Logarithmic Functions - IB Mathematics Analysis and Approaches

Simplify the expression $2\log_3(6) - \log_3(4)$ .
Answer: 2 Using the power rule for logarithms, $n\log_b(a) = \log_b(a^n)$ , we can rewrite the first term: $2\log_3(6) = \log_3(6^2) = \log_3(36)$ . The expression becomes $\log_3(36) - \log_3(4)$ . Now, using the quotient rule, $\log_b(a) - \log_b(c) = \log_b(a/c)$ , we get $\log_3(36/4) = \log_3(9)$ . Since $3^2 = 9$ , $\log_3(9) = 2$ . Distractor A comes from the error $2\log_3(6) = \log_3(12)$ , which leads to $\log_3(12) - \log_3(4) = \log_3(3) = 1$ .
Find the solution set for the inequality $\log_{0.5}(x-1) > -2$ .
Answer: $(1, 5)$ First, the domain requires the argument of the logarithm to be positive: $x-1 > 0$ , so $x > 1$ . Next, to solve the inequality, we exponentiate both sides. Since the base is $0.5$ , which is between 0 and 1, we must reverse the inequality sign. $x-1 < (0.5)^{-2}$ . We calculate $(0.5)^{-2} = (1/2)^{-2} = 2^2 = 4$ . So the inequality becomes $x-1 < 4$ , which gives $x < 5$ . Combining this with the domain constraint $x > 1$ , the solution set is $1 < x < 5$ , or $(1, 5)$ . Distractor C comes from forgetting to reverse the inequality sign.
The equation $\ln(x^2 - 4) - \ln(x-2) = \ln(5)$ has a solution $x=k$ . What is the value of $k$ ?
Answer: 3 First, consider the domain. We need $x^2-4>0$ , which means $x < -2$ or $x > 2$ . We also need $x-2>0$ , which means $x>2$ . The overall domain is $x>2$ . Using the quotient rule for logarithms, the equation becomes $\ln\left(\frac{x^2-4}{x-2}\right) = \ln(5)$ . Since $x^2-4 = (x-2)(x+2)$ , we can simplify the argument: $\frac{(x-2)(x+2)}{x-2} = x+2$ . The equation is now $\ln(x+2) = \ln(5)$ . This implies $x+2 = 5$ , so $x=3$ . The solution $x=3$ is in the domain $x>2$ , so it is a valid solution. Thus, $k=3$ .
If $e^{2x} = 9$ and $e^y = \sqrt{5}$ , what is the exact value of $x + y$ ?
Answer: $\ln(3\sqrt{5})$ From $e^{2x} = 9$ , take the natural logarithm of both sides to get $2x = \ln(9)$ . Since $9=3^2$ , this is $2x = \ln(3^2) = 2\ln(3)$ , so $x = \ln(3)$ . From $e^y = \sqrt{5}$ , take the natural logarithm of both sides to get $y = \ln(\sqrt{5})$ . Then, $x+y = \ln(3) + \ln(\sqrt{5})$ . Using the product rule for logarithms, $x+y = \ln(3\sqrt{5})$ .
If $\log_c(2)=x$ and $\log_c(5)=y$ , which expression is equivalent to $\log_c(0.8)$ ?
Answer: $2x-y$ Convert the decimal to a fraction: $0.8 = \frac{8}{10} = \frac{4}{5}$ . Using the quotient rule for logarithms: $\log_c(0.8) = \log_c\left(\frac{4}{5}\right) = \log_c(4) - \log_c(5)$ . We know $\log_c(5) = y$ . Since $4 = 2^2$ , we have $\log_c(4) = \log_c(2^2) = 2\log_c(2) = 2x$ . Therefore, $\log_c(0.8) = 2x - y$ .
Find the sum of all valid solutions to the equation $\log_3(x+7) + \log_3(x-1) = 2$ .
Answer: 2 First, combine the logarithms using the product rule: $\log_3((x+7)(x-1)) = 2$ . Convert this to exponential form: $(x+7)(x-1) = 3^2$ . Expanding the left side gives $x^2 + 6x - 7 = 9$ , which simplifies to the quadratic equation $x^2 + 6x - 16 = 0$ . Factoring gives $(x+8)(x-2) = 0$ , so the potential solutions are $x = -8$ and $x = 2$ . However, the domain of the original logarithmic expressions must be considered. For $\log_3(x+7)$ , we need $x+7 > 0$ , so $x > -7$ . For $\log_3(x-1)$ , we need $x-1 > 0$ , so $x > 1$ . Both conditions must be met, so we require $x > 1$ . The solution $x = -8$ is extraneous. The only valid solution is $x = 2$ . The sum of all valid solutions is just 2.
Given $a = \log_{10}(2)$ and $b = \log_{10}(3)$ , express $\log_{10}(7.2)$ in terms of $a$ and $b$ .
