Exponential and Log Equations – A Level Maths Year 1

When the unknown is in the index, take logarithms of both sides. The power law $\log\left(a^x\right) = x\log a$ brings the unknown down to where you can reach it. DfE content area F asks you to solve equations of the form $a^x = b$ in this way.

The key step

$a^x = b \;\Rightarrow\; x\ln a = \ln b \;\Rightarrow\; x = \dfrac{\ln b}{\ln a}$

Any base of logarithm works — $\ln$ or $\log_{10}$ give the same answer. Use $\ln$ whenever $e$ appears in the question, because $\ln e^{k} = k$.

Worked Example 1 — A basic exponential equation

Solve $3^x = 20$, giving your answer to 3 significant figures.

①Take $\ln$: $\ln 3^x = \ln 20$

②Power law: $x\ln 3 = \ln 20$

③$x = \dfrac{\ln 20}{\ln 3} = 2.73$ (3 s.f.)

Sense check: $3^2 = 9$ and $3^3 = 27$, so the answer must lie between $2$ and $3$, closer to $3$ ✓

Worked Example 2 — An expression in the index

Solve (a) $5^{2x-1} = 40$ (b) $e^{0.3t} = 5$.

①(a) $(2x - 1)\ln 5 = \ln 40$, so $2x - 1 = \dfrac{\ln 40}{\ln 5} = 2.2920\ldots$

②$2x = 3.2920\ldots$, so $x = 1.65$ (3 s.f.)

③(b) $\ln$ undoes $e$ directly: $0.3t = \ln 5$, so $t = \dfrac{\ln 5}{0.3} = 5.36$ (3 s.f.)

Bracket the whole index: $(2x-1)\ln 5$, not $2x - 1\ln 5$. Keep full calculator values until the last line.

Take logs of both sides and use the power law on each.
Expand the brackets.
Collect the $x$ terms on one side and factorise out $x$.
Divide. Leave the answer in terms of logs if an exact answer is wanted.

Worked Example 3 — Exact answer with different bases

Solve $2^{x+1} = 3^x$, giving $x$ exactly and to 3 significant figures.

①$(x + 1)\ln 2 = x\ln 3$

②$x\ln 2 + \ln 2 = x\ln 3$

③$\ln 2 = x\ln 3 - x\ln 2 = x(\ln 3 - \ln 2)$

④$x = \dfrac{\ln 2}{\ln 3 - \ln 2} = \dfrac{\ln 2}{\ln 1.5} \approx 1.71$

Check: $2^{2.71} \approx 6.54$ and $3^{1.71} \approx 6.54$ ✓

Because $e^{2x} = \left(e^x\right)^2$ and $2^{2x} = \left(2^x\right)^2$, an equation with terms in $e^{2x}$, $e^x$ and a constant is really a quadratic in $e^x$. Substitute $y = e^x$ (or $y = 2^x$), solve the quadratic, then undo the substitution.

Rewriting the terms

$a^{2x} = \left(a^x\right)^2 \qquad a^{x+1} = a \times a^x \qquad a^{2x+1} = a\left(a^x\right)^2 \qquad a^{-x} = \dfrac{1}{a^x}$

Worked Example 4 — A quadratic in $e^x$

Solve $e^{2x} - 5e^x + 6 = 0$, giving exact answers.

①Let $y = e^x$: $y^2 - 5y + 6 = 0$

②$(y - 2)(y - 3) = 0$, so $y = 2$ or $y = 3$.

③$e^x = 2 \Rightarrow x = \ln 2$; $e^x = 3 \Rightarrow x = \ln 3$.

Worked Example 5 — A quadratic in $2^x$

Solve $2^{2x} - 6 \times 2^x + 8 = 0$.

①Let $y = 2^x$: $y^2 - 6y + 8 = 0$

②$(y - 2)(y - 4) = 0$, so $y = 2$ or $y = 4$.

③$2^x = 2 \Rightarrow x = 1$; $2^x = 4 \Rightarrow x = 2$.

Worked Example 6 — Rejecting an impossible root

Solve $e^x - 2 = 15e^{-x}$.

①Multiply through by $e^x$ (never zero): $e^{2x} - 2e^x = 15$

②Let $y = e^x$: $y^2 - 2y - 15 = 0$, so $(y - 5)(y + 3) = 0$.

③$y = 5$ or $y = -3$. But $e^x > 0$ for all $x$, so $e^x = -3$ has no solution.

④$e^x = 5 \Rightarrow x = \ln 5$.

Always say why you reject a root. "$e^x = -3$ is impossible since $e^x > 0$" earns the final mark; silently dropping it may not.

Use the laws to combine everything into one log on each side (or one log and a number).
Remove the logs: $\log_a P = \log_a Q \Rightarrow P = Q$, or $\log_a P = c \Rightarrow P = a^c$.
Solve the resulting equation.
Check every answer in the original equation — the log of a negative number or zero is undefined.

