IB Mathematics: Analysis and Approaches Quiz: Exponents And Logarithms

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Exponents And LogarithmsQuestion 1 of 20

The solution to $2^{x+1} = 5^x$ is given by $x =$

\frac{\ln 2}{\ln 5 - \ln 2}

\frac{\ln 2}{\ln 5 + \ln 2}

\frac{\ln 5}{\ln 2}

1

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Exponents And Logarithms

Practice Exponents And Logarithms 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 Exponents And Logarithms, 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

The solution to $2^{x+1} = 5^x$ is given by $x =$

$\frac{\ln 2}{\ln 5 - \ln 2}$ (correct answer)
$\frac{\ln 2}{\ln 5 + \ln 2}$
$\frac{\ln 5}{\ln 2}$
$1$

Explanation: To solve for

x

, we take the natural logarithm of both sides of the equation:

\ln(2^{x+1}) = \ln(5^x)

Using the power rule for logarithms,

\ln(a^b) = b \ln a

(x+1)\ln 2 = x\ln 5

Distribute

\ln 2

on the left side:

x\ln 2 + \ln 2 = x\ln 5

Rearrange the terms to isolate

x

\ln 2 = x\ln 5 - x\ln 2

Factor out

x

on the right side:

\ln 2 = x(\ln 5 - \ln 2)

Finally, divide to solve for

x

x = \frac{\ln 2}{\ln 5 - \ln 2}

Question 2

Find the sum of the solutions to the equation $9^x - 10 \cdot 3^x + 9 = 0$ .

0
2 (correct answer)
3
10

Explanation: The equation can be rewritten as

(3^2)^x - 10 \cdot 3^x + 9 = 0

, which is

(3^x)^2 - 10(3^x) + 9 = 0

. Let

u = 3^x

. The equation becomes a quadratic in

u

u^2 - 10u + 9 = 0

. Factoring this gives

(u-1)(u-9) = 0

. So,

u=1

u=9

. If

u=1

, then

3^x = 1

, which means

x=0

. If

u=9

, then

3^x = 9 = 3^2

, which means

x=2

. The solutions are

x=0

and

x=2

. The sum of the solutions is

0 + 2 = 2

Question 3

Which of the following is equivalent to $\frac{2\ln(6) - \ln(9)}{\ln(2)}$ ?

1
2 (correct answer)
$\ln(2)$
$\log_2(3)$

Explanation: First, simplify the numerator using the laws of logarithms.

2\ln(6) - \ln(9) = \ln(6^2) - \ln(9) = \ln(36) - \ln(9)

Using the quotient rule,

\ln(a) - \ln(b) = \ln(a/b)

\ln(36) - \ln(9) = \ln\left(\frac{36}{9}\right) = \ln(4)

Now the entire expression is:

\frac{\ln(4)}{\ln(2)}

Using the change of base formula,

\frac{\ln(a)}{\ln(b)} = \log_b(a)

, this becomes:

\log_2(4) = 2

Question 4

Which of the following is equivalent to $\frac{2\ln(6) - \ln(9)}{\ln(2)}$ ?

1
2 (correct answer)
$\ln(2)$
$\log_2(3)$

Explanation: First, simplify the numerator using the laws of logarithms.

2\ln(6) - \ln(9) = \ln(6^2) - \ln(9) = \ln(36) - \ln(9)

Using the quotient rule,

\ln(a) - \ln(b) = \ln(a/b)

\ln(36) - \ln(9) = \ln\left(\frac{36}{9}\right) = \ln(4)

Now the entire expression is:

\frac{\ln(4)}{\ln(2)}

Using the change of base formula,

\frac{\ln(a)}{\ln(b)} = \log_b(a)

, this becomes:

\log_2(4) = 2

Question 5

Solve for $x$ in the equation $\log_4(x+2) - \log_4(x-1) = \frac{1}{2}$ .

