IB Mathematics: Analysis and Approaches Quiz: Exponential And Logarithmic Functions

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Exponential And Logarithmic FunctionsQuestion 1 of 20

A population of bacteria is modelled by $P(t) = P_0 e^{0.15t}$ , where $t$ is the time in hours. Determine the time it takes for the population to triple, giving the answer in terms of natural logarithms.

\frac{3 \ln 3}{20}

\frac{20 \ln 2}{3}

\frac{20 \ln 3}{3}

\ln(3) - \ln(0.15)

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Exponential And Logarithmic Functions

Practice Exponential And 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 Exponential And Logarithmic Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

How to use this quiz

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

$\frac{3 \ln 3}{20}$
$\frac{20 \ln 2}{3}$
$\frac{20 \ln 3}{3}$ (correct answer)
$\ln(3) - \ln(0.15)$

Explanation: We need to find the time

t

when the population

P(t)

is three times the initial population

P_0

. So, we set

P(t) = 3P_0

3P_0 = P_0 e^{0.15t}

. Divide by

P_0

3 = e^{0.15t}

. Take the natural logarithm of both sides:

\ln 3 = 0.15t

. Solve for

t

t = \frac{\ln 3}{0.15}

. To express this in the form given in the options, convert the decimal to a fraction:

0.15 = \frac{15}{100} = \frac{3}{20}

. So,

t = \frac{\ln 3}{3/20} = \frac{20 \ln 3}{3}

Question 2

Given that $\log_3(a) = p$ and $\log_3(b) = q$ , express $\log_3(9a^2/b)$ in terms of $p$ and $q$ .

$2+2p-q$ (correct answer)
$2+2p+q$
$9+2p-q$
$2(p-q)$

Explanation: We use the laws of logarithms to expand the expression. Using the quotient rule:

\log_3(9a^2/b) = \log_3(9a^2) - \log_3(b)

. Using the product rule on the first term:

\log_3(9) + \log_3(a^2) - \log_3(b)

. Using the power rule on the second term:

\log_3(9) + 2\log_3(a) - \log_3(b)

. We know that

\log_3(9) = 2

, and we are given

\log_3(a) = p

and

\log_3(b) = q

. Substituting these values gives:

2 + 2p - q

Question 3

The intensity $I$ of a light beam after passing through a substance of thickness $x$ cm is given by $I(x) = I_0 (10^{-kx})$ , where $I_0$ is the initial intensity and $k$ is a constant. If the intensity is halved after passing through 2 cm of the substance, what is the value of $k$ ?

$\frac{1}{2}$
$\log 2$
$2\log 2$
$\frac{1}{2}\log 2$ (correct answer)

Explanation: When you encounter exponential decay problems involving logarithms, you're dealing with situations where a quantity decreases by a constant percentage over equal intervals. The key is translating the given condition into an equation you can solve. Given that the intensity is halved after passing through 2 cm, you know that

I(2) = \frac{I_0}{2}

. Substituting into the formula:

\frac{I_0}{2} = I_0(10^{-2k})

Dividing both sides by

I_0

\frac{1}{2} = 10^{-2k}

To solve for

k

, take the logarithm base 10 of both sides:

\log\left(\frac{1}{2}\right) = \log(10^{-2k}) = -2k

Since

\log\left(\frac{1}{2}\right) = -\log 2

, you get:

-\log 2 = -2k

Therefore:

k = \frac{\log 2}{2} = \frac{1}{2}\log 2

Looking at the wrong answers: Choice (A)

\frac{1}{2}

ignores the logarithmic relationship entirely. Choice (B)

\log 2

comes from forgetting to divide by 2 when solving

-\log 2 = -2k

. Choice (C)

2\log 2

results from incorrectly multiplying instead of dividing by 2. The correct answer is (D)

\frac{1}{2}\log 2

. Study tip: In exponential decay problems, always set up your equation using the given condition first, then use logarithms to solve for unknown constants. Remember that

\log(a^n) = n\log(a)

is crucial for extracting variables from exponents.

