16}\). 5. \(24\cdot2^{-4}=24\cdot\frac{1}{16}=\frac{3}{2}\).

a) \(\frac{1}{36}\) b) \(\frac{27}{8}\) c) \(25\) d) \(-\frac{1}{16}\) e) \(\frac{3}{2}\)

- Treat a sum or difference in a denominator as one grouped base. - Change the sign of an exponent when moving a factor across the fraction bar. - Preserve multiplication before subtraction in part 3).

1. Move \(c^4\) to the numerator with exponent \(-4\): \(a^3b^2c^{-4}\). 2. Treat the entire sum as one base: \((x^2+y^2)^{-1}\). 3. Rewrite each reciprocal while preserving the order of operations: \(r^{-3}-s^{-3}t^{-2}\). 4. Move both denominator factors to the numerator: \(u^{-2}v^{-3}\).

1) \(a^3b^2c^{-4}\) 2) \((x^2+y^2)^{-1}\) 3) \(r^{-3}-s^{-3}t^{-2}\) 4) \(u^{-2}v^{-3}\)

- Move denominator factors to the numerator by changing exponent signs. - Handle numerical coefficients separately. - A negative exponent in the denominator becomes positive in the numerator.

1. \(\frac{7x^2}{y^3}=7x^2y^{-3}\). 2. Move \(b^{-2}\) to the numerator and invert the coefficient: \(\frac{a^4}{5b^{-2}}=0.2a^4b^2\). 3. Move the denominator factors to the numerator: \(u^{-1}v^2w^{-3}z^4\). 4. Invert the coefficient and denominator powers: \(0.125p^{-2}q^3\).

1) \(7x^2y^{-3}\) 2) \(0.2a^4b^2\) 3) \(u^{-1}v^2w^{-3}z^4\) 4) \(0.125p^{-2}q^3\)

- Simplify coefficients before removing the fraction bar. - Apply a negative exponent to every factor inside parentheses. - Change exponent signs when moving denominator factors.

1. Divide coefficients and subtract exponents: \(2x^{-2}y^{-2}\). 2. Since \((2a)^{-2}=0.25a^{-2}\), dividing by \(a^2c\) gives \(0.25a^{-4}b^3c^{-1}\). 3. Move the denominator factors to the numerator: \(0.25u^2v^{-1}w^{-2}z^2\).

1) \(2x^{-2}y^{-2}\) 2) \(0.25a^{-4}b^3c^{-1}\) 3) \(0.25u^2v^{-1}w^{-2}z^2\)

- Negate the entire exponent expression when moving a power from the denominator. - Treat coefficients and variable factors separately. - Distribute the leading negative sign across \(m+3\).

1. Move the denominator power to the numerator: \(y^{-(m+3)}=y^{-m-3}\). 2. Moving the denominator factors changes their exponent signs: \(15a^4\cdot5^1b^nc^{-2}\). 3. Multiply the coefficients: \(15\cdot5=75\), so the result is \(75a^4b^nc^{-2}\).

1) \(y^{-m-3}\) 2) \(75a^4b^nc^{-2}\)

- Move the negative power from the denominator to the numerator. - Find the reciprocal of \(0.5\). - Compare the two rewritten expressions.

1. Move \((x+y)^{-3}\) from the denominator: \(T_1=2(x+y)^3\). 2. Since the reciprocal of \(0.5\) is \(2\), \(T_2=2(x+y)^3\). 3. Both expressions simplify to the same result, so they are equivalent for every allowed pair.

Yes. Both expressions simplify to \(2(x+y)^3\) for \(x\ne-y\).

- Change the sign of a denominator exponent when moving the factor to the numerator. - Simplify numerical coefficients separately. - Keep each binomial in parentheses and treat it as one base.

1. Move \((x+y)^{-2}\) from the denominator: \(12(x+y)^2\). 2. Since \(5\div2=2.5\), moving \((a-b)^{-3}\) gives \(2.5ab(a-b)^3\). 3. The coefficient is \(3^{-1}\div6^{-1}=\frac13\div\frac16=2\). Move the remaining negative-power factors from the denominator to obtain \(2p^2(m+n)(m-n)\).

1) \(12(x+y)^2\) 2) \(2.5ab(a-b)^3\) 3) \(2p^2(m+n)(m-n)\)

- Compare the decimal coefficient with the power of five. - Interpret each negative exponent as a reciprocal. - Simplify before substituting the given values.