Answer: $3a + 2b - 1$ First, rewrite 7.2 as a fraction: $7.2 = \frac{72}{10}$ . Then $\log_{10}(7.2) = \log_{10}\left(\frac{72}{10}\right) = \log_{10}(72) - \log_{10}(10) = \log_{10}(72) - 1$ . Next, find the prime factorization of 72: $72 = 8 \times 9 = 2^3 \times 3^2$ . So, $\log_{10}(72) = \log_{10}(2^3 \times 3^2) = \log_{10}(2^3) + \log_{10}(3^2) = 3\log_{10}(2) + 2\log_{10}(3) = 3a + 2b$ . Substituting this back gives $\log_{10}(7.2) = 3a + 2b - 1$ .
Find the value of $x$ that satisfies the equation $\log_2(x) - \log_4(x) = 3$ .
Answer: 64 To solve this equation, first express all logarithms in the same base. Using the change of base formula, $\log_4(x) = \frac{\log_2(x)}{\log_2(4)} = \frac{\log_2(x)}{2}$ . Substitute this into the original equation: $\log_2(x) - \frac{1}{2}\log_2(x) = 3$ . This simplifies to $\frac{1}{2}\log_2(x) = 3$ , so $\log_2(x) = 6$ . Converting to exponential form gives $x = 2^6 = 64$ . Distractor A could result from the error $\log_4(x) = 2\log_2(x)$ , which would lead to $-\log_2(x)=3$ and $x=2^{-3}=1/8$ .
Given that $\log_4(x) = y$ , find an expression for $\log_8(x)$ in terms of $y$ .
Answer: $\frac{2y}{3}$ Use the change of base formula, $\log_b(a) = \frac{\log_c(a)}{\log_c(b)}$ . We can convert $\log_8(x)$ to base 4: $\log_8(x) = \frac{\log_4(x)}{\log_4(8)}$ . We are given $\log_4(x) = y$ . We need to evaluate $\log_4(8)$ . Let $z = \log_4(8)$ , then $4^z = 8$ . Expressing both sides with base 2: $(2^2)^z = 2^3$ , which means $2^{2z} = 2^3$ . Therefore, $2z=3$ and $z = \frac{3}{2}$ . Substituting this back into the formula gives $\log_8(x) = \frac{y}{3/2} = \frac{2y}{3}$ .
The function $f(x) = \ln(3-x)$ is defined on the domain $(-\infty, k)$ . What is the value of $k$ and the equation of the vertical asymptote of the graph of $y=f(x)$ ?
Answer: $k=3$ , asymptote $x=3$ The natural logarithm function $\ln(u)$ is defined only for $u > 0$ . For the function $f(x) = \ln(3-x)$ , the argument is $3-x$ . Therefore, we must have $3-x > 0$ , which implies $x < 3$ . The domain of the function is $(-\infty, 3)$ , so $k=3$ . The vertical asymptote of a logarithmic function occurs where its argument is equal to zero. Setting the argument of the logarithm to zero gives $3-x=0$ , which solves to $x=3$ . Thus, the vertical asymptote is the line $x=3$ .
The graph of $y = \log_a x$ passes through the point $(8, -\frac{3}{2})$ . Find the value of $a$ .
Answer: $1/4$ Substitute the coordinates of the point into the equation: $-\frac{3}{2} = \log_a(8)$ . Convert this logarithmic equation into exponential form: $a^{-3/2} = 8$ . To solve for $a$ , we can raise both sides to the power of $-2/3$ : $(a^{-3/2})^{-2/3} = 8^{-2/3}$ . This simplifies to $a = 8^{-2/3}$ . Now evaluate the right side: $8^{-2/3} = \frac{1}{8^{2/3}} = \frac{1}{(\sqrt[3]{8})^2} = \frac{1}{2^2} = \frac{1}{4}$ .
Let $f(x) = e^{2x-1} + 3$ . Find the inverse function $f^{-1}(x)$ .
Answer: $f^{-1}(x) = \frac{\ln(x-3)+1}{2}$ To find the inverse function, we start with $y = e^{2x-1} + 3$ and solve for $x$ in terms of $y$ . First, swap $x$ and $y$ : $x = e^{2y-1} + 3$ . Isolate the exponential term: $x - 3 = e^{2y-1}$ . Take the natural logarithm of both sides: $\ln(x-3) = 2y-1$ . Solve for $y$ : $\ln(x-3) + 1 = 2y$ , so $y = \frac{\ln(x-3)+1}{2}$ . Therefore, $f^{-1}(x) = \frac{\ln(x-3)+1}{2}$ .