Question 4

The value of a car, $V$ dollars, after $t$ years is modelled by the function $V(t) = 25000 e^{-kt}$ . After 2 years, the car's value is $18000.

Find the value of the car after 5 years, to the nearest dollar.

$7500
$10724
$11664 (correct answer)
$15275

Explanation: First, find the decay constant

k

. We have

V(2) = 18000

, so

18000 = 25000 e^{-2k}

. This gives

\frac{18}{25} = e^{-2k}

, so

e^{-k} = \sqrt{\frac{18}{25}} = \frac{3\sqrt{2}}{5}

. Now find

V(5) = 25000 e^{-5k} = 25000(e^{-k})^5 = 25000\left(\frac{3\sqrt{2}}{5}\right)^5 = 25000 \cdot \frac{243 \cdot 4\sqrt{2}}{3125} = 25000 \cdot \frac{972\sqrt{2}}{3125} \approx 11664

. To the nearest dollar, the value is $11664.

Question 5

$\frac{3 \ln 3}{20}$
$\frac{20 \ln 2}{3}$
$\frac{20 \ln 3}{3}$ (correct answer)
$\ln(3) - \ln(0.15)$

Explanation: We need to find the time

t

when the population

P(t)

is three times the initial population

P_0

. So, we set

P(t) = 3P_0

3P_0 = P_0 e^{0.15t}

. Divide by

P_0

3 = e^{0.15t}

. Take the natural logarithm of both sides:

\ln 3 = 0.15t

. Solve for

t

t = \frac{\ln 3}{0.15}

. To express this in the form given in the options, convert the decimal to a fraction:

0.15 = \frac{15}{100} = \frac{3}{20}

. So,

t = \frac{\ln 3}{3/20} = \frac{20 \ln 3}{3}

Question 6

If $\log_x 81 = 4$ and $\log_3 y = -1$ , find the value of $xy$ .

-3
1 (correct answer)
3
9

Explanation: We need to solve for

x

and

y

separately. From

\log_x 81 = 4

, we can write this in exponential form as

x^4 = 81

. Since

81 = 3^4

, we have

x^4 = 3^4

. As the base

x

of a logarithm must be positive (and not equal to 1), we have

x=3

. From

\log_3 y = -1

, we can write this in exponential form as

y = 3^{-1} = \frac{1}{3}

. Now, we find the product

xy

xy = 3 \cdot \frac{1}{3} = 1

Question 7

Given that $y = \log_a x$ , which expression is equivalent to $\log_a (ax)^3$ ?

$3y$
$3+y$
$3+3y$ (correct answer)
$1+3y$

Explanation: We use the properties of logarithms. First, the power rule:

\log_a (ax)^3 = 3 \log_a (ax)

. Next, the product rule:

3 \log_a (ax) = 3 (\log_a a + \log_a x)

. We know that

\log_a a = 1

and we are given that

y = \log_a x

. Substituting these in gives:

3 (1 + y) = 3 + 3y

Question 8

Find the solution to the equation $e^x - 12e^{-x} = 1$ .

$\ln 3$
$\ln 4$ (correct answer)
$\ln(1/3)$ and $\ln 4$
$\ln 3$ and $\ln 4$

Explanation: Let

u = e^x

. Since

e^{-x} = 1/e^x = 1/u

, the equation becomes

u - \frac{12}{u} = 1

. To eliminate the fraction, multiply the entire equation by

u

(note that

u=e^x

is always positive):

u^2 - 12 = u

. Rearrange into a standard quadratic form:

u^2 - u - 12 = 0

. Factor the quadratic:

(u-4)(u+3) = 0

. The solutions for

u

are

u=4

and

u=-3

. Since

u = e^x

u

must be positive. Therefore,

u=-3

is not a valid solution. We are left with

e^x = 4

. Solving for

x

gives

x = \ln 4

Question 9

Find the sum of the solutions of the equation $4^x - 6 \cdot 2^x + 8 = 0$ .

