- A logarithm asks for the exponent on the base. - Think of a power of \(5\) that equals \(125\).

Since \(5^3=125\), \(\log_5(125)=3\). The equivalent exponential equation is \(5^3=125\).

\(\log_5(125)=3\), equivalently \(5^3=125\).

- Recall which arithmetic operation inside a logarithm becomes addition outside it. - Logarithm properties apply to products, quotients, and powers, not ordinary sums or differences.

The product property of logarithms is \(\log(ab)=\log(a)+\log(b)\) for positive \(a\) and \(b\). There is no corresponding sum or difference property.

b) \(\log(ab)=\log(a)+\log(b)\)

- Neither logarithm has an easy whole-number value by itself. Can you rewrite their difference? - Look for a quotient that is a power of the logarithm's base.

1. The quotient property gives \(\log_3(18)-\log_3(2)=\log_3\!\left(\frac{18}{2}\right)=\log_3(9)\). 2. Since \(3^2=9\), the value is \(2\).

\(2\)

- Which part of the logarithm is raised to a power? - How does the exponent affect the coefficient when you expand a logarithm?

1. The power property moves an exponent on the logarithm's input in front of the logarithm. 2. Therefore, \(\log_5(x^3)=3\log_5(x)\) for \(x>0\).

\(3\log_5(x)\)

- Can the input to a real logarithm be zero or negative? - Check the sign of the expression inside \(\ln\) for each choice.

1. The input of a real logarithm must be positive, so \(x-1>0\), or \(x>1\). 2. Of the listed values, only \(2\) and \(3\) satisfy this condition.

\(2\) and \(3\).

- Work out what is inside the logarithm on the left. - Which operation between logarithms is covered by the product property? - Compare the two logarithm inputs, not just their appearances.

1. The left side is \(\log_{10}(5)\). 2. By the product property, the right side is \(\log_{10}(2\cdot3)=\log_{10}(6)\). 3. Since \(5\ne6\), the logarithms are different. The claim is false.

False; the two sides are \(\log_{10}(5)\) and \(\log_{10}(6)\).

- Which number is the base of the logarithm on the left? - In the change-of-base formula, where does the original base appear?

1. The change-of-base formula is \(\log_b(a)=\frac{\ln(a)}{\ln(b)}\). 2. Here the base is \(8\), so the missing denominator is \(\ln(8)\).

\(\ln(8)\)

- Read the marked point's input and output from the coordinate grid. - Rewrite the resulting logarithmic equation in exponential form.

1. The marked point is \((2, 1)\), so \(\log_b(2)=1\). 2. In exponential form, this says \(b^1=2\). Therefore \(b=2\).

\(b=2\)

- What values are not allowed as the argument of a real logarithm? - Factor the quadratic expression to find its zeros. - Use the sign of the quadratic on the intervals determined by its zeros.

1. A real logarithm is undefined when its argument is less than or equal to \(0\), so solve \(x^2 - 7x + 10 \le 0\). 2. Factor the quadratic: \(x^2 - 7x + 10 = (x - 2)(x - 5)\). 3. The product is nonpositive between the zeros, including the zeros. Therefore, \(2 \le x \le 5\).

\(x \in [2, 5]\)

- When is a natural logarithm defined over the real numbers? - Test a negative value of \(x\). - Compare the domains of both sides before applying a logarithm property. - What happens to a negative number when it is squared?

1. The left side, \(\ln(x^2)\), is defined when \(x^2>0\), which means \(x\ne0\). 2. The right side, \(2\ln(x)\), is defined only when \(x>0\). 3. The domains are not the same. For example, when \(x=-2\), \(\ln(x^2)=\ln(4)\) is defined, but \(2\ln(-2)\) is not real. 4. Therefore, Beatriz's statement is false. The identity \(\ln(x^2)=2\ln(x)\) is valid only for \(x>0\). For all \(x\ne0\), the correct identity is \(\ln(x^2)=2\ln|x|\).

The claim is false. The left side is defined for \(x\ne0\), while the right side is defined only for \(x>0\).

- Move each coefficient inside its logarithm as an exponent. - A sum of logarithms becomes a product. - A difference of logarithms becomes a quotient.

1. Use the power property: \(3\ln(x)=\ln(x^3)\) and \(\frac{1}{2}\ln(y)=\ln(\sqrt{y})\). 2. Use the product and quotient properties: \(\ln(x^3)+\ln(\sqrt{y})-\ln(z^2)=\ln\!\left(\frac{x^3\sqrt{y}}{z^2}\right)\).