$x = -4$
$x = 0$
$x = 4$ (correct answer)
No solution

Explanation: Using the quotient rule for logarithms,

\log_b(M) - \log_b(N) = \log_b(M/N)

, we can combine the terms on the left:

\log_4\left(\frac{x+2}{x-1}\right) = \frac{1}{2}

By the definition of logarithms, this is equivalent to:

\frac{x+2}{x-1} = 4^{1/2}

Since

4^{1/2} = \sqrt{4} = 2

\frac{x+2}{x-1} = 2

Now, solve for

x

x+2 = 2(x-1)

x+2 = 2x-2

4 = x

We must check that this solution is in the domain of the original logarithms. For

\log_4(x+2)

, we need

x+2 > 0

, so

x > -2

. For

\log_4(x-1)

, we need

x-1 > 0

, so

x > 1

. The solution

x=4

satisfies both conditions.

Question 6

Find all values of $x$ that satisfy the equation $\log_x(5x-6)=2$ .

$x=2$ only
$x=3$ only
$x=2$ and $x=3$ (correct answer)
$x=1$ and $x=6$

Explanation: Convert the logarithmic equation to its exponential form:

x^2 = 5x-6

. Rearrange this into a quadratic equation:

x^2 - 5x + 6 = 0

. Factor the quadratic:

(x-2)(x-3) = 0

. The potential solutions are

x=2

and

x=3

. We must check these solutions against the conditions for logarithms. The base of a logarithm,

x

, must be positive and not equal to 1 (

x>0, x \neq 1

). Both

x=2

and

x=3

satisfy this. The argument of the logarithm,

5x-6

, must be positive. For

x=2

5(2)-6 = 10-6 = 4 > 0

. So,

x=2

is a valid solution. For

x=3

5(3)-6 = 15-6 = 9 > 0

. So,

x=3

is a valid solution. Both values satisfy the equation and the domain constraints.

Question 7

Express $\log_9(12)$ in terms of $a$ , where $a = \log_3(2)$ .

$\frac{a+1}{2}$ (correct answer)
$a + \frac{1}{2}$
$2a + 1$
$a + 2$

Explanation: We use the change of base formula to convert

\log_9(12)

to base 3.

\log_9(12) = \frac{\log_3(12)}{\log_3(9)}

The denominator is

\log_3(9) = \log_3(3^2) = 2

. For the numerator, we use the product rule:

\log_3(12) = \log_3(4 \cdot 3) = \log_3(4) + \log_3(3)

. This gives us

\log_3(2^2) + 1 = 2\log_3(2) + 1 = 2a + 1

. Therefore:

\log_9(12) = \frac{2a+1}{2} = \frac{a+1}{2}

Question 8

If $0 < a < 1$ and $b > 1$ , which of the following statements must be true about $\log_a b$ ?

$\log_a b < -1$
$\log_a b < 0$ (correct answer)
$0 < \log_a b < 1$
$\log_a b > 1$

Explanation: Let

y = \log_a b

. By definition, this means

a^y = b

. We are given that

0 < a < 1

and

b > 1

. Since the base

a

is a positive fraction less than 1, raising it to a positive power

y>0

would result in a value less than 1 (e.g.,

(1/2)^2 = 1/4

). To get a result

b

that is greater than 1, we must raise

a

to a negative power (e.g.,

(1/2)^{-2} = 4

). Therefore,

y

must be negative. Thus,

\log_a b < 0

. It is not necessarily less than -1, for example

\log_{1/4}(2) = -1/2

Question 9

Given that $x, y > 0$ and $x, y \neq 1$ , if $\log_x(y) + \log_y(x) = 2$ , what is the relationship between $x$ and $y$ ?

$x = y$ (correct answer)
$xy = 1$
$x = y^2$
$y = x^2$

Explanation: Let

u = \log_x(y)

. Using the change of base formula,

\log_y(x) = \frac{1}{\log_x(y)} = \frac{1}{u}

. Substituting into the given equation:

u + \frac{1}{u} = 2

Multiply the entire equation by

u

to eliminate the fraction:

u^2 + 1 = 2u

Rearrange into a quadratic equation:

u^2 - 2u + 1 = 0

This is a perfect square trinomial:

(u-1)^2 = 0

. The only solution is

u=1

. Substituting back

u = \log_x(y)

\log_x(y) = 1

By the definition of logarithms, this means

y = x^1

, so

y=x

Question 10

Find the value of $e^{3 \ln 2} - \log(100)$ .