1
2
3 (correct answer)
6

Explanation: Rewrite the equation using the base 2. Since

4^x = (2^2)^x = (2^x)^2

, the equation is

(2^x)^2 - 6 \cdot 2^x + 8 = 0

. Let

u = 2^x

. The equation becomes a quadratic:

u^2 - 6u + 8 = 0

. Factoring gives

(u-2)(u-4) = 0

, so the solutions for

u

are

u=2

and

u=4

. Let the solutions for

x

x_1

and

x_2

. If

u=2

, then

2^{x_1} = 2 \implies x_1 = 1

. If

u=4

, then

2^{x_2} = 4 = 2^2 \implies x_2 = 2

. The sum of the solutions is

x_1 + x_2 = 1 + 2 = 3

Question 10

The equation $9^x - 5 \cdot 3^{x+1} + 54 = 0$ has two solutions, $x_1$ and $x_2$ . Find the value of $x_1 + x_2$ .

2
3
$\log_3 15$
$\log_3 54$ (correct answer)

Explanation: First, rewrite the equation in terms of

3^x

9^x = (3^2)^x = (3^x)^2

and

3^{x+1} = 3 \cdot 3^x

. The equation becomes

(3^x)^2 - 5(3 \cdot 3^x) + 54 = 0

, which simplifies to

(3^x)^2 - 15(3^x) + 54 = 0

. Let

u = 3^x

. The equation is

u^2 - 15u + 54 = 0

. Factoring the quadratic gives

(u-6)(u-9)=0

. The solutions for

u

are

u=6

and

u=9

. So,

3^{x_1} = 6

and

3^{x_2} = 9

. This gives

x_1 = \log_3 6

and

x_2 = \log_3 9 = 2

. The sum of the solutions is

x_1 + x_2 = \log_3 6 + \log_3 9

. Using the product rule for logarithms, this is

\log_3(6 \cdot 9) = \log_3 54

Question 11

Find the set of values of $x$ for which $2 \log_5(x-1) > \log_5(x+5)$ .

$\{x \in \mathbb{R} \mid 1 < x < 4\}$
$\{x \in \mathbb{R} \mid x > 4\}$ (correct answer)
$\{x \in \mathbb{R} \mid x < -1 \text{ or } x > 4\}$
$\{x \in \mathbb{R} \mid -5 < x < -1 \text{ or } x > 4\}$

Explanation: First, establish the domain for the logarithms. We need

x-1 > 0 \Rightarrow x > 1

and

x+5 > 0 \Rightarrow x > -5

. The combined domain is

x > 1

. Now, solve the inequality:

\log_5((x-1)^2) > \log_5(x+5)

. Since

\log_5 x

is an increasing function, we can remove the logarithms:

(x-1)^2 > x+5

x^2 - 2x + 1 > x+5

x^2 - 3x - 4 > 0

(x-4)(x+1) > 0

. The roots of the quadratic are

x=4

and

x=-1

. The inequality holds when

x < -1

x > 4

. Finally, we must find the intersection of this solution set with the domain

x > 1

. The only values that satisfy both conditions are

x > 4

Question 12

The value of a car, $V$ dollars, after $t$ years is modelled by the function $V(t) = 25000 e^{-kt}$ . After 2 years, the car's value is $18000.

Find the value of the car after 5 years, to the nearest dollar.