\(\ln\!\left(\frac{x^3\sqrt{y}}{z^2}\right)\)

- Separate the quotient first. - Expand the products in the numerator and denominator. - Move exponents in front of logarithms. - Rewrite \(\sqrt{e}\) as a power of \(e\).

1. Apply the quotient and product properties: \(\ln(\sqrt{e})+\ln(a^4)-\ln(b)-\ln(c^2)\). 2. Apply the power property and simplify \(\ln(\sqrt{e})=\ln(e^{1/2})=\frac{1}{2}\). 3. The result is \(\frac{1}{2}+4\ln(a)-\ln(b)-2\ln(c)\).

\(\frac{1}{2}+4\ln(a)-\ln(b)-2\ln(c)\)

- Write \(x\) as a product of two equal square-root factors. - Apply the product property of logarithms. - Combine the two identical terms.

1. Since \(x>0\), write \(x=\sqrt{x}\sqrt{x}\). 2. Take natural logarithms: \(\ln(x)=\ln(\sqrt{x}\sqrt{x})\). 3. By the product property, \(\ln(x)=\ln(\sqrt{x})+\ln(\sqrt{x})=2\ln(\sqrt{x})\). 4. Divide by \(2\) to obtain \(\ln(\sqrt{x})=\frac{1}{2}\ln(x)\).

\(\ln(x)=2\ln(\sqrt{x})\), so \(\ln(\sqrt{x})=\frac{1}{2}\ln(x)\).

- What needs to be true about each logarithm input? - How can a difference of logarithms with the same base be written as one logarithm? - Check the domain using the expression before combining the logarithms.

1. Both logarithm inputs must be positive: \(x+4>0\) and \(x>0\). Together these give \(x>0\). 2. Apply the quotient property to obtain \(\log_5\!\left(\frac{x+4}{x}\right)\).

\(\log_5\!\left(\frac{x+4}{x}\right)\), with domain \(x>0\).

- Read the coordinates of the marked point. - What happens to the input and output when you reverse a function? - Consider the line across which inverse-function graphs reflect.

1. The marked point on \(f\) is \(P=(2, 4)\), meaning \(f(2)=4\). 2. For inverse functions the input and output switch, so \(g(4)=2\). 3. The corresponding point on \(g\) is \((4, 2)\), a reflection of \(P\) across the line \(y=x\).

\((4, 2)\); the inverse graph swaps the input and output coordinates.

- Check whether each logarithm input is positive at the suggested negative value. - Which positive expression has the same square as a negative \(x\)? - Apply the power property only after making the logarithm input positive.

1. At \(x=-3\), the left side is \(\ln(9)\), which is defined, but \(\ln(-3)\) on the right is not a real number. 2. For \(x\ne0\), \(x^2=|x|^2\) and \(|x|>0\), so \(\ln(x^2)=\ln(|x|^2)=2\ln|x|\).

The right side is undefined at \(x=-3\). The correct expression for \(x\ne0\) is \(2\ln|x|\).

- Convert the logarithm into a power equation. - Express the fractional base as a negative power of \(3\). - Which sign of exponent makes a base below \(1\) produce a value above \(1\)?

1. Let \(x=\log_{\frac{1}{3}}(9)\). Then \(\left(\frac{1}{3}\right)^x=9=3^2\). 2. Rewrite the base: \(3^{-x}=3^2\), so \(-x=2\) and \(x=-2\). 3. A negative exponent is needed because the base is between \(0\) and \(1\), while the result is greater than \(1\).

\(-2\)

- Recall that the natural logarithm and the exponential function are inverses. - Rewrite radicals as rational exponents. - Apply logarithm rules for products and quotients. - Use exponent rules for powers with the same base.

1. \(\sqrt[3]{e}=e^{1/3}\), so \(\ln(e^{1/3})=\frac{1}{3}\). 2. \(3\ln(2)=\ln(2^3)=\ln(8)\). Therefore, \(3\ln(2)-\ln(4)=\ln\!\left(\frac{8}{4}\right)=\ln(2)\), and \(e^{\ln(2)}=2\). 3. \(\frac{e^x}{e^{x-2}}=e^{x-(x-2)}=e^2\), so \(\ln(e^2)=2\). 4. Since \(x>0\), \(e^{\ln(x^2)}=x^2\) and \(e^{-\ln(x)}=\frac{1}{x}\). Their product is \(x\).

