Free Math Worksheets

- 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, the student'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 \(\ln(\sqrt{x})=\frac{1}{2}\ln(x)\).

- Compare the vertical asymptotes. - Track the location of the \(x\)-intercept \((1, 0)\). - A horizontal shift changes the input; a vertical shift changes the output.

1. The graph of \(f\) has vertical asymptote \(x=0\) and \(x\)-intercept \((1, 0)\). 2. The graph of \(g\) has vertical asymptote \(x=2\) and \(x\)-intercept \((3, 0)\). It is the graph of \(f\) shifted right \(2\) units, so \(g(x)=\ln(x-2)\). 3. The graph of \(h\) has the same vertical asymptote as \(f\), and the point \((1, 0)\) has moved to \((1, 2)\). It is the graph of \(f\) shifted up \(2\) units, so \(h(x)=\ln(x)+2\).

\(g(x)=\ln(x-2)\), a shift right \(2\) units \(h(x)=\ln(x)+2\), a shift up \(2\) units

- Compare the horizontal distance from a point to the reflection line before and after reflection. - How does reflection across a vertical line change the x-coordinate? - Does the y-coordinate change under this reflection? - Substitute the reflected point into the new function.

1. Reflection across a vertical line \(x=a\) maps an x-coordinate to \(2a-x\). For \(a=5\), replace \(x\) with \(10-x\). 2. Therefore, \(g(x)=f(10-x)=\ln(10-x)\), with domain \(x<10\). 3. The reflected x-coordinate of \(P(1, 0)\) is \(2\cdot5-1=9\), while the y-coordinate stays unchanged. Thus, \(P'=(9, 0)\). 4. Since \(g(9)=\ln(10-9)=\ln(1)=0\), the point \(P'\) lies on the graph of \(g\).

\(g(x)=\ln(10-x)\), with domain \((-\infty, 10)\); \(P'=(9, 0)\). Since \(g(9)=0\), the point lies on the graph.

- How is the number of digits of a positive integer related to its common logarithm? - Which logarithm properties let you rewrite \(\log(6^{60})\)? - What does the integer part of a common logarithm tell you about the size of a number?

1. Use the power property: \(\log(6^{60})=60\log(6)\). 2. Use the product property: \(\log(6)=\log(2\cdot3)=\log(2)+\log(3)\). 3. Substitute the given approximations: \(\log(6)\approx0.3010+0.4771=0.7781\). 4. Then \(\log(6^{60})\approx60\cdot0.7781=46.686\). 5. For a positive integer \(N\), the number of decimal digits is \(\lfloor\log(N)\rfloor+1\). Therefore, \(6^{60}\) has \(46+1=47\) digits.

\(6^{60}\) has \(47\) digits.

- What condition must the common logarithm satisfy for a number to have at most \(100\) digits? - Write an inequality involving \(n\). - Use the power property of logarithms.

1. A positive integer has at most \(100\) digits when its common logarithm is less than \(100\), so \(\log(2^n)<100\). 2. Apply the power property: \(n\log(2)<100\). 3. Then \(n<\frac{100}{\log(2)}\approx\frac{100}{0.30103}\approx332.19\). 4. Since \(n\) is a natural number, the greatest possible value is \(n=332\). 5. Check: \(\log(2^{332})\approx99.94\), so \(2^{332}\) has \(100\) digits, while \(\log(2^{333})\approx100.24\), so \(2^{333}\) has \(101\) digits.

\(n=332\)

- 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 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 \(\ln(a^2)=2\ln|a|\). - 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 \(x\neq2\), the domain is \(\mathbb{R}\setminus\{2\}\). The vertical asymptote is \(x=2\). 2. Rewrite \(p(x)=2\ln|x-2|\). Starting with \(y=\ln(x)\), reflect its right-hand branch across the \(y\)-axis to create \(y=\ln|x|\), stretch vertically by a factor of \(2\), and shift 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) Reflect the right-hand branch across the \(y\)-axis, stretch vertically by a factor of \(2\), and shift right \(2\) units. c) \((1, 0)\) and \((3, 0)\)

- Use the inverse relationship between \(e^x\) and \(\ln(x)\). - For which bases do powers decrease as the exponent increases? - What is the sign of \(\ln(b)\) when \(0<b<1\)? - Can a positive base raised to a real power equal zero?

1. Since \(b=e^{\ln(b)}\), \(b^x=\left(e^{\ln(b)}\right)^x=e^{x\ln(b)}\). Therefore, \(k=\ln(b)\). 2. The function \(b^x\) is strictly decreasing when \(0<b<1\). On this interval, \(\ln(b)<0\), so the equivalent condition is \(k<0\). 3. For every allowed base and every real \(x\), \(b^x>0\). Therefore, the range is \((0,\infty)\). The graph may approach the x-axis as an asymptote, but it cannot reach \(y=0\).

