AS & A Level · Practice questions · Solving exponential equations and inequalities

Logs to the rescue.

Papers 2 and 3. Six original Cambridge-style questions on equations and inequalities with the unknown in the index, ending with a compound interest comparison.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Isolate, take logs, solve, check the sign.

Isolate the power $a^{f(x)}$ on one side.
Take ln of both sides: $f(x)\ln a = \ln b$.
Both sides: expand, collect $x$ terms, factorise out $x$.
Inequalities: reverse the sign when dividing by $\ln a$ with $0 < a < 1$.
Hidden quadratic: $u = a^x$, reject $u \le 0$.
Integers: for 'smallest $n$', round in the direction the inequality demands.

Recall

[2 marks]

Solve, giving answers to 3 significant figures: (a) $3^x = 20$ (b) $7^x = 0.5$.

(a) $x = \dfrac{\ln 20}{\ln 3} = 2.73$ ✓
(b) $x = \dfrac{\ln 0.5}{\ln 7} = -0.356$ ✓

Calculation

[3 marks]

Solve $2^{3x + 1} = 5^x$, giving your answer to 3 significant figures.

$(3x + 1)\ln 2 = x\ln 5$, so $3x\ln 2 + \ln 2 = x\ln 5$. ✓
$\ln 2 = x(\ln 5 - 3\ln 2)$, so $x = \dfrac{\ln 2}{\ln 5 - \ln 8} = \dfrac{\ln 2}{\ln 0.625}$. ✓
$x = -1.47$ (3 s.f.). ✓

Calculation

[3 marks]

Solve the inequality $4 \times 3^{2x} > 50$.

$3^{2x} > 12.5$, so $2x\ln 3 > \ln 12.5$. ✓
$\ln 3 > 0$, so the sign stays: $x > \dfrac{\ln 12.5}{2\ln 3}$. ✓
$x > 1.15$ (3 s.f.). ✓

Analysis

[4 marks]

Solve the equation $2^{2x + 1} - 7(2^x) + 3 = 0$, giving any non-exact answer to 3 significant figures.

Let $u = 2^x$: $2u^2 - 7u + 3 = 0$, so $(2u - 1)(u - 3) = 0$. ✓
$2^x = \dfrac{1}{2}$ gives $x = -1$. ✓
$2^x = 3$ gives $x = \dfrac{\ln 3}{\ln 2} = 1.58$. ✓

Analysis

[4 marks]

Solve the inequality $|2^x - 6| < 2$.

$-2 < 2^x - 6 < 2$, so $4 < 2^x < 8$. ✓
$2^x$ is increasing, so $2^2 < 2^x < 2^3$ gives $2 < x < 3$. ✓

Analysis

[6 marks]

Ana invests $2000 at 4% compound interest per year, so after $n$ years her account holds $2000 \times 1.04^n$ dollars. Ben invests $2500 at 3% per year, so his holds $2500 \times 1.03^n$ dollars.

(a) Find the smallest whole number of years after which Ana's account exceeds $3500.

(b) Find the smallest whole number of years after which Ana's account holds more than Ben's.

(a) $1.04^n > 1.75$, so $n > \dfrac{\ln 1.75}{\ln 1.04} = 14.27\dots$ Smallest $n = 15$. (Check: $2000 \times 1.04^{14} = 3463$, $2000 \times 1.04^{15} = 3602$.) ✓
(b) $2000 \times 1.04^n > 2500 \times 1.03^n$, so $\left(\dfrac{1.04}{1.03}\right)^n > 1.25$. ✓
$n\ln \dfrac{1.04}{1.03} > \ln 1.25$, and $\ln \dfrac{1.04}{1.03} > 0$, so $n > \dfrac{\ln 1.25}{\ln (1.04/1.03)} = 23.09\dots$ ✓
Smallest $n = 24$. (Check: after 23 years 4929 < 4934; after 24 years 5127 > 5082.) ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.