1. Since \(5^{-1}=0.2\), the numerical factors cancel. 2. Moving the negative powers out of the denominator gives \(T=c^2(a+b)(a-b)^2\). 3. Substitute: \(T=(0.1)^2(7+3)(7-3)^2=0.01\cdot10\cdot16=1.6\).

a) \(T=c^2(a+b)(a-b)^2\) b) \(1.6\)

- Simplify the numerical coefficient first. - Subtract exponents for each variable separately. - Rewrite the negative exponent as a reciprocal to compare the answers.

1. The numerical quotient is \(2^{-3}\div8^{-1}=\frac{1}{8}\div\frac{1}{8}=1\). 2. For \(x\), \(x^2\div x^{-1}=x^{2-(-1)}=x^3\). 3. For \(y\), \(y^{-2}\div y=y^{-3}\). 4. Thus \(S=x^3y^{-3}\), so Leon is correct. 5. Since \(y^{-3}=\frac{1}{y^3}\), \(x^3y^{-3}=\frac{x^3}{y^3}\), so Sophie is also correct.

Both students are correct because \(x^3y^{-3}=\frac{x^3}{y^3}\).

- Treat variable expressions in exponents like numerical expressions. - Apply an outside exponent to every factor. - Subtract negative exponents carefully.

1. Divide coefficients and subtract exponents: \(5a^{2k-k}b^{(k-1)-2k}=5a^kb^{-k-1}=\frac{5a^k}{b^{k+1}}\). 2. Expand the power: \(\left(\frac{1}{2}x^{-n}y^2\right)^{-2}=4x^{2n}y^{-4}\). Multiplying by \(4x^{-2n}\) gives \(16x^0y^{-4}=\frac{16}{y^4}\).

a) \(\frac{5a^k}{b^{k+1}}\) b) \(\frac{16}{y^4}\)

- Separate the numerical coefficient from the variable power. - Treat the entire binomial \(x+2\) as the base. - Evaluate the parentheses before applying the negative exponent.

1. For a), \(\frac{1}{4x^2}=\frac{1}{4}\cdot\frac{1}{x^2}=0.25x^{-2}\). 2. For b), \(\frac{3}{5}=0.6\), so \(T=0.6(x+2)^{-2}\). 3. For c), \(T=0.6(3+2)^{-2}=0.6\cdot5^{-2}=0.6\cdot0.04=0.024\).

a) \(\frac{1}{4x^2}=0.25x^{-2}\) b) \(0.6(x+2)^{-2}\) c) \(0.024\)

- Move a negative power across the fraction bar to make its exponent positive. - Treat a binomial in parentheses as one base. - Preserve all restrictions from the original expression.

1. Move the negative powers across the fraction bar: \(\frac{6b^2}{a^3}\). 2. Move \(2^{-3}\) and \(z^{-4}\) to the numerator: \(\frac{2^3z^4}{x+y}=\frac{8z^4}{x+y}\). 3. Move the negative powers: \(\frac{q^3r(p-q)^2}{p^2}\). 4. The original restrictions are \(a,b\ne0\); \(x+y\ne0\) and \(z\ne0\); and \(p\ne0\), \(r\ne0\), and \(p-q\ne0\).

1) \(\frac{6b^2}{a^3}\), where \(a,b\ne0\) 2) \(\frac{8z^4}{x+y}\), where \(x+y\ne0\) and \(z\ne0\) 3) \(\frac{q^3r(p-q)^2}{p^2}\), where \(p\ne0\), \(r\ne0\), and \(p\ne q\)

- Rewrite the numerator and denominator without negative exponents. - Apply the exponent to both factors in \(2x\). - Divide by multiplying by the reciprocal.

1. Rewrite the numerator: \((2x)^{-3}=\frac{1}{(2x)^3}=\frac{1}{8x^3}\). 2. Rewrite the denominator: \(y^{-2}=\frac{1}{y^2}\). 3. Simplify the complex fraction: \(A=\frac{1}{8x^3}\div\frac{1}{y^2}=\frac{1}{8x^3}\cdot y^2=\frac{y^2}{8x^3}\). 4. This is exactly \(B\), so the expressions are equivalent.

Yes. Both expressions equal \(\frac{y^2}{8x^3}\) for \(x\ne0\) and \(y\ne0\).

- Handle the numerical coefficient and each letter base separately. - Multiplying like bases and dividing like bases change exponents in different ways. - In part b), apply the outside power before the quotient rule.