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

- Expand the logarithm of a product. - Equal powers with base \(e\) have equal exponents. - Apply the exponential function after isolating \(\ln(x)\).

1. Use the product property: \(\ln(e^3x)=\ln(e^3)+\ln(x)=3+\ln(x)\). 2. Therefore, \(f(x)=3+\ln(x)-\ln(x)=3\) for every \(x>0\). 3. Since the exponential expressions in part b have the same base, set the exponents equal: \(2\ln(x)+1=3\). 4. Then \(2\ln(x)=2\), so \(\ln(x)=1\) and \(x=e\).

a) \(f(x)=3\) b) \(x=e\)

- What condition must the logarithm's argument satisfy for every real input? - When is an upward-opening quadratic strictly above the x-axis? - Use the discriminant to determine when a quadratic has no real zeros. - Remember that an argument equal to \(0\) is not allowed.

1. The logarithm is defined for every real \(x\) only if \(x^2 + 6x + c > 0\) for every real \(x\). 2. The quadratic opens upward. It is always positive exactly when it has no real zeros, so its discriminant must be negative. 3. The discriminant is \(6^2 - 4(1)(c) = 36 - 4c\). Solve \(36 - 4c < 0\) to obtain \(c > 9\). 4. When \(c = 9\), the quadratic equals \((x + 3)^2\) and is \(0\) at \(x = -3\), so the inequality must be strict.

\(c > 9\)

- What restriction does \(\ln(x)\) place on \(x\)? - Which values make the denominator equal to \(0\)? - A value that makes both numerator and denominator zero is still excluded from the original function's domain. - Use the zero-product property on the numerator.

1. The natural logarithm requires \(x > 0\). The denominator must also be nonzero: \(x^2 - 1 \ne 0\), so \(x \ne -1\) and \(x \ne 1\). Because \(-1\) is already excluded by \(x > 0\), the domain is \((0, 1) \cup (1, \infty)\). 2. An x-intercept can occur when the numerator is zero: \((x - 5)\ln(x) = 0\). Thus, \(x - 5 = 0\) or \(\ln(x) = 0\), giving candidates \(x = 5\) and \(x = 1\). 3. The value \(x = 1\) is not in the domain because the denominator is zero there. The only x-intercept is \((5, 0)\).

The domain is \((0, 1) \cup (1, \infty)\), and the only x-intercept is \((5, 0)\).

- State the condition that makes the logarithm defined. - Also require the entire denominator to be nonzero. - Find the zeros of the numerator. - Check each candidate zero against the domain.

1. The logarithm requires \(x + 3 > 0\), so \(x > -3\). 2. The denominator must be nonzero. Since \(\ln(x + 3) = 0\) when \(x + 3 = 1\), exclude \(x = -2\). Therefore, the domain is \((-3, -2) \cup (-2, \infty)\). 3. For an x-intercept, the numerator must be zero: \(x^2 - 4 = 0\). Thus, \(x = -2\) or \(x = 2\). 4. The value \(x = -2\) is not in the domain, while \(x = 2\) is. Therefore, the only x-intercept is \((2, 0)\).

Domain: \((-3, -2) \cup (-2, \infty)\) X-intercept: \((2, 0)\)

- Express each logarithm input as a product or quotient involving \(2\) and \(5\). - Rewrite decimals such as \(2.5\) and \(0.4\) as fractions. - Express a square root as a power. - Use the power property to move an exponent in front of a logarithm.

1. \(\ln(10)=\ln(2\cdot5)=\ln(2)+\ln(5)\approx0.69+1.61=2.30\). 2. \(\ln(2.5)=\ln\!\left(\frac{5}{2}\right)=\ln(5)-\ln(2)\approx1.61-0.69=0.92\). 3. \(\ln(80)=\ln(2^4\cdot5)=4\ln(2)+\ln(5)\approx4(0.69)+1.61=4.37\). 4. \(\ln(0.4)=\ln\!\left(\frac{2}{5}\right)=\ln(2)-\ln(5)\approx-0.92\). 5. \(\ln(\sqrt{5})=\frac{1}{2}\ln(5)\approx0.805\).

a) \(\approx2.30\) b) \(\approx0.92\) c) \(\approx4.37\) d) \(\approx-0.92\) e) \(\approx0.805\)

- Determine when a quotient is positive by comparing the signs of its numerator and denominator. - Each logarithm on the right must have a positive input. - Compare the two domain conditions. - Look for x-values where the left side exists but the right side does not.