$7500
$10724
$11664 (correct answer)
$15275

Explanation: First, find the decay constant

k

. We have

V(2) = 18000

, so

18000 = 25000 e^{-2k}

. This gives

\frac{18}{25} = e^{-2k}

, so

e^{-k} = \sqrt{\frac{18}{25}} = \frac{3\sqrt{2}}{5}

. Now find

V(5) = 25000 e^{-5k} = 25000(e^{-k})^5 = 25000\left(\frac{3\sqrt{2}}{5}\right)^5 = 25000 \cdot \frac{243 \cdot 4\sqrt{2}}{3125} = 25000 \cdot \frac{972\sqrt{2}}{3125} \approx 11664

. To the nearest dollar, the value is $11664.

Question 13

$\frac{1}{2}$
$\log 2$
$2\log 2$
$\frac{1}{2}\log 2$ (correct answer)

I(2) = \frac{I_0}{2}

. Substituting into the formula:

\frac{I_0}{2} = I_0(10^{-2k})

Dividing both sides by

I_0

\frac{1}{2} = 10^{-2k}

To solve for

k

, take the logarithm base 10 of both sides:

\log\left(\frac{1}{2}\right) = \log(10^{-2k}) = -2k

Since

\log\left(\frac{1}{2}\right) = -\log 2

, you get:

-\log 2 = -2k

Therefore:

k = \frac{\log 2}{2} = \frac{1}{2}\log 2

Looking at the wrong answers: Choice (A)

\frac{1}{2}

ignores the logarithmic relationship entirely. Choice (B)

\log 2

comes from forgetting to divide by 2 when solving

-\log 2 = -2k

. Choice (C)

2\log 2

results from incorrectly multiplying instead of dividing by 2. The correct answer is (D)

\frac{1}{2}\log 2

. Study tip: In exponential decay problems, always set up your equation using the given condition first, then use logarithms to solve for unknown constants. Remember that

\log(a^n) = n\log(a)

is crucial for extracting variables from exponents.

Question 14

The numbers $\ln(x-2)$ , $\ln(x)$ , and $\ln(x+6)$ are three consecutive terms of an arithmetic sequence. Find the value of $x$ .

-3
$-1+\sqrt{13}$
3 (correct answer)
4

Explanation: For an arithmetic sequence, the common difference between consecutive terms is constant. So,

\ln(x) - \ln(x-2) = \ln(x+6) - \ln(x)

. This can be rewritten as

2\ln(x) = \ln(x-2) + \ln(x+6)

. The domain requires

x-2>0

x>0

, and

x+6>0

, which simplifies to

x>2

. Using logarithm laws, the equation becomes

\ln(x^2) = \ln((x-2)(x+6))

. Equating the arguments gives

x^2 = (x-2)(x+6) = x^2 + 4x - 12

. Subtracting

x^2

from both sides gives

0 = 4x - 12

, which means

4x = 12

and

x=3

. This solution is valid as it satisfies the domain condition

x>2

Question 15

Given that $y = \log_a x$ , which expression is equivalent to $\log_a (ax)^3$ ?

$3y$
$3+y$
$3+3y$ (correct answer)
$1+3y$

Explanation: We use the properties of logarithms. First, the power rule:

\log_a (ax)^3 = 3 \log_a (ax)

. Next, the product rule:

3 \log_a (ax) = 3 (\log_a a + \log_a x)

. We know that

\log_a a = 1

and we are given that

y = \log_a x

. Substituting these in gives:

3 (1 + y) = 3 + 3y

Question 16

The equation $9^x - 5 \cdot 3^{x+1} + 54 = 0$ has two solutions, $x_1$ and $x_2$ . Find the value of $x_1 + x_2$ .

2
3
$\log_3 15$
$\log_3 54$ (correct answer)

Explanation: First, rewrite the equation in terms of

3^x

9^x = (3^2)^x = (3^x)^2

and

3^{x+1} = 3 \cdot 3^x

. The equation becomes

(3^x)^2 - 5(3 \cdot 3^x) + 54 = 0

, which simplifies to

(3^x)^2 - 15(3^x) + 54 = 0

. Let

u = 3^x

. The equation is

u^2 - 15u + 54 = 0

. Factoring the quadratic gives

(u-6)(u-9)=0

. The solutions for

u

are

u=6

and

u=9

. So,

3^{x_1} = 6

and

3^{x_2} = 9

. This gives

x_1 = \log_3 6

and

x_2 = \log_3 9 = 2

. The sum of the solutions is

x_1 + x_2 = \log_3 6 + \log_3 9

. Using the product rule for logarithms, this is

\log_3(6 \cdot 9) = \log_3 54

Question 17

Solve the equation $\log_2(x+1) - \log_2(x-1) = 3$ .