1. In a), the coefficient remains \(2\), and the exponent of \(x\) is \(2+(-2)-(-3)=3\). The result is \(2x^3\). 2. In b), the numerator is \(9m^4n^{-2}\). 3. Divide each like base: \(\frac{9m^4n^{-2}}{9m^{-1}n}=m^{4-(-1)}n^{-2-1}=m^5n^{-3}=\frac{m^5}{n^3}\).

a) \(2x^3\) b) \(\frac{m^5}{n^3}\)

- How would you write the outside cube as a product of equal factors? - Use the exponent rule on that product before testing a particular value. - Does checking one value prove two expressions always agree?

1. For any \(y\ne0\), \((y^{-2})^3=y^{-2}\cdot y^{-2}\cdot y^{-2}=y^{-6}\). Talia is correct because there are three factors with exponent \(-2\). 2. Rewrite \(y^{-6}=\frac{1}{y^6}\). 3. At \(y=2\), both \((2^{-2})^3\) and \(\frac{1}{2^6}\) equal \(\frac{1}{64}\). Checking one input alone would not prove equivalence.

Talia is correct for all \(y\ne0\): \((y^{-2})^3=y^{-2}\cdot y^{-2}\cdot y^{-2}=y^{-6}=\frac{1}{y^6}\). At \(y=2\), both forms equal \(\frac{1}{64}\).

- Apply an outside exponent to every factor. - Subtract exponents when dividing like bases. - A negative exponent takes the reciprocal.

1. \((-x^{-4})^{-2}=(-1)^{-2}x^8=x^8\). 2. \(\frac{(-1)^{k+2}}{(-1)^k}=(-1)^2=1\). 3. First, \(\frac{2a^2}{3b^{-1}}=\frac{2a^2b}{3}\). Taking the reciprocal and cubing gives \(\left(\frac{3}{2a^2b}\right)^3=\frac{27}{8a^6b^3}\).

a) \(x^8\) b) \(1\) c) \(\frac{27}{8a^6b^3}\)

- Determine the sign of a negative base with an odd or even exponent. - Multiply exponents for a power of a power. - Rewrite \(25\) using base \(5\).

1. Since \((-a)^3=-a^3\), the quotient is \(-a^{3-5-(-2)}=-a^0=-1\). 2. The power of a power gives \(\left(-\frac{x}{y}\right)^{-2n}\). Since \(-2n\) is even, this equals \(\left(\frac{x}{y}\right)^{-2n}=\frac{y^{2n}}{x^{2n}}\). 3. Rewrite \(25^n=(5^2)^n=5^{2n}\). Then \(\frac{5^{2n+1}}{5^{2n}}=5\).

a) \(-1\) b) \(\frac{y^{2n}}{x^{2n}}\) c) \(5\)

- Work from the innermost powers outward. - Keep coefficients and like letter bases separate. - Replace negative exponents with reciprocals only after simplifying. - For an identity with an unknown exponent, match the power of the same base on both sides.

1. Inside the parentheses, \(b^{-4}b^3=b^{-1}\). Raising this to \(-1\) gives \(b\). 2. Since \(x^0=1\), the coefficient is \(\frac{15\cdot5}{5}=15\), and the exponent is \(2+0-(-2)=4\). The result is \(15x^4\). 3. The numerator becomes \(a^{-4}b^2\). Dividing gives \(a^{-4-(-3)}b^{2-1}=a^{-1}b=\frac{b}{a}\). 4. The left side is \(2a^{n+2}b^{-5}\). To match \(2a^5b^{-5}\) for every nonzero input, \(n+2=5\), so \(n=3\).

1) \(b\) 2) \(15x^4\) 3) \(\frac{b}{a}\) 4) \(n=3\)

- Treat each binomial as one base. - Rewrite both negative exponents as reciprocals. - Dividing by a fraction means multiplying by its reciprocal.

1. Rewrite the negative powers: \((a+b)^{-2}=\frac{1}{(a+b)^2}\) and \((a-b)^{-2}=\frac{1}{(a-b)^2}\). 2. Divide by multiplying by the reciprocal: \(B=\frac{1}{(a+b)^2}\cdot(a-b)^2=\frac{(a-b)^2}{(a+b)^2}\). 3. For \(a=5\) and \(b=3\), \(B=\frac{(5-3)^2}{(5+3)^2}=\frac{4}{64}=\frac{1}{16}\).

The simplified expression is \(\frac{(a-b)^2}{(a+b)^2}\). For \(a=5\) and \(b=3\), the value is \(\frac{1}{16}\).

- Interpret an exponent of \(-1\) using reciprocals. - Compare the product forms after expressing each negative power as a fraction. - In part c), evaluate the two sides independently rather than assuming a power distributes over addition.

