Home / A-Level Maths 9709 / Logarithmic and exponential functions / Solving exponential equations and inequalities
Cambridge International AS & A Level · Topic 3.2
A-Level Maths 9709 / Topic 3.2 / AS & A Level

Bring the power down.

Papers 2 and 3. When the unknown sits in an index, take logarithms of both sides so the power law brings it down. Then solve a linear equation, a quadratic in \(a^x\), or an inequality, watching for the sign flip when you divide by the log of a number below 1.

Topic 3.2All logarithmic and exponential functions lessons →
The key idea

Take ln (or lg) of both sides: \(a^{f(x)} = b \Rightarrow f(x)\ln a = \ln b\). For an inequality, dividing by \(\ln a\) keeps the sign when \(a > 1\) but REVERSES it when \(0 < a < 1\), because then \(\ln a < 0\).

x y O (0, 1) 2.32 y = 5 y = 2x 2x < 5 here
Fig. 1: y = 2^x crosses y = 5 where x = ln 5 / ln 2 = 2.32, and because 2^x is increasing, 2^x < 5 exactly when x < 2.32
Section 01

Take logs of both sides.

Taking logarithms turns an index equation into an ordinary one, because \(\ln a^{k} = k\ln a\). Any base works; ln and lg are on every calculator. Keep \(\ln 2\), \(\ln 3\) as exact numbers until the last line, then round to 3 significant figures.

Worked example Simple index equation
Solve \(2^x = 5\), giving your answer to 3 significant figures.
  1. Take ln: \(x\ln 2 = \ln 5\).
  2. \(x = \dfrac{\ln 5}{\ln 2} = 2.3219\dots\)
Answer: \(x = 2.32\) (Fig. 1)
Worked example Unknown on both sides
Solve \(3^{x + 1} = 4^{2x - 1}\), giving your answer to 3 significant figures.
  1. Take ln: \((x + 1)\ln 3 = (2x - 1)\ln 4\).
  2. Expand: \(x\ln 3 + \ln 3 = 2x\ln 4 - \ln 4\).
  3. Collect \(x\) terms: \(\ln 3 + \ln 4 = 2x\ln 4 - x\ln 3 = x(2\ln 4 - \ln 3)\).
  4. \(x = \dfrac{\ln 3 + \ln 4}{2\ln 4 - \ln 3} = \dfrac{\ln 12}{\ln \frac{16}{3}} = 1.4844\dots\)
Answer: \(x = 1.48\)
Your turn
Solve \(6^{x - 2} = 4^x\), giving your answer to 3 significant figures.
Examiner trap

\(\dfrac{\ln 5}{\ln 2}\) is NOT \(\ln \dfrac{5}{2}\) and not \(\ln 5 - \ln 2\). A quotient of logs does not simplify: work it out as one number divided by another.

Section 02

Inequalities: watch the sign.

Solve an inequality exactly like the equation, but every time you divide by a logarithm, ask whether it is negative. \(\ln a > 0\) when \(a > 1\), and \(\ln a < 0\) when \(0 < a < 1\). Dividing by a negative number reverses the inequality.

Worked example Positive base
Solve \(3 \times 2^{3x - 1} < 5\).
  1. Isolate the power: \(2^{3x - 1} < \dfrac{5}{3}\).
  2. Take ln (ln is increasing, so the sign stays): \((3x - 1)\ln 2 < \ln \dfrac{5}{3}\).
  3. \(\ln 2 > 0\), so divide without reversing: \(3x - 1 < \dfrac{\ln(5/3)}{\ln 2} = 0.7369\dots\)
  4. \(3x < 1.7369\dots\), so \(x < 0.5789\dots\)
Answer: \(x < 0.579\)
Worked example Base less than 1: the sign flips
Find the smallest integer \(n\) for which \(0.8^n < 0.1\).
  1. Take ln: \(n\ln 0.8 < \ln 0.1\).
  2. \(\ln 0.8 = -0.223\dots\) is NEGATIVE, so dividing by it reverses the sign: \(n > \dfrac{\ln 0.1}{\ln 0.8}\).
  3. \(\dfrac{\ln 0.1}{\ln 0.8} = 10.318\dots\), so \(n > 10.318\dots\)
  4. Check: \(0.8^{10} = 0.107 > 0.1\) and \(0.8^{11} = 0.0859 < 0.1\).
Answer: \(n = 11\)
Examiner trap

In \(n\ln 0.8 < \ln 0.1\), both logs are negative. Forgetting to reverse gives \(n < 10.3\), which says the SMALL values work, the opposite of what happens as \(0.8^n\) shrinks.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

The solution of \(2^x = 9\) is

Quick check+10 XP

Solving \(0.5^x < 0.2\) gives \(x\ln 0.5 < \ln 0.2\). What comes next?

Quick check+10 XP

If \(u = 2^x\), then \(4^x\) equals

Quick check+10 XP

How many real solutions does \(3^x = -2\) have?

Section 03

Quadratics in \(a^x\) and moduli.

\(4^x = (2^2)^x = (2^x)^2\) and \(2^{x + 1} = 2 \times 2^x\). Spot these to rewrite an equation as a quadratic in \(u = 2^x\). Solve for \(u\), reject any \(u \le 0\), then take logs.

Worked example Hidden quadratic
Solve \(2^{2x + 1} - 7(2^x) + 3 = 0\).
  1. \(2^{2x + 1} = 2 \times (2^x)^2\). Let \(u = 2^x\): \(2u^2 - 7u + 3 = 0\).
  2. \((2u - 1)(u - 3) = 0\), so \(u = \dfrac{1}{2}\) or \(u = 3\).
  3. \(2^x = \dfrac{1}{2} = 2^{-1}\) gives \(x = -1\).
  4. \(2^x = 3\) gives \(x = \dfrac{\ln 3}{\ln 2} = 1.5849\dots\)
Answer: \(x = -1\) or \(x = 1.58\)
Worked example Modulus with an exponential
Solve \(|3^x - 4| = 2\).
  1. \(3^x - 4 = 2\) or \(3^x - 4 = -2\), so \(3^x = 6\) or \(3^x = 2\).
  2. \(x = \dfrac{\ln 6}{\ln 3} = 1.6309\dots\) or \(x = \dfrac{\ln 2}{\ln 3} = 0.6309\dots\)
Answer: \(x = 1.63\) or \(x = 0.631\)
Your turn
Solve \(4^x - 3(2^{x + 1}) + 8 = 0\).
Examiner trap

\(2^{2x + 1}\) is \(2 \times 2^{2x} = 2u^2\), not \(u^2 + 1\) and not \(2u\). Split the index with \(a^{m + n} = a^m \times a^n\) before you substitute.

Stage 2 · Exam

Exam-style questions

Unlocks once the four checks above are done. Worth more XP, written to AS & A Level Paper 1 to 6 standard.

Finish the four checks above to unlock the exam questions
Exam style+20 XP

Solve \(2^{x + 1} = 3^x\), giving your answer to 3 significant figures.

Exam style+20 XP

Find the smallest integer \(n\) such that \(0.9^n < 0.05\).

Exam style+20 XP

Solve \(9^x - 4(3^x) + 3 = 0\).

Exam style+20 XP

Solve \(5^{2x - 1} = 7\), giving your answer to 3 significant figures.

Skill unlocked

Index equations, solved.

This skill is now recorded on your profile. Keep the chain going.

-Rank -Level -Score -Topics
Go deeper · practice
Six original Cambridge-style questions
Solving \(a^x = b\), unknowns on both sides, inequalities with the sign flip, quadratics in \(a^x\), a modulus and a savings problem.