What are integer exponents and how do they work in mathematics?

Integer exponents are a way to express repeated multiplication of a base number. For example, 3^4 means multiplying 3 by itself four times (3 * 3 * 3 * 3). In the IB MYP Mathematics curriculum, understanding integer exponents is crucial as they simplify expressions and solve equations efficiently.

How do you apply the laws of exponents in mathematical problems?

The laws of exponents, such as the product of powers, power of a power, and quotient of powers, help simplify expressions. For instance, the product of powers law states that when multiplying two powers with the same base, you add the exponents: a^m * a^n = a^(m+n). These laws are essential in the IB MYP curriculum for simplifying complex expressions and solving equations.

What is a logarithm and how is it related to exponents?

A logarithm is the inverse operation of exponentiation. It answers the question: 'To what exponent must the base be raised, to produce a given number?' For example, if 2^3 = 8, then log base 2 of 8 is 3. In the IB MYP Mathematics curriculum, understanding logarithms is important for solving exponential equations and understanding exponential growth and decay.

How do you solve exponential equations using logarithms?

To solve exponential equations using logarithms, you can take the logarithm of both sides of the equation, which allows you to bring down the exponent as a coefficient. For example, to solve 2^x = 8, you can take log base 2 of both sides to get x = log base 2 of 8, which equals 3. This method is a key skill in the IB MYP Mathematics curriculum for solving real-world problems involving exponential growth.

What are some common mistakes to avoid when working with integer exponents and logarithms?

Common mistakes include confusing the laws of exponents, such as incorrectly applying the product of powers law, and misunderstanding the properties of logarithms, like assuming log(a + b) equals log(a) + log(b). In the IB MYP Mathematics curriculum, it's important to carefully apply the correct rules and understand the properties of these operations to avoid errors in calculations.

Why is "Writing $\log(x + y) = \log x + \log y$." a common mistake in Integer Exponents and Logarithm?

Why it happens: Confusing it with the product rule. How to avoid it: The product rule applies to (xy), NOT (x+y). (x+y) has no simplification.

Why is "Assuming $\log$ always means $\log_e$." a common mistake in Integer Exponents and Logarithm?

Why it happens: Different conventions in different texts. How to avoid it: In MYP / Cambridge, means \log_10. means \log_e. Always check the convention.

Why is "Computing $\log$ of a negative or zero number." a common mistake in Integer Exponents and Logarithm?

Why it happens: Forgetting domain restrictions. How to avoid it: \log_a x is only defined for x > 0. You CAN'T take the log of 0 or a negative.

NumberIB MYP Maths Extended Level

Integer Exponents and Logarithm

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Integer Exponents and Logarithm — IB MYP Mathematics (Extended): the inverse of exponentiation

Logarithms answer the question 'what power?'. This MYP Mathematics Extended note covers logs as inverses of exponentiation, common bases (10, $e$ , 2), and the laws of logarithms.

What you’ll learn

Mapped to the IB MYP Mathematics subject guide (2026 onwards).

MYP Mathematics A — Convert between exponential and logarithmic form.
MYP Mathematics A — Apply the three log laws.
MYP Mathematics C — Use logs to solve exponential equations.
MYP Mathematics D — Apply logs in growth/decay contexts.

What is a logarithm?

The inverse of an exponent.

If $a^x = y$ , then we say ' $x$ is the logarithm of $y$ to base $a$ ': $\log_a y = x.$

So a logarithm answers: 'to what power must I raise the base to get this number?'

Examples:

$10^2 = 100$ , so $\log_{10} 100 = 2$ .
$2^5 = 32$ , so $\log_2 32 = 5$ .
$3^0 = 1$ , so $\log_3 1 = 0$ .
$5^1 = 5$ , so $\log_5 5 = 1$ .

Key facts (true for any base $a > 0$ , $a \ne 1$ ):

$\log_a 1 = 0$ (because $a^0 = 1$ ).
$\log_a a = 1$ (because $a^1 = a$ ).
$\log_a (a^x) = x$ (logs undo exponentials).
$a^{\log_a x} = x$ (and vice versa).