1. \(k=\ln(b)\) 2. \(g\) is strictly decreasing for \(0<b<1\), which is equivalent to \(k<0\). 3. The range is \((0,\infty)\), and \(b^x\) is never \(0\).

- 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 \((e, 1)\)

- Rewrite both functions using the same base. - Compare \(\ln(4)\) with \(1.4\). - How does a larger positive coefficient in the exponent affect exponential growth?

1. Rewrite the base: \(4=e^{\ln(4)}\). Thus, \(f(x)=e^{x\ln(4)}\). 2. Since \(\ln(4)\approx1.3863\) and \(1.4>1.3863\), the exponent of \(g\) grows faster than the exponent of \(f\) for \(x>0\). 3. Therefore, \(g(x)>f(x)\) for every \(x>0\). Also, \(g'(x)=1.4e^{1.4x}\) is greater than \(f'(x)=\ln(4)e^{x\ln(4)}\) for \(x>0\), so \(g\) has the greater growth rate.

\(g(x)=e^{1.4x}\) grows faster because \(f(x)=e^{x\ln(4)}\), \(\ln(4)\approx1.3863\), and \(1.4>1.3863\).

- Identify where the graph of \(f\) lies above the \(x\)-axis. - Recall the behavior of \(\ln(u)\) as \(u\to0^+\) and as \(u\to\infty\). - Consider the composition of two increasing functions.

1. From the graph, \(f(x)=0.5x+1\). The logarithm requires \(f(x)>0\), so \(0.5x+1>0\), which gives \(x>-2\). Thus, the domain is \((-2, \infty)\). 2. As \(x\to-2^+\), \(f(x)\to0^+\), so \(g(x)\to-\infty\). As \(x\to\infty\), \(f(x)\to\infty\), so \(g(x)\to\infty\). 3. The linear function \(f\) is strictly increasing, and \(\ln(x)\) is strictly increasing. Therefore, their composition \(g\) is strictly increasing on \((-2, \infty)\).

a) \((-2, \infty)\) b) \(\lim_{x\to-2^+}g(x)=-\infty\) and \(\lim_{x\to\infty}g(x)=\infty\) c) \(g\) is strictly increasing.

- Compare the outputs at \(x=e\). - Check whether the graphs share the same \(x\)-intercept. - Look for a constant scale factor or a sign change.

1. All three graphs pass through \((1, 0)\), so there is no translation. 2. At \(x=e\), \(f(e)=1\), while the graph of \(g\) has value \(2\). Thus, \(g(x)=2\ln(x)\), a vertical stretch by a factor of \(2\). 3. The graph of \(h\) has outputs opposite those of \(f\). Thus, \(h(x)=-\ln(x)\), a reflection across the \(x\)-axis.

a) \(g(x)=2\ln(x)\) and \(h(x)=-\ln(x)\) b) \(g\) is a vertical stretch by a factor of \(2\); \(h\) is a reflection across the \(x\)-axis.

- Find where the graph of \(f\) is above the \(x\)-axis. - Read the monotonicity of \(f\) on those intervals. - An increasing outer function preserves the monotonicity of the inner function.

1. The logarithm requires \(f(x)>0\). From the graph, \(f(x)>0\) for \(x<-1\) and \(x>1\). Thus, the domain is \((-\infty, -1)\cup(1, \infty)\). 2. The graph shows that \(f\) is strictly decreasing on both \((-\infty, -1)\) and \((1, \infty)\). 3. Since the natural logarithm is strictly increasing, composing with it preserves the direction of monotonicity. Therefore, \(g\) is strictly decreasing on each domain interval.

Domain: \((-\infty, -1)\cup(1, \infty)\) The function is strictly decreasing on \((-\infty, -1)\) and on \((1, \infty)\).

- 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 each function for symmetry. - Locate intercepts and extrema. - Identify where a denominator becomes zero.

1. Graph a is \(h_1\). Its domain is \(x>-3\), and its \(x\)-intercept is \((-2, 0)\). 2. Graph b is \(h_2\). It is even, so it is symmetric about the \(y\)-axis, and its minimum is \((0, 0)\). 3. Graph c is \(h_3\). Its domain is \(x>0\). Since \(h_3'(x)=1-\frac{1}{x}\), it has a critical point at \(x=1\), where its minimum value is \(1\). 4. Graph d is \(h_4\). It is undefined at \(x=1\), producing a vertical asymptote there. 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) \(h_4(x)=\frac{1}{\ln(x)}\)