$\frac{9}{7}$ (correct answer)
$\frac{7}{5}$
$\frac{5}{3}$
$\frac{3}{2}$

Explanation: The domain requires

x+1>0

and

x-1>0

, which simplifies to

x>1

. Using the quotient rule for logarithms, the equation becomes

\log_2\left(\frac{x+1}{x-1}\right) = 3

. Converting to exponential form:

\frac{x+1}{x-1} = 2^3 = 8

. Now, solve for

x

x+1 = 8(x-1)

x+1 = 8x - 8

9 = 7x

x = \frac{9}{7}

. We check if this solution is in the domain. Since

\frac{9}{7} \approx 1.28

, it is greater than 1, so the solution is valid.

Question 18

Given that $\log_3(a) = p$ and $\log_3(b) = q$ , express $\log_3(9a^2/b)$ in terms of $p$ and $q$ .

$2+2p-q$ (correct answer)
$2+2p+q$
$9+2p-q$
$2(p-q)$

Explanation: We use the laws of logarithms to expand the expression. Using the quotient rule:

\log_3(9a^2/b) = \log_3(9a^2) - \log_3(b)

. Using the product rule on the first term:

\log_3(9) + \log_3(a^2) - \log_3(b)

. Using the power rule on the second term:

\log_3(9) + 2\log_3(a) - \log_3(b)

. We know that

\log_3(9) = 2

, and we are given

\log_3(a) = p

and

\log_3(b) = q

. Substituting these values gives:

2 + 2p - q

Question 19

If $e^{2x} = 9$ and $e^{3y} = 8$ , what is the value of $e^{x-y}$ ?

$\frac{2}{3}$
$\frac{3}{2}$ (correct answer)
$1$
$6$

Explanation: First, solve for

e^x

and

e^y

. From

e^{2x} = 9

, we have

(e^x)^2 = 3^2

. Since

e^x

must be positive,

e^x = 3

. From

e^{3y} = 8

, we have

(e^y)^3 = 2^3

. Taking the cube root, we get

e^y = 2

. The expression

e^{x-y}

can be written as

\frac{e^x}{e^y}

. Substituting the values we found:

e^{x-y} = \frac{3}{2}

Question 20

Find the set of values of $x$ for which $2 \log_5(x-1) > \log_5(x+5)$ .

$\{x \in \mathbb{R} \mid 1 < x < 4\}$
$\{x \in \mathbb{R} \mid x > 4\}$ (correct answer)
$\{x \in \mathbb{R} \mid x < -1 \text{ or } x > 4\}$
$\{x \in \mathbb{R} \mid -5 < x < -1 \text{ or } x > 4\}$

Explanation: First, establish the domain for the logarithms. We need

x-1 > 0 \Rightarrow x > 1

and

x+5 > 0 \Rightarrow x > -5

. The combined domain is

x > 1

. Now, solve the inequality:

\log_5((x-1)^2) > \log_5(x+5)

. Since

\log_5 x

is an increasing function, we can remove the logarithms:

(x-1)^2 > x+5

x^2 - 2x + 1 > x+5

x^2 - 3x - 4 > 0

(x-4)(x+1) > 0

. The roots of the quadratic are

x=4

and

x=-1

. The inequality holds when

x < -1

x > 4

. Finally, we must find the intersection of this solution set with the domain

x > 1

. The only values that satisfy both conditions are

x > 4