1. For the left side, require \(\frac{x-2}{x-5}>0\). The numerator and denominator have the same sign when \(x<2\) or \(x>5\). Thus, the left-side domain is \((-\infty, 2)\cup(5, \infty)\). 2. For the right side, both logarithm inputs must be positive: \(x-2>0\) and \(x-5>0\). Together, these conditions give \(x>5\). 3. Therefore, the equation is valid only for \(x>5\). When \(x<2\), the quotient on the left is positive, but each individual logarithm on the right has a negative input and is not real.

No. The left side is defined for \(x<2\) or \(x>5\), but the right side is defined only for \(x>5\).

- Find the minimum possible value of \(x^2+e\). - Is the natural logarithm increasing or decreasing? - What happens as \(|x|\) becomes large? - Recall the value of \(\ln(e)\).

1. Since \(x^2\geq0\), the logarithm input satisfies \(x^2+e\geq e\). 2. The natural logarithm is strictly increasing, so \(f(x)\geq\ln(e)=1\). 3. Equality occurs at \(x=0\). As \(|x|\to\infty\), \(x^2+e\to\infty\), so \(f(x)\to\infty\). 4. Because \(f\) is continuous, it takes every value from \(1\) upward. Thus, its range is \([1, \infty)\).

\([1, \infty)\)

- Find the minimum possible value of the denominator. - Determine the resulting range of the logarithm input. - What happens to \(\ln(u)\) as \(u\to0^+\)? - Evaluate the function at \(x=0\).

1. Since \(x^4+1\geq1\), \(0<\frac{e^2}{x^4+1}\leq e^2\). The maximum input to the logarithm is \(e^2\), attained at \(x=0\). 2. Because the natural logarithm is strictly increasing, the maximum value is \(g(0)=\ln(e^2)=2\). 3. As \(|x|\to\infty\), the fraction \(\frac{e^2}{x^4+1}\to0^+\), so \(g(x)\to-\infty\). 4. The function is continuous, so it takes every value less than or equal to \(2\). Therefore, the range is \((-\infty, 2]\).

\((-\infty, 2]\)

- The input of a logarithm must be positive. - Combine the expression inside the logarithm into one fraction. - Identify where the numerator is zero and where the denominator is zero. - Use a sign chart to determine where the fraction is positive.

1. The logarithm input must be positive: \(1+\frac{4}{x-2}>0\). 2. Combine the terms: \(1+\frac{4}{x-2}=\frac{x+2}{x-2}\). 3. The critical values are \(x=-2\), where the numerator is zero, and \(x=2\), where the expression is undefined. 4. A sign analysis shows that \(\frac{x+2}{x-2}>0\) when \(x<-2\) or \(x>2\). 5. Therefore, the domain is \((-\infty, -2)\cup(2, \infty)\).

\((-\infty, -2)\cup(2, \infty)\)

- Determine when \((x-2)^2\) is positive. - Use \(\ln(a^2)=2\ln|a|\) on its real domain. - How can the single branch of \(\ln x\) be used to create both branches of \(\ln|x|\)? - Read the horizontal shift from \(x-2\). - A logarithm equals \(0\) when its input equals \(1\).

1. The logarithm input must be positive. Since \((x-2)^2>0\) for every \(x\ne2\), the domain is \(\mathbb{R}\setminus\{2\}\). The vertical asymptote is \(x=2\). 2. Rewrite \(p(x)=2\ln|x-2|\). Begin with the branch \(y=\ln x\) for \(x>0\). Copy that branch and reflect the copy across the y-axis; together the two branches form \(y=\ln|x|\). Stretch both branches vertically by a factor of \(2\), then shift the result right \(2\) units. 3. For the \(x\)-intercepts, solve \(\ln((x-2)^2)=0\). Then \((x-2)^2=1\), so \(x=1\) or \(x=3\).

a) Domain: \(\mathbb{R}\setminus\{2\}\); vertical asymptote: \(x=2\) b) Copy the \(\ln x\) branch, reflect the copy across the y-axis to form \(\ln|x|\), stretch vertically by \(2\), and shift right \(2\) units. c) \((1, 0)\) and \((3, 0)\)

- Rewrite a logarithmic equation in exponential form. - Take natural logarithms of both sides and use the power property. - Express exact logarithm inputs as powers of their bases when possible. - Recall that \(\ln(e)=1\).