1. By the definition of a negative exponent, \((xy)^{-1}=\frac{1}{xy}\). 2. Also, \(x^{-1}y^{-1}=\frac{1}{x}\cdot\frac{1}{y}=\frac{1}{xy}\). Therefore, \((xy)^{-1}=x^{-1}y^{-1}\). 3. When \(x=2\) and \(y=5\), both forms equal \(\frac{1}{10}=0.1\). 4. For Maja's claim, \((2+5)^{-1}=\frac{1}{7}\), while \(2^{-1}+5^{-1}=\frac{1}{2}+\frac{1}{5}=\frac{7}{10}\). The values are different, so the claim is false.

a) \((xy)^{-1}=x^{-1}y^{-1}\) b) Both forms equal \(\frac{1}{10}=0.1\). c) The claim is false: \((2+5)^{-1}=\frac{1}{7}\), but \(2^{-1}+5^{-1}=\frac{7}{10}\).

- A negative exponent on a fraction takes the reciprocal. - Apply the outside exponent to both numerator and denominator. - Simplify the quotient of the substituted values before evaluating the power.

1. A negative exponent reverses the fraction: \(T=\left(\frac{b^n}{a^n}\right)^2\). 2. Apply the outside exponent: \(T=\frac{b^{2n}}{a^{2n}}=\left(\frac{b}{a}\right)^{2n}\). 3. For \(a=2\), \(b=4\), and \(n=2\), \(T=\left(\frac42\right)^4=2^4=16\).

\(T=\frac{b^{2n}}{a^{2n}}\). The numerical value is \(16\).

- Simplify the coefficient and each variable base separately. - Use the quotient rule to write an equation for each exponent. - Match the resulting exponents to those in the target expression.

1. Simplify the numerical coefficient: \(4\div2=2\). 2. For base \(a\), the quotient rule gives \(a^n\div a^2=a^{n-2}\). Set \(n-2=3\), so \(n=5\). 3. For base \(b\), the quotient rule gives \(b^{-3}\div b^m=b^{-3-m}\). Set \(-3-m=-5\), so \(m=2\). 4. With \(n=5\) and \(m=2\), the expression simplifies to \(2a^3b^{-5}\).

\(n=5\) and \(m=2\)

- Rewrite each numerical fraction as a power with the same base as the other side. - For a product of powers, combine the exponents before matching them. - A reciprocal reverses the base and changes the sign of the exponent.

1. For a), write \(\frac{1}{125}=5^{-3}\). Then \(z-3=-1\), so \(z=2\). 2. For b), \((x^{-2})^z x^5=x^{-2z+5}\). Set \(-2z+5=-1\), so \(z=3\). 3. For c), \(\frac{9}{4}=\left(\frac{3}{2}\right)^2=\left(\frac{2}{3}\right)^{-2}\). Therefore, \(z=-2\).

a) \(z=2\) b) \(z=3\) c) \(z=-2\)

- Expand powers of products and quotients first. - Track the sign when a negative base has an odd exponent. - Look for factors that cancel after exponent properties are applied.

1. The numerator in part a is \(9p^{-4}q^2\). Dividing gives \(p^0q^{-1}=\frac{1}{q}\). 2. \((-x^2)^{-3}=-x^{-6}\). Then \(\frac{-x^{-6}\cdot x^8}{x^{-2}}=-x^{2-(-2)}=-x^4\). 3. \(\left(\frac{2a^{-3}}{b^2}\right)^{-2}=\frac{a^6b^4}{4}\). Multiplying by \(\frac{a^{-6}}{b^4}\) gives \(\frac{1}{4}\).

a) \(\frac{1}{q}\) b) \(-x^4\) c) \(\frac{1}{4}\)

- Simplify the numerator and denominator separately using exponent properties. - A nonzero base raised to the zero power equals \(1\). - Substitute into the simplified expression for part c.

1. Use the power-of-a-power property: \((y^n)^2=y^{2n}\). Combine the denominator: \(y^4\cdot y^2=y^6\). Therefore, \(T=\frac{y^{2n}}{y^6}=y^{2n-6}\). 2. For \(T\) to equal \(1\) for every nonzero \(y\), the exponent must be \(0\). Solve \(2n-6=0\) to get \(n=3\). 3. When \(n=4\), \(T=y^{2(4)-6}=y^2\). For \(y=3\), \(T=3^2=9\).

a) \(y^{2n-6}\) b) \(n=3\) c) \(9\)

- For part 1, consider the numerical coefficient and each variable factor separately. - For part 2, apply the power of a product and power of a power properties before canceling common factors. - For part 3, test how the sign changes for a few even and odd values of \(k\).