4
5
6 (correct answer)
7

Explanation: We evaluate each term separately. For the first term, use the logarithm power rule

n \ln a = \ln(a^n)

and the identity

e^{\ln x} = x

e^{3 \ln 2} = e^{\ln(2^3)} = e^{\ln 8} = 8

For the second term,

\log(100)

means

\log_{10}(100)

\log_{10}(100) = \log_{10}(10^2) = 2

Now, subtract the second value from the first:

8 - 2 = 6

Question 11

Find the solution to the equation $2 \ln x = \ln(2x+3)$ .

$x = -1$
$x = 3$ (correct answer)
$x = -1, 3$
No solution

Explanation: Using the power rule for logarithms, the left side can be rewritten:

\ln(x^2) = \ln(2x+3)

. If

\ln A = \ln B

, then

A = B

. So, we can set the arguments equal:

x^2 = 2x+3

This is a quadratic equation:

x^2 - 2x - 3 = 0

. Factoring gives

(x-3)(x+1) = 0

, leading to potential solutions

x=3

and

x=-1

. We must check these solutions in the original equation. The domain of

\ln x

x > 0

. Therefore,

x=-1

is an extraneous solution. The domain of

\ln(2x+3)

2x+3>0

x > -3/2

. The only solution that satisfies both domain constraints is

x=3

Question 12

Given that $e^x + e^{-x} = 4$ , find the value of $e^{2x} + e^{-2x}$ .

8
12
14 (correct answer)
16

Explanation: We are given

e^x + e^{-x} = 4

. To find

e^{2x} + e^{-2x}

, we can square the given expression.

(e^x + e^{-x})^2 = 4^2

(e^x)^2 + 2(e^x)(e^{-x}) + (e^{-x})^2 = 16

Using exponent rules,

(e^x)^2 = e^{2x}

(e^{-x})^2 = e^{-2x}

, and

e^x \cdot e^{-x} = e^{x-x} = e^0 = 1

. So the equation becomes:

e^{2x} + 2(1) + e^{-2x} = 16

e^{2x} + e^{-2x} + 2 = 16

Subtracting 2 from both sides gives:

e^{2x} + e^{-2x} = 14

Question 13

Find all values of $x$ that satisfy the equation $\log_x(5x-6)=2$ .

$x=2$ only
$x=3$ only
$x=2$ and $x=3$ (correct answer)
$x=1$ and $x=6$

Explanation: Convert the logarithmic equation to its exponential form:

x^2 = 5x-6

. Rearrange this into a quadratic equation:

x^2 - 5x + 6 = 0

. Factor the quadratic:

(x-2)(x-3) = 0

. The potential solutions are

x=2

and

x=3

. We must check these solutions against the conditions for logarithms. The base of a logarithm,

x

, must be positive and not equal to 1 (

x>0, x \neq 1

). Both

x=2

and

x=3

satisfy this. The argument of the logarithm,

5x-6

, must be positive. For

x=2

5(2)-6 = 10-6 = 4 > 0

. So,

x=2

is a valid solution. For

x=3

5(3)-6 = 15-6 = 9 > 0

. So,

x=3

is a valid solution. Both values satisfy the equation and the domain constraints.

Question 14

Given that $e^x + e^{-x} = 4$ , find the value of $e^{2x} + e^{-2x}$ .

8
12
14 (correct answer)
16

Explanation: We are given

e^x + e^{-x} = 4

. To find

e^{2x} + e^{-2x}

, we can square the given expression.

(e^x + e^{-x})^2 = 4^2

(e^x)^2 + 2(e^x)(e^{-x}) + (e^{-x})^2 = 16

Using exponent rules,

(e^x)^2 = e^{2x}

(e^{-x})^2 = e^{-2x}

, and

e^x \cdot e^{-x} = e^{x-x} = e^0 = 1

. So the equation becomes:

e^{2x} + 2(1) + e^{-2x} = 16

e^{2x} + e^{-2x} + 2 = 16

Subtracting 2 from both sides gives:

e^{2x} + e^{-2x} = 14

Question 15

Given that $\ln a = 3$ and $\ln b = 4$ , find the value of $\ln\left(\frac{a^2\sqrt{b}}{e}\right)$ .