1. If \(y=\log_b(x)\), then \(b^y=x\). Taking natural logarithms gives \(\ln(b^y)=\ln(x)\), so \(y\ln(b)=\ln(x)\). Therefore, \(y=\frac{\ln(x)}{\ln(b)}\). 2. \(\log_5(125)=\frac{\ln(125)}{\ln(5)}=3\). 3. \(\log_2(10)=\frac{\ln(10)}{\ln(2)}\approx3.3219\). 4. \(\log_{0.5}(4)=\frac{\ln(4)}{\ln(0.5)}=-2\). 5. By change of base, \(\log_5(e)=\frac{\ln(e)}{\ln(5)}=\frac{1}{\ln(5)}\). Therefore, \(\ln(5)\log_5(e)=1\).

a) \(\log_b(x)=\frac{\ln(x)}{\ln(b)}\) b) \(\log_5(125)=3\); \(\log_2(10)\approx3.3219\); \(\log_{0.5}(4)=-2\) c) \(\log_5(e)=\frac{1}{\ln(5)}\), so the product is \(1\).

- Take a logarithm when the variable is in an exponent. - Apply the change-of-base formula to both logarithms in part b. - Use the inverse relationship between logarithms and exponential functions. - Determine the exponent that takes \(e^2\) to \(e\).

1. Taking natural logarithms of \(5^x=20\) gives \(x\ln(5)=\ln(20)\). Thus, \(x=\frac{\ln(20)}{\ln(5)}\approx1.861\). 2. By change of base, \(\log_a(x)=\frac{\ln(x)}{\ln(a)}\) and \(\log_x(a)=\frac{\ln(a)}{\ln(x)}\). Therefore, \(\frac{1}{\log_x(a)}=\frac{\ln(x)}{\ln(a)}=\log_a(x)\). 3. Since \(\ln(e^3)=3\) and \(\log_{e^2}(e)=\frac{1}{2}\), the value is \(3+\frac{1}{2}=\frac{7}{2}\).

a) \(x=\frac{\ln(20)}{\ln(5)}\approx1.861\) b) \(\log_a(x)=\frac{1}{\log_x(a)}\) c) \(\frac{7}{2}\)

- Apply the change-of-base formula with base \(e^{-1}\). - Recall that every basic logarithmic graph passes through \((1, 0)\). - Compare the graphs of \(y=\ln(x)\) and \(y=-\ln(x)\). - Relate the base interval to the monotonicity of the corresponding exponential function.

1. By change of base, \(g(x)=\frac{\ln(x)}{\ln(e^{-1})}=\frac{\ln(x)}{-1}=-\ln(x)\). 2. Since \(\ln(x)<0\) for \(0<x<1\), \(f\) is negative on \((0, 1)\). Since \(g(x)=-\ln(x)\), \(g\) is negative when \(x>1\). 3. A logarithmic function with base greater than \(1\) is strictly increasing. A logarithmic function with base between \(0\) and \(1\) is strictly decreasing. Here, \(e>1\), so \(f\) increases, while \(1/e<1\), so \(g\) decreases.

1. \(g(x)=-\ln(x)\) 2. \(f(x)<0\) on \((0, 1)\); \(g(x)<0\) on \((1, \infty)\) 3. Bases greater than \(1\) produce increasing logarithmic functions; bases between \(0\) and \(1\) produce decreasing logarithmic functions.

- Apply the product property of logarithms. - Consider what happens after doubling the input many times. - Rewrite each logarithmic equation in exponential form.

1. By the product property, \(g(2x)=\log_2(2x)=\log_2(2)+\log_2(x)=1+g(x)\). 2. Repeatedly doubling the input adds \(1\) to the output each time. Therefore, the outputs can exceed any fixed bound, so \(g\) is unbounded above. 3. If \(\log_2(x)=20\), then \(x=2^{20}=1{,}048{,}576\). 4. If \(\log_2(x)=30\), then \(x=2^{30}=1{,}073{,}741{,}824\).

a) \(g(2x)=g(x)+1\) b) Repeated doubling makes the outputs increase without bound. c) \(x=1{,}048{,}576\) and \(x=1{,}073{,}741{,}824\), respectively.

- At a common point, the output must not change with \(n\). - Set \(u=\ln(x)\) and compare the cases \(n=1\) and \(n=2\). - Convert the resulting values of \(u\) back to values of \(x\).