1. Find a factor whose cube gives each part of the expression: \((-2)^3 = -8\), \((x^2)^3 = x^6\), and \((y^n)^3 = y^{3n}\). Therefore, \(T = -2x^2y^n\). 2. Simplify the left side: \(\frac{(a^n b)^m}{a^{nm}} = \frac{(a^n)^m b^m}{a^{nm}} = \frac{a^{nm}b^m}{a^{nm}} = b^m\). The equation is true under the stated conditions. 3. \((-x^2)^k = (-1)^k(x^2)^k = (-1)^kx^{2k}\). If \(k\) is even, then \((-1)^k = 1\), so the identity holds for every real \(x\). If \(k\) is odd, it fails for every \(x \ne 0\). Therefore, it holds for all real \(x\) exactly when \(k\) is even.

1. \(T = -2x^2y^n\) 2. The equation is true because the left side simplifies to \(b^m\). 3. The identity holds for all real \(x\) exactly when \(k\) is even.

- Expand every power of a product first. - Subtract exponents when dividing like bases. - Move factors with negative exponents across the fraction bar.

1. Expand the powers: \((x^2y^{-3})^2=x^4y^{-6}\), \((x^4z^2)^3=x^{12}z^6\), and \((x^2yz)^4=x^8y^4z^4\). 2. The first fraction simplifies to \(x^{16-8}y^{-6-4}z^{6-4}=x^8y^{-10}z^2\). 3. Since \(\frac{y^2}{z^{-1}}=y^2z\), the full product is \(x^8y^{-8}z^3=\frac{x^8z^3}{y^8}\).

\(\frac{x^8z^3}{y^8}\)

- Expand each power of a product first. - Combine the numerator before dividing. - Subtract negative exponents carefully.

1. Expand the numerator: \((5a^2b^{-1})^2=25a^4b^{-2}\) and \((2a^{-2}c)^3=8a^{-6}c^3\). Their product is \(200a^{-2}b^{-2}c^3\). 2. Expand the denominator: \((10a^{-1}b^{-2}c^2)^2=100a^{-2}b^{-4}c^4\). 3. Divide coefficients and subtract exponents: \(2a^0b^2c^{-1}=2b^2c^{-1}\).

\(2b^2c^{-1}\)

- Track the sign carefully when a negative base is raised to an even or odd power. - Simplify the innermost powers first. - Use \((a^m)^n = a^{mn}\). - After simplifying, check whether all terms contain the same power of \(x\).

1. Simplify the first term: \([(-2x)^2]^3 = (4x^2)^3 = 64x^6\). 2. Simplify the second term: \((-4x^3)^2 = 16x^6\). 3. Simplify the third term: \([-(2x)^3]^2 = (-8x^3)^2 = 64x^6\). 4. Simplify the factor in the fourth term: \(2(-2x^2)^3 = 2(-8x^6) = -16x^6\). 5. Combine like terms, including the subtraction before the fourth term: \(64x^6 - 16x^6 + 64x^6 - (-16x^6) = 128x^6\).

\(128x^6\)

- Rewrite decimals as fractions. - Evaluate the numerator and denominator separately. - Dividing by a fraction means multiplying by its reciprocal.

1. For part a, \((1.5)^{-2}=\left(\frac{3}{2}\right)^{-2}=\frac{4}{9}\), \(\left(\frac{1}{3}\right)^{-1}=3\), and \(4^{-1}=\frac{1}{4}\). Thus \(\frac{4}{9}+\frac{3}{4}=\frac{43}{36}\). 2. For part b, the numerator is \(\frac{1}{3}+\frac{9}{4}=\frac{31}{12}\). The denominator is \(4-\frac{7}{2}=\frac{1}{2}\). Therefore, \(\frac{31}{12}\div\frac{1}{2}=\frac{31}{6}\).

a) \(\frac{43}{36}\) b) \(\frac{31}{6}\)

- Combine powers with the same base first. - Evaluate the main numerator and denominator separately. - Track the sign of the denominator carefully.

1. The numerator is \(2^{-2}+\frac{1}{9}=\frac{1}{4}+\frac{1}{9}=\frac{13}{36}\). 2. The denominator is \(\frac{1}{9}-\frac{1}{4}=-\frac{5}{36}\). 3. Divide: \(\frac{13}{36}\div\left(-\frac{5}{36}\right)=-\frac{13}{5}=-2.6\).

\(-\frac{13}{5}\), or \(-2.6\)