Let $f(x) = \log_3(x)$ . The graph of $g(x)$ is obtained by reflecting $f(x)$ in the $y$ -axis and then translating it by 2 units to the right. Which of the following is the equation for $g(x)$ ?
Answer: $g(x) = \log_3(2-x)$ Let's perform the transformations step-by-step. First, reflecting $f(x) = \log_3(x)$ in the $y$ -axis means replacing $x$ with $-x$ . This gives an intermediate function $h(x) = \log_3(-x)$ . Second, translating this function by 2 units to the right means replacing $x$ with $(x-2)$ . Applying this to $h(x)$ gives $g(x) = h(x-2) = \log_3(-(x-2))$ . Distributing the negative sign inside the argument gives $g(x) = \log_3(-x+2)$ or $g(x) = \log_3(2-x)$ .
Which of the following statements is true for all $x, y > 0$ ?
Answer: $\log_2(x) = \frac{\ln x}{\ln 2}$ This question tests knowledge of logarithm properties. Choice A is a common misconception; the correct rule is $\ln(xy) = \ln(x) + \ln(y)$ . Choice B is also a common error; the correct rule is $\ln(x^2) = 2\ln x$ . Choice C is incorrect; the correct rule is $\ln(e^x \cdot e^y) = \ln(e^{x+y}) = x+y$ , but $\ln(e^x + e^y)$ cannot be simplified this way. Choice D is the change of base formula, $\log_b a = \frac{\log_c a}{\log_c b}$ , applied with $b=2$ , $a=x$ , and $c=e$ . This formula is universally true for all valid bases and arguments.
Find the solution set for the equation $\log_2(x) + \log_2(x-2) = 3$ .
Answer: $\{4\}$ The domain of the equation requires $x > 0$ and $x-2 > 0$ , which simplifies to $x > 2$ . Using the logarithm product rule, the equation becomes $\log_2(x(x-2)) = 3$ . Converting to exponential form gives $x(x-2) = 2^3 = 8$ . Expanding this gives $x^2 - 2x = 8$ , or $x^2 - 2x - 8 = 0$ . Factoring the quadratic yields $(x-4)(x+2) = 0$ , with potential solutions $x=4$ and $x=-2$ . Checking against the domain $x > 2$ , we see that $x=-2$ is an extraneous solution. The only valid solution is $x=4$ . Distractor D comes from the common error of thinking $\log(a)+\log(b)=\log(a+b)$ , which would give $\log_2(x+x-2)=3 \Rightarrow 2x-2=8 \Rightarrow x=5$ .
What is the domain of the function $f(x) = \log_{10}\left(\frac{x-2}{5-x}\right)$ ?
Answer: $(2, 5)$ The argument of a logarithm must be strictly positive. Thus, we must solve the inequality $\frac{x-2}{5-x} > 0$ . A rational expression is positive if the numerator and denominator have the same sign. Case 1: Both are positive. $x-2 > 0$ and $5-x > 0$ , which means $x > 2$ and $x < 5$ . This gives the interval $(2, 5)$ . Case 2: Both are negative. $x-2 < 0$ and $5-x < 0$ , which means $x < 2$ and $x > 5$ . There is no value of $x$ that satisfies both conditions. Therefore, the domain is $(2, 5)$ . Distractor B results from incorrectly solving the inequality, perhaps by multiplying by $(5-x)$ without considering its sign, leading to $(x-2)(x-5) > 0$ . Distractor C incorrectly allows the argument to be zero.
Given that $\log_x 25 = 2$ and $\log_3 y = 4$ , find the value of $x - \log_9 y$ .
Answer: 3 First, solve for $x$ from $\log_x 25 = 2$ . In exponential form, this is $x^2 = 25$ . Since the base of a logarithm must be positive (and not 1), we have $x=5$ . Next, solve for $y$ from $\log_3 y = 4$ . In exponential form, this is $y = 3^4 = 81$ . Now calculate $x - \log_9 y$ . Substitute $x=5$ and $y=81$ : $5 - \log_9(81)$ . Since $9^2 = 81$ , we have $\log_9(81) = 2$ . Therefore, $x - \log_9 y = 5 - 2 = 3$ .
Find the sum of the $x$ -intercept and $y$ -intercept of the function $f(x) = \log_2(x+4) - 1$ .
Answer: -1 To find the $x$ -intercept, set $f(x)=0$ : $0 = \log_2(x+4) - 1$ . This gives $\log_2(x+4) = 1$ . In exponential form, this is $x+4 = 2^1 = 2$ , so $x = -2$ . The $x$ -intercept is -2. To find the $y$ -intercept, set $x=0$ : $y = \log_2(0+4) - 1 = \log_2(4) - 1$ . Since $\log_2(4) = 2$ , the $y$ -intercept is $y = 2 - 1 = 1$ . The sum of the intercepts is $-2 + 1 = -1$ .

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