Worked Example 7 — Two logs equal to a number

Solve $\log_2 x + \log_2(x + 3) = 2$.

①Combine: $\log_2\left(x(x+3)\right) = 2$

②Index form: $x(x + 3) = 2^2 = 4$

③$x^2 + 3x - 4 = 0 \Rightarrow (x + 4)(x - 1) = 0$

④$x = -4$ would need $\log_2(-4)$, which is undefined. Reject it.

⑤$x = 1$. Check: $\log_2 1 + \log_2 4 = 0 + 2 = 2$ ✓

Worked Example 8 — A log on each side

Solve $\ln x + \ln(x + 2) = \ln 15$.

①$\ln\left(x(x+2)\right) = \ln 15$, so $x^2 + 2x = 15$.

②$x^2 + 2x - 15 = 0 \Rightarrow (x + 5)(x - 3) = 0$

③$x = -5$ makes $\ln x$ undefined, so $x = 3$ only.

Worked Example 9 — Both roots valid

Solve $2\log_3 x - \log_3(x - 2) = 2$.

①Power law: $\log_3 x^2 - \log_3(x-2) = 2$

②Division law: $\log_3\left(\dfrac{x^2}{x-2}\right) = 2$, so $\dfrac{x^2}{x - 2} = 9$.

③$x^2 = 9x - 18 \Rightarrow x^2 - 9x + 18 = 0 \Rightarrow (x - 3)(x - 6) = 0$

④Both $x = 3$ and $x = 6$ satisfy $x > 2$, so both are valid.

Check $x = 6$: $2\log_3 6 - \log_3 4 = \log_3 \dfrac{36}{4} = \log_3 9 = 2$ ✓

Inequalities such as "after how many years will the value fall below …" are solved the same way, with one danger.

The log of a number between $0$ and $1$ is negative. If you divide by $\ln 0.8$ (which is about $-0.223$), the inequality sign reverses.

Worked Example 10 — Decay below a threshold

Find the smallest integer $n$ for which $0.8^n < 0.1$.

①Take $\ln$ (an increasing function, so the sign is kept): $n\ln 0.8 < \ln 0.1$

②Divide by $\ln 0.8 < 0$ and reverse: $n > \dfrac{\ln 0.1}{\ln 0.8} = 10.32\ldots$

③Smallest integer: $n = 11$.

Check: $0.8^{10} \approx 0.107$ (not yet below $0.1$) and $0.8^{11} \approx 0.0859$ ✓

$a^x = b$

$x = \dfrac{\ln b}{\ln a}$

$e^{kx} = b$

$x = \dfrac{\ln b}{k}$

Both sides

Take logs, expand, collect $x$, factorise.

Hidden quadratic

$e^{2x} = \left(e^x\right)^2$; let $y = e^x$.

Reject

$e^x$ and $a^x$ are never zero or negative.

Log equations

Combine to one log, then switch to index form.

Check

Every argument of a log must be positive.

Inequalities

Dividing by $\ln$ of a number less than 1 flips the sign.

Exact answers

Leave as $\ln 5$, $\dfrac{\ln 2}{\ln 1.5}$, etc.

Question 1

Solve $4^x = 50$, giving your answer to 3 significant figures.

▶ Show solution

$x\ln 4 = \ln 50$, so $x = \dfrac{\ln 50}{\ln 4} = 2.82$ (3 s.f.)

Question 2

Solve $3e^{2x} = 24$, giving an exact answer in the form $k\ln 2$.

▶ Show solution

$e^{2x} = 8 \Rightarrow 2x = \ln 8 = 3\ln 2$

$x = \tfrac32\ln 2$ $(\approx 1.04)$

Question 3

Solve $7^{x-2} = 3^x$, giving your answer to 3 significant figures.

▶ Show solution

$(x - 2)\ln 7 = x\ln 3$

$x\ln 7 - 2\ln 7 = x\ln 3$

$x(\ln 7 - \ln 3) = 2\ln 7$

$x = \dfrac{2\ln 7}{\ln 7 - \ln 3} = 4.59$ (3 s.f.)

Question 4

Solve $e^{2x} - 7e^x + 12 = 0$, giving exact answers.

▶ Show solution

Let $y = e^x$: $y^2 - 7y + 12 = 0 \Rightarrow (y-3)(y-4) = 0$.

$e^x = 3 \Rightarrow x = \ln 3$; $e^x = 4 \Rightarrow x = \ln 4 = 2\ln 2$.

Question 5

Solve $3^{2x} - 4 \times 3^x + 3 = 0$.

▶ Show solution

Let $y = 3^x$: $y^2 - 4y + 3 = 0 \Rightarrow (y - 1)(y - 3) = 0$.

$3^x = 1 \Rightarrow x = 0$; $3^x = 3 \Rightarrow x = 1$.

Question 6

Solve $\log_3(x