IB Mathematics: Analysis and Approaches Quiz: Logarithmic Functions

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Logarithmic FunctionsQuestion 1 of 20

Find the sum of the $x$ -intercept and $y$ -intercept of the function $f(x) = \log_2(x+4) - 1$ .

-2

-1

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Logarithmic Functions

Practice Logarithmic Functions in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Logarithmic Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Find the sum of the $x$ -intercept and $y$ -intercept of the function $f(x) = \log_2(x+4) - 1$ .

-2
-1 (correct answer)
1
3

Explanation: 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

Question 2

The equation $\ln(x^2 - 4) - \ln(x-2) = \ln(5)$ has a solution $x=k$ . What is the value of $k$ ?

-3
3 (correct answer)
5
7

Explanation: First, consider the domain. We need

x^2-4>0

, which means

x < -2

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

Question 3

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 $g(x)$ ?

$g(x) = \log_3(2-x)$ (correct answer)
$g(x) = \log_3(-x-2)$
$g(x) = -\log_3(x-2)$
$g(x) = \log_3(-(x+2))$

Explanation: 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)

g(x) = \log_3(2-x)

Question 4

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 $y=f(x)$ ?

$k=-3$ , asymptote $x=-3$
$k=3$ , asymptote $x=-3$
$k=3$ , asymptote $x=3$ (correct answer)
$k=\infty$ , asymptote $y=3$

Explanation: 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

Question 5

Find the sum of the $x$ -intercept and $y$ -intercept of the function $f(x) = \log_2(x+4) - 1$ .

-2
-1 (correct answer)
1
3

Explanation: 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

Question 6

If $\log_c(2)=x$ and $\log_c(5)=y$ , which expression is equivalent to $\log_c(0.8)$ ?

$x-y$
$\frac{x}{y}$
$2x-y$ (correct answer)
$3x-y$

Explanation: 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

Question 7

Simplify the expression $2\log_3(6) - \log_3(4)$ .

1
2 (correct answer)
$\log_3(8)$
$\log_3(32)$

Explanation: 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

Question 8

Given that $\log_x 25 = 2$ and $\log_3 y = 4$ , find the value of $x - \log_9 y$ .

3 (correct answer)
4
9
16

Explanation: 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

Question 9

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 $y=f(x)$ ?

$k=-3$ , asymptote $x=-3$
$k=3$ , asymptote $x=-3$
$k=3$ , asymptote $x=3$ (correct answer)
$k=\infty$ , asymptote $y=3$

Explanation: 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

Question 10

Find the solution set for the equation $\log_2(x) + \log_2(x-2) = 3$ .

$\{-2, 4\}$
$\{-2\}$
$\{4\}$ (correct answer)
$\{5\}$

Explanation: 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

Question 11

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 $g(x)$ ?

$g(x) = \log_3(2-x)$ (correct answer)
$g(x) = \log_3(-x-2)$
$g(x) = -\log_3(x-2)$
$g(x) = \log_3(-(x+2))$

Explanation: 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)

g(x) = \log_3(2-x)

Question 12

Given $a = \log_{10}(2)$ and $b = \log_{10}(3)$ , express $\log_{10}(7.2)$ in terms of $a$ and $b$ .

$2a + 3b - 1$
$3a + 2b - 1$ (correct answer)
$3a + 2b + 1$
$a^3 + b^2 - 1$

Explanation: 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

Question 13

Given that $\log_4(x) = y$ , find an expression for $\log_8(x)$ in terms of $y$ .

$\frac{y}{2}$
$\frac{2y}{3}$ (correct answer)
$\frac{3y}{2}$
$2y$

Explanation: 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}

Question 14

Find the solution set for the inequality $\log_{0.5}(x-1) > -2$ .

$(1, 1.25)$
$(1, 5)$ (correct answer)
$(5, \infty)$
The empty set $\emptyset$

Explanation: 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.

Question 15

What is the domain of the function $f(x) = \log_{10}\left(\frac{x-2}{5-x}\right)$ ?

$(2, 5)$ (correct answer)
$(-\infty, 2) \cup (5, \infty)$
$[2, 5)$
$(-\infty, 2] \cup (5, \infty)$

Explanation: 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.

Question 16

Which of the following statements is true for all $x, y > 0$ ?

$\ln(x+y) = \ln(x) + \ln(y)$
$(\ln x)^2 = 2\ln x$
$\ln(e^x + e^y) = x+y$
$\log_2(x) = \frac{\ln x}{\ln 2}$ (correct answer)

Explanation: 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.

Question 17

Let $f(x) = e^{2x-1} + 3$ . Find the inverse function $f^{-1}(x)$ .

$f^{-1}(x) = \frac{\ln(x-3)+1}{2}$ (correct answer)
$f^{-1}(x) = \frac{\ln(x+3)-1}{2}$
$f^{-1}(x) = \ln(2x-1) - 3$
$f^{-1}(x) = \frac{\ln(x)-2}{2}$

Explanation: 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}

Question 18

Given that $\log_4(x) = y$ , find an expression for $\log_8(x)$ in terms of $y$ .

$\frac{y}{2}$
$\frac{2y}{3}$ (correct answer)
$\frac{3y}{2}$
$2y$

Explanation: 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}

Question 19

Let $f(x) = e^{2x-1} + 3$ . Find the inverse function $f^{-1}(x)$ .

$f^{-1}(x) = \frac{\ln(x-3)+1}{2}$ (correct answer)
$f^{-1}(x) = \frac{\ln(x+3)-1}{2}$
$f^{-1}(x) = \ln(2x-1) - 3$
$f^{-1}(x) = \frac{\ln(x)-2}{2}$

Explanation: 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}

Question 20

What is the domain of the function $f(x) = \log_{10}\left(\frac{x-2}{5-x}\right)$ ?

$(2, 5)$ (correct answer)
$(-\infty, 2) \cup (5, \infty)$
$[2, 5)$
$(-\infty, 2] \cup (5, \infty)$

Explanation: 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.