1. Let \(u=\ln(x)\). A common point requires \(u^n\) to have the same value for every positive integer \(n\). 2. In particular, \(u=u^2\), so \(u(u-1)=0\). Thus, \(u=0\) or \(u=1\). 3. If \(u=0\), then \(x=1\) and every output is \(0\), giving \((1, 0)\). 4. If \(u=1\), then \(x=e\) and every output is \(1\), giving \((e, 1)\). 5. Both values work for all \(n\), and the equation \(u=u^2\) shows there are no others.

\((1, 0)\) and \((e, 1)\)

- Use the zero and asymptotes to construct \(k\). - The input of a natural logarithm must be positive. - Make a sign chart using the numerator zero and denominator zero.

1. The graph has vertical asymptote \(x=2\), zero \(x=5\), and horizontal asymptote \(y=1\). A matching rule is \(k(x)=\frac{x-5}{x-2}\). It also gives \(k(0)=\frac{-5}{-2}=2.5\), matching the graph. 2. The logarithm requires \(k(x)>0\), so solve \(\frac{x-5}{x-2}>0\). 3. The critical values are \(2\) and \(5\). The numerator and denominator have the same sign when \(x<2\) or \(x>5\). 4. Therefore, the domain is \((-\infty, 2)\cup(5, \infty)\), which is exactly \(\mathbb{R}\setminus[2, 5]\).

One possible rule is \(k(x)=\frac{x-5}{x-2}\). The domain of \(g\) is \((-\infty, 2)\cup(5, \infty)=\mathbb{R}\setminus[2, 5]\).

- The logarithm is defined where its input is positive. - Identify the graph that lies above the x-axis only between \(-4\) and \(2\). - Use the zero and vertical asymptote to construct the rational function. - Substitute a clear point to determine the scale factor.

1. The logarithm is defined exactly where \(k(x)>0\). Therefore, the correct graph must lie above the x-axis only for \(-4<x<2\). 2. Graph A has a zero at \(x=-4\), a vertical asymptote at \(x=2\), and positive values only between these two values. Graph B is positive outside that interval. Therefore, Graph A is correct. 3. A rule with zero \(-4\) and vertical asymptote \(2\) is \(k(x)=a\frac{x+4}{2-x}\). 4. Graph A passes through \((0, 2)\), so \(2=a\frac{4}{2}=2a\). Thus, \(a=1\), and \(k(x)=\frac{x+4}{2-x}\).

Graph A; \(k(x)=\frac{x+4}{2-x}\)

- A substitution can turn an expression involving \(\log x\) into a quadratic inequality. - Use the product property to combine a sum of logarithms. - State the domain before solving, especially when the argument contains a square. - Use a sign chart or the graph of a quadratic to solve the resulting polynomial inequality.

1. The domain requires \(x^2 > 0\), so \(x \ne 0\). Since the common logarithm is increasing, \(\log(x^2) \le 2\) gives \(x^2 \le 100\), or \(-10 \le x \le 10\). Excluding \(0\) gives \([-10, 0) \cup (0, 10]\). 2. The domain requires \(x > 0\). Let \(u = \log x\). Then \(u^2 - 3u + 2 \le 0\), or \((u - 1)(u - 2) \le 0\). Thus, \(1 \le u \le 2\). Substituting back gives \(1 \le \log x \le 2\), so \(10 \le x \le 100\). 3. The domain requires \(x > 9\). Use the product property: \(\log(x(x - 9)) > 1\). Since the logarithm is increasing, \(x(x - 9) > 10\), so \(x^2 - 9x - 10 > 0\). Factoring gives \((x - 10)(x + 1) > 0\), so \(x < -1\) or \(x > 10\). Intersecting with \(x > 9\) gives \(x > 10\).

1) \([-10, 0) \cup (0, 10]\) 2) \([10, 100]\) 3) \((10, \infty)\)

- Use the change-of-base formula in part a. - Choose familiar powers when testing the claim. - Compare doubling a base with squaring a base. - Determine what change to the denominator in the change-of-base formula would halve the result.