Conventional shorthand:

$\log x$ usually means $\log_{10} x$ (common log).
$\ln x$ means $\log_e x$ (natural log, base $e \approx 2.71828$ ).
In Extended Level computer-science contexts, $\log_2$ is common.

A logarithm has a SMALLER VALUE than the number itself for $a > 1$ . $\log_{10} 1000 = 3$ , $\log_{10} 1\,000\,000 = 6$ . Logs compress huge numbers into small ones.

$\log_a y = x \iff a^x = y$ .
$\log_a 1 = 0$ and $\log_a a = 1$ .
Common log: base 10. Natural log: base $e$ .
Logs UNDO exponentials.

The three laws of logarithms

Mirror images of the index laws.

Because logs are inverse exponentials, the log laws follow directly from the index laws:

Index law	Log law
$a^m \cdot a^n = a^{m+n}$	$\log(xy) = \log x + \log y$
$a^m / a^n = a^{m-n}$	$\log(x/y) = \log x - \log y$
$(a^m)^n = a^{mn}$	$\log(x^n) = n \log x$

(These hold for any base, so I've omitted the subscript.)

Worked example. Expand $\log(7x^2)$ .

Product rule: $\log 7 + \log(x^2)$ .
Power rule: $\log 7 + 2 \log x$ .

Worked example. Combine $3 \log 2 + \log 5 - \log 4$ into a single log.

$3 \log 2 = \log(2^3) = \log 8$ .
$\log 8 + \log 5 = \log(8 \times 5) = \log 40$ .
$\log 40 - \log 4 = \log(40 / 4) = \log 10 = 1$ (using base 10).

Change of base (Extended-level extension): $\log_a x = \dfrac{\log_b x}{\log_b a}$ . Lets you compute any-base log on a calculator that only does $\log_{10}$ or $\ln$ .

Example: $\log_2 100 = \dfrac{\log_{10} 100}{\log_{10} 2} = \dfrac{2}{0.301} \approx 6.64$ .

Product: $\log(xy) = \log x + \log y$ .
Quotient: $\log(x/y) = \log x - \log y$ .
Power: $\log(x^n) = n \log x$ .
Change of base: $\log_a x = \log_b x / \log_b a$ .

Solving exponential equations with logs

Take logs of both sides.

When the unknown is in the EXPONENT, take logs of both sides to bring it down.

Worked example. Solve $2^x = 50$ .

Take log (base 10) of both sides: $\log(2^x) = \log 50$ .
Power law: $x \log 2 = \log 50$ .
$x = \dfrac{\log 50}{\log 2} \approx \dfrac{1.699}{0.301} \approx 5.64$ .

So $2^{5.64} \approx 50$ . (Check: $2^5 = 32$ , $2^6 = 64$ . Yes — between.)

Worked example. Solve $5 \cdot 3^x = 200$ .

Divide by 5: $3^x = 40$ .
Take log: $x \log 3 = \log 40$ .
$x = \dfrac{\log 40}{\log 3} \approx \dfrac{1.602}{0.477} \approx 3.36$ .

Worked example (compound interest, criterion D). $1000 invested at $5\%$ annual interest. How many years until it doubles?

Growth equation: $A = 1000 \times 1.05^t$ . Want $A = 2000$ .
$1.05^t = 2$ .
$t \log 1.05 = \log 2$ .
$t = \dfrac{\log 2}{\log 1.05} \approx \dfrac{0.301}{0.0212} \approx 14.2$ years.

(Famous 'rule of 72': time to double $\approx 72 / \text{interest rate \%}$ . Here: $72 / 5 = 14.4$ . Matches.)

Unknown in the exponent → take logs.
Use the power rule to bring the exponent down.
Divide to isolate.

How it’s examined

Criterion A: convert and apply log laws. Criterion C: solve exponential equations using logs. Criterion D: growth / decay applications.