5
7 (correct answer)
8
11

Explanation: Using the laws of logarithms, the expression can be broken down as follows:

\ln\left(\frac{a^2\sqrt{b}}{e}\right) = \ln(a^2) + \ln(\sqrt{b}) - \ln(e)

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

Now, substitute the given values

\ln a = 3

and

\ln b = 4

= 2(3) + \frac{1}{2}(4) - 1

= 6 + 2 - 1 = 7

Question 16

Find the sum of the solutions to the equation $9^x - 10 \cdot 3^x + 9 = 0$ .

0
2 (correct answer)
3
10

Explanation: The equation can be rewritten as

(3^2)^x - 10 \cdot 3^x + 9 = 0

, which is

(3^x)^2 - 10(3^x) + 9 = 0

. Let

u = 3^x

. The equation becomes a quadratic in

u

u^2 - 10u + 9 = 0

. Factoring this gives

(u-1)(u-9) = 0

. So,

u=1

u=9

. If

u=1

, then

3^x = 1

, which means

x=0

. If

u=9

, then

3^x = 9 = 3^2

, which means

x=2

. The solutions are

x=0

and

x=2

. The sum of the solutions is

0 + 2 = 2

Question 17

Find the exact value of $\log_3 \left( \frac{1}{27\sqrt{3}} \right)$ .

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

Explanation: First, express the argument of the logarithm as a power of 3. We have

27 = 3^3

and

\sqrt{3} = 3^{1/2}

. Therefore,

27\sqrt{3} = 3^3 \cdot 3^{1/2} = 3^{3 + 1/2} = 3^{7/2}

. The expression becomes

\log_3 \left( \frac{1}{3^{7/2}} \right)

. Using the rule for negative exponents, this is

\log_3(3^{-7/2})

. By the definition of logarithms,

\log_a(a^x) = x

, so the value is

-\frac{7}{2}

Question 18

If $x = \log_4(P)$ and $y = \log_{16}(P)$ , which expression is equivalent to $y$ ?

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

Explanation: We can use the change of base formula. A convenient base to use is 4.

y = \log_{16}(P) = \frac{\log_4(P)}{\log_4(16)}

We are given that

x = \log_4(P)

. We can calculate

\log_4(16) = \log_4(4^2) = 2

. Substituting these into the equation for

y

y = \frac{x}{2}

Question 19

The solution to $3^{2x-1} = 4$ is $x = \log_b(c)$ . Find $b$ and $c$ .

$b=3, c=2$
$b=3, c=\sqrt{12}$
$b=9, c=4$
$b=9, c=12$ (correct answer)

Explanation: We start by isolating the term with the exponent

x

3^{2x-1} = 4

\frac{3^{2x}}{3^1} = 4

3^{2x} = 12

Using the exponent rule

(a^m)^n = a^{mn}

, we can rewrite the left side:

(3^2)^x = 12

9^x = 12

Now, we convert this exponential equation to logarithmic form

y = \log_b(a) \iff b^y=a

x = \log_9(12)

Comparing this to the form

x = \log_b(c)

, we have

b=9

and

c=12

Question 20

The solution to $2^{x+1} = 5^x$ is given by $x =$

$\frac{\ln 2}{\ln 5 - \ln 2}$ (correct answer)
$\frac{\ln 2}{\ln 5 + \ln 2}$
$\frac{\ln 5}{\ln 2}$
$1$

Explanation: To solve for

x

, we take the natural logarithm of both sides of the equation:

\ln(2^{x+1}) = \ln(5^x)

Using the power rule for logarithms,

\ln(a^b) = b \ln a

(x+1)\ln 2 = x\ln 5

Distribute

\ln 2

on the left side:

x\ln 2 + \ln 2 = x\ln 5

Rearrange the terms to isolate

x

\ln 2 = x\ln 5 - x\ln 2

Factor out

x

on the right side:

\ln 2 = x(\ln 5 - \ln 2)

Finally, divide to solve for

x

x = \frac{\ln 2}{\ln 5 - \ln 2}