1.7 Rational exponents, laws of logarithms and exponential equations (IB Maths: Analysis and Approaches HL): Revision notes

Section 1

Rational exponents

A rational exponent combines a root and a power: $a^{\frac{1}{m}} = \sqrt[m]{a}, \qquad a^{\frac{n}{m}} = \left(\sqrt[m]{a}\right)^n.$ If $m$ is even, $\sqrt[m]{a}$ means the positive root, so $16^{\frac{3}{4}} = \left(\sqrt[4]{16}\right)^3 = 2^3 = 8$ . A negative exponent means a reciprocal: $16^{-\frac{3}{4}} = \frac{1}{8}$ .

Take the root first — the numbers stay small: $27^{\frac{2}{3}} = 3^2 = 9$ .

Key termsrational exponent

Common mistake

$27^{\frac{2}{3}} \ne 18$ . The exponent is not a multiplier: find $\sqrt[3]{27} = 3$ , then square.

Common mistake

A negative exponent does not make the value negative: $8^{-\frac13} = \frac12$ , not $-2$ .

Section 2

Laws of logarithms

For $a, x, y > 0$ (and $a\ne1$ ):

product law: $\log_a xy = \log_a x + \log_a y$
quotient law: $\log_a \frac{x}{y} = \log_a x - \log_a y$
power law: $\log_a x^m = m\log_a x$

These follow from the laws of exponents because $\log_a x$ is the exponent to which $a$ is raised to give $x$ . For example $\log_2\left(8x^3\right) = \log_2 8 + 3\log_2 x = 3 + 3\log_2 x$ .

Key termsproduct lawquotient lawpower law

Common mistake

$\log(x+y)$ is not $\log x + \log y$ , and $\frac{\log x}{\log y}$ is not $\log x - \log y$ .

Exam tip

Write roots as powers first: $\log\sqrt{x} = \log x^{\frac12} = \frac12\log x$ .

Section 3

Change of base

To rewrite a logarithm in a different base, use the change of base formula $\log_a x = \frac{\log_b x}{\log_b a}, \quad a, b, x > 0.$ It lets you evaluate any logarithm on a calculator using $\ln$ or $\log_{10}$ , and link logarithms whose bases are powers of each other: $\log_8 y = \frac{\log_2 y}{\log_2 8} = \frac{1}{3}\log_2 y$ .

Key termschange of base

Example

$\log_9 27 = \frac{\log_3 27}{\log_3 9} = \frac{3}{2}$ .

Section 4

Solving exponential equations

Two approaches:

Same base: write both sides as powers of one base and equate exponents. $\left(\frac14\right)^x = 8^{x+2}$ becomes $2^{-2x} = 2^{3x+6}$ , so $-2x = 3x+6$ and $x = -\frac{6}{5}$ .
Take logarithms: when the bases cannot be matched. $2^{x-1} = 10$ gives $x - 1 = \log_2 10$ , so $x = 1 + \frac{\ln10}{\ln2}$ .

With different bases on both sides, take logs, expand the brackets and collect the $x$ terms: $2^{x+1} = 3^{x-1}$ gives $x(\ln3-\ln2) = \ln3+\ln2$ , so $x = \frac{\ln6}{\ln1.5}$ .

Key termsexponential equation

Common mistake

Taking logs of only part of a side. In $500\times2^{t/3} = 3000$ , divide by 500 first, then take logs.

Exam tip

In models, check the answer makes sense: a time must be positive, and a population must reach the target after, not before, it starts.

Section 5

Exponential models

Growth models such as $N = 500\times2^{\frac{t}{3}}$ lead straight to exponential equations. Here the population doubles every 3 hours, so it reaches $4000 = 500\times2^3$ after $9$ hours.

For a non-power target, isolate the power and take logarithms: $2^{\frac{t}{3}} = 6$ gives $t = 3\log_2 6 = 3 + 3\log_2 3 \approx 7.75$ hours. Give an exact form when asked and a 3 s.f. value otherwise.

Key termsdoubling time

Must know

$a^{\frac{n}{m}} = \left(\sqrt[m]{a}\right)^n$ ; even roots are positive; negative exponent means reciprocal.
$\log_a xy = \log_a x+\log_a y$ , $\log_a\frac{x}{y} = \log_a x-\log_a y$ , $\log_a x^m = m\log_a x$ .
$\log_a x = \frac{\log_b x}{\log_b a}$ .
Solve exponential equations by matching bases or by taking logarithms; isolate the power first.

That's the notes covered.

Carry on to the next subtopic.