Sources: IB MYP Mathematics subject guide (IBO, official). Last reviewed 2026-05-25.

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Step-by-step worked examples — Integer Exponents and Logarithm

Step-by-step solutions to past-paper-style questions on integer exponents and logarithm, written exactly the way a tutor would explain them at the board.

1Convert between log and exponential form
Getting started• conversion
Question
Rewrite (a) $\log_4 64 = 3$ in exponential form, and (b) $5^3 = 125$ in log form.
Step-by-step solution
1. Step 1
  (a) $\log_a y = x \iff a^x = y$ . So $\log_4 64 = 3 \Rightarrow 4^3 = 64$ .
2. Step 2
  (b) $5^3 = 125 \Rightarrow \log_5 125 = 3$ .
Answer
(a) $4^3 = 64$ . (b) $\log_5 125 = 3$ .
2Evaluate logarithms
Building confidence• evaluate
Question
Evaluate without a calculator: (a) $\log_2 32$ , (b) $\log_{10} 0.01$ , (c) $\log_3 1$ .
Step-by-step solution
1. Step 1
  (a) Need power of 2 giving 32. $2^5 = 32$ , so $\log_2 32 = 5$ .
2. Step 2
  (b) $0.01 = 10^{-2}$ , so $\log_{10} 0.01 = -2$ .
3. Step 3
  (c) $3^0 = 1$ , so $\log_3 1 = 0$ . (True for any base.)
Answer
(a) 5; (b) −2; (c) 0.
3Apply log laws
Building confidence• laws
Question
Express $2\log 5 + 3\log 2 - \log 4$ as a single log.
Step-by-step solution
1. Step 1
  Power rule: $2 \log 5 = \log 25$ and $3 \log 2 = \log 8$ .
2. Step 2
  Sum: $\log 25 + \log 8 = \log(25 \times 8) = \log 200$ .
3. Step 3
  Quotient: $\log 200 - \log 4 = \log(200/4) = \log 50$ .
Answer
$\log 50$
4Solve an exponential equation
Stretch• solve
Question
Solve $3 \cdot 2^x = 96$ . Give your answer to 3 s.f.
Step-by-step solution
1. Step 1
  Divide by 3: $2^x = 32$ .
2. Step 2
  Recognise: $32 = 2^5$ , so $x = 5$ exactly.
3. Step 3
  (If using logs in general: $x \log 2 = \log 32$ , so $x = \dfrac{\log 32}{\log 2} = 5$ .)
Answer
$x = 5$

Key Definitions and Keywords — Integer Exponents and Logarithm

Definitions to memorise and the exact keywords mark schemes credit for integer exponents and logarithm answers — sharpened from recent examiner reports for the 2026 IB MYP Mathematics (Extended) sitting.

Logarithm $\log_a y$
Examiner keyword
The power to which $a$ must be raised to give $y$ . $\log_a y = x \iff a^x = y$ .
Common logarithm
Examiner keyword
$\log_{10}$ , often written $\log$ (no subscript) or $\lg$ .
Natural logarithm
$\ln = \log_e$ , where $e \approx 2.71828$ . The base in calculus.
Logarithm laws
Examiner keyword
$\log(xy) = \log x + \log y$ ; $\log(x/y) = \log x - \log y$ ; $\log(x^n) = n \log x$ .

Common Mistakes and Misconceptions — Integer Exponents and Logarithm

The traps other students keep falling into on integer exponents and logarithm questions — taken from recent IB MYP Mathematics (Extended) examiner reports and mark schemes — and how to avoid them.

✕Writing $\log(x + y) = \log x + \log y$ .
Why it happens
Confusing it with the product rule.
How to avoid it
The product rule applies to $\log(xy)$ , NOT $\log(x+y)$ . $\log(x+y)$ has no simplification.
✕Assuming $\log$ always means $\log_e$ .
Why it happens
Different conventions in different texts.
How to avoid it
In MYP / Cambridge, $\log$ means $\log_{10}$ . $\ln$ means $\log_e$ . Always check the convention.
✕Computing $\log$ of a negative or zero number.
Why it happens
Forgetting domain restrictions.
How to avoid it
$\log_a x$ is only defined for $x > 0$ . You CAN'T take the log of 0 or a negative.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

Past paper style quiz

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Integer Exponents and Logarithm — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Index law

Log law

a^m \cdot a^n = a^{m+n}

\log(xy) = \log x + \log y

a^m / a^n = a^{m-n}

\log(x/y) = \log x - \log y

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

\log(x^n) = n \log x

When the unknown is in the EXPONENT, take logs of both sides to bring it down.

Worked example. Solve $2^x = 50$ .

Take log (base 10) of both sides: $\log(2^x) = \log 50$ .
Power law: $x \log 2 = \log 50$ .
$x = \dfrac{\log 50}{\log 2} \approx \dfrac{1.699}{0.301} \approx 5.64$ .

So $2^{5.64} \approx 50$ . (Check: $2^5 = 32$ , $2^6 = 64$ . Yes — between.)

Worked example. Solve $5 \cdot 3^x = 200$ .

Divide by 5: $3^x = 40$ .
Take log: $x \log 3 = \log 40$ .
$x = \dfrac{\log 40}{\log 3} \approx \dfrac{1.602}{0.477} \approx 3.36$ .

Worked example (compound interest, criterion D). $1000 invested at $5\%$ annual interest. How many years until it doubles?

Growth equation: $A = 1000 \times 1.05^t$ . Want $A = 2000$ .
$1.05^t = 2$ .
$t \log 1.05 = \log 2$ .
$t = \dfrac{\log 2}{\log 1.05} \approx \dfrac{0.301}{0.0212} \approx 14.2$ years.

(Famous 'rule of 72': time to double $\approx 72 / \text{interest rate \%}$ . Here: $72 / 5 = 14.4$ . Matches.)

Integer Exponents and Logarithm

Short Study Notes in the form of Slides

Short Study Notes — Integer Exponents and Logarithm

Detailed Notes

What you’ll learn

How it’s examined

1Convert between log and exponential form

2Evaluate logarithms

3Apply log laws

4Solve an exponential equation

Logarithm log⁡ay\log_a yloga​y

Common logarithm

Natural logarithm

Logarithm laws

✕Writing log⁡(x+y)=log⁡x+log⁡y\log(x + y) = \log x + \log ylog(x+y)=logx+logy.

✕Assuming log⁡\loglog always means log⁡e\log_eloge​.

✕Computing log⁡\loglog of a negative or zero number.

Practice questions

Past paper style quiz

Assess your understanding

Integer Exponents and Logarithm — frequently asked questions

Integer Exponents and Logarithm

Short Study Notes in the form of Slides

Short Study Notes — Integer Exponents and Logarithm

Detailed Notes

What you’ll learn

How it’s examined

1Convert between log and exponential form

2Evaluate logarithms

3Apply log laws

4Solve an exponential equation

Logarithm log⁡ay\log_a yloga​y

Common logarithm

Natural logarithm

Logarithm laws

✕Writing log⁡(x+y)=log⁡x+log⁡y\log(x + y) = \log x + \log ylog(x+y)=logx+logy.

✕Assuming log⁡\loglog always means log⁡e\log_eloge​.

✕Computing log⁡\loglog of a negative or zero number.

Practice questions

Past paper style quiz

Assess your understanding

Integer Exponents and Logarithm — frequently asked questions

Logarithm $\log_a y$

✕Writing $\log(x + y) = \log x + \log y$ .

✕Assuming $\log$ always means $\log_e$ .

✕Computing $\log$ of a negative or zero number.

Logarithm $\log_a y$

✕Writing $\log(x + y) = \log x + \log y$ .

✕Assuming $\log$ always means $\log_e$ .

✕Computing $\log$ of a negative or zero number.