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Exponents and logarithms
Exponents and logarithms: inverse operations
Exponents and logarithms: inverse operations Practice
Exponents and logarithms: inverse operations Test
Solving exponential and logarithmic equations
Solving exponential and logarithmic equations Practice
Solving exponential and logarithmic equations Test
CatalogueDPMathematicsMathematics: Analysis and Approaches HLNumber and algebra

Exponents and logarithms: inverse operations

Read powers and logarithms as two directions of the same operation.

  • AA HL
  • Papers 1 and 2
  • No GDC needed
  • About 20–25 min

By the end of this lesson

  • simplify powers with stated conditions
  • combine and expand logarithms correctly
  • change logarithm base

Understand the idea

Compare $3^4=81$ with $\log_3 81=4$. The first asks for the result of repeated scaling; the second asks which power produces that result. A logarithm returns an exponent, not a reciprocal.

Because multiplying powers of the same base joins their factors, $3^2\times3^4=3^6$. Reversing that statement explains why a logarithm turns a product into a sum.

Exponent → value

$4\xrightarrow{\ 3^{(\cdot)}\ }81$

Raise 3 to the fourth power.

Value → exponent

$81\xrightarrow{\ \log_3\ }4$

Ask which power of 3 gives 81.

The same relationship read in opposite directions.

Checkpoint

Which expression equals $\log_4(uv)$ for positive $u,v$?
Show explanation

Products become sums because exponents add when same-base powers multiply.

Put the idea into notation

For positive $a$, $a^m a^n=a^{m+n}$, $a^m/a^n=a^{m-n}$ and $(a^m)^n=a^{mn}$. Also $a^0=1$, $a^{-n}=1/a^n$, and $a^{p/q}=\sqrt[q]{a^p}$. In the fractional exponent, p is an integer and q is a positive integer. Positive bases keep these real-power rules unambiguous.

For $b\gt 0$, $b\ne1$ and $u,v\gt 0$: $$\log_b(uv)=\log_b u+\log_b v,\qquad \log_b(u/v)=\log_b u-\log_b v.$$ The power rule is $\log_b(u^k)=k\log_b u$. There is no corresponding sum rule for $\log_b(u+v)$.

Changing base uses $$\log_b u=\frac{\ln u}{\ln b}.$$ Both logs in the ratio must use the same base.

Example 1

Keep track of the exponent

Simplify $\dfrac{(2^3)^4}{2^5}$.

  1. Power of a power: $(2^3)^4=2^{12}$.
  2. Divide same-base powers: $2^{12-5}=2^7=128$.
Why multiply these exponents but subtract those?

Multiplication of exponents happens inside a power of a power; subtraction happens when dividing.

Example 2

Combine before evaluating

Evaluate $\log_2 12-\log_2 3$.

  1. Both arguments are positive, so combine as $\log_2(12/3)$.
  2. $\log_2 4=2$, since $2^2=4$.
Why divide the logarithm arguments?

Taking the difference of the arguments would ask a different question.

Example 3

Derive change of base

Write $\log_5 7$ using natural logarithms.

  1. Let $y=\log_5 7$. Then $5^y=7$.
  2. Take natural logs: $y\ln5=\ln7$.
  3. $y=\ln7/\ln5$.
Why must the base differ from 1?

The denominator is nonzero because the base is not 1. This reasoning also works for bases between 0 and 1.

In the exam

Quick check

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  1. What does $\log_4 64=3$ tell us?
    Show explanation

    A logarithm reports the exponent needed to reach its argument.

  2. Evaluate $27^{2/3}$.
    Show explanation

    Take the cube root, then square: 3² = 9.

  3. Is $\log_2(4+4)=\log_2 4+\log_2 4$?
    Show explanation

    A logarithm has a product rule, not a sum rule.

  4. Given $\ln a=2$ and $\ln b=1$, find $\ln(a^2/b)$.
    Show explanation

    The power and quotient rules give 2 ln a − ln b = 4 − 1.

  5. For $b\gt1$, $\log_b u=3$. If u is replaced by $b^2u$, what is the new logarithm?
    Show explanation

    Multiplying the argument by b² adds log base b of b², which is 2.

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