1. a) By change of base, \(\ln(x) = \frac{\log(x)}{\log(e)}\). Thus, \(k = \frac{1}{\log(e)} = \ln(10) \approx 2.3026\). 2. b) \(\log_2(16) = 4\) and \(\log_4(16) = 2\), so the claim happens to be true in this case. 3. For a counterexample, \(\log_3(9) = 2\), but doubling the base gives \(\log_6(9) = \frac{\ln(9)}{\ln(6)} \approx 1.2263\), not \(1\). 4. In general, squaring the base halves the logarithm because \(\log_{b^2}(a) = \frac{\ln(a)}{\ln(b^2)} = \frac{1}{2}\log_b(a)\). Doubling the base does not generally have that effect.

a) \(k \approx 2.3026\) b) The claim works for \(\log_2(16) = 4\) and \(\log_4(16) = 2\), but it is false in general. For example, \(\log_3(9) = 2\), while \(\log_6(9) \approx 1.2263 \ne 1\).

- Require the logarithm input to be positive. - Use the quotient and power properties of logarithms. - What does \(f(-x)=f(x)\) imply about the graph? - Separate the effects of the coefficient, negative sign, and vertical shift.

1. The expression \(\frac{e}{x^2}\) is defined and positive whenever \(x\ne0\). Thus, the domain is \(\mathbb{R}\setminus\{0\}\). 2. Apply the quotient property: \(f(x)=\ln(e)-\ln(x^2)=1-\ln(x^2)\). For \(x\ne0\), \(\ln(x^2)=2\ln|x|\), so \(f(x)=1-2\ln|x|\). 3. Begin with the right-hand branch of \(y=\ln(x)\). Reflect it across the y-axis to create the left-hand branch, producing \(y=\ln|x|\). 4. Stretch vertically by a factor of \(2\), reflect across the x-axis, and shift up \(1\) unit. Because \(f(-x)=f(x)\), the final graph is symmetric about the y-axis and has two branches.

a) \(\mathbb{R}\setminus\{0\}\) b) \(f(x)=1-2\ln|x|\) c) Create \(\ln|x|\) by reflecting the right branch across the y-axis, stretch vertically by \(2\), reflect across the x-axis, and shift up \(1\) unit. The graph has two branches and y-axis symmetry.

- Set \(y=\log_a(x)\) and rewrite in exponential form. - Compare \(a^y\) with \(a^0\) when \(y>0\). - Treat increasing and decreasing exponential functions separately. - Locate \(0.2\) and \(5\) relative to \(1\).

1. Let \(y=\log_a(x)\), so \(a^y=x\). 2. If \(a>1\), the function \(a^y\) is increasing. Thus, \(y>0\) exactly when \(a^y>a^0=1\), which means \(x>1\). 3. If \(0<a<1\), the function \(a^y\) is decreasing. Thus, \(y>0\) exactly when \(a^y<a^0=1\), which means \(0<x<1\). 4. Therefore, \(\log_a(x)>0\) exactly when \(a\) and \(x\) are on the same side of \(1\). 5. For \(\log_{0.2}(5)\), the base is less than \(1\) and the argument is greater than \(1\), so the value is negative.

1. The logarithm is positive exactly when the base and argument are both greater than \(1\) or both lie between \(0\) and \(1\). 2. \(\log_{0.2}(5)<0\)

- Compare domains and vertical asymptotes. - Check which functions are symmetric about the y-axis. - Compare values and end behavior near important domain boundaries. - Identify where a denominator becomes zero.

1. Graph a is \(h_1\). Its domain is \(x>-3\), it has vertical asymptote \(x=-3\), and its x-intercept is \((-2,0)\). 2. Graph b is \(h_2\). Because \(x^2+1\ge1\), \(h_2(x)\ge0\), with equality at \(x=0\). Also, \(h_2\) is even, so the graph is symmetric about the y-axis. 3. Graph c is \(h_3\). Its domain is \(x>0\). As \(x\to0^+\), \(-\ln(x)\to\infty\), so \(h_3(x)\to\infty\). Also, \(h_3(1)=1\), and as \(x\to\infty\), the linear term \(x\) eventually dominates \(\ln(x)\), so \(h_3(x)\to\infty\). These features match graph c. 4. Graph d is \(h_4\). Its domain is \(x>0\) with \(x\ne1\). Since \(\ln(1)=0\), the graph has vertical asymptote \(x=1\). As \(x\to0^+\), its values approach \(0\) from below.

a) \(h_1(x)=\ln(x+3)\) b) \(h_2(x)=\ln(x^2+1)\) c) \(h_3(x)=x-\ln(x)\) d) \(h_4(x)=\frac{1}{\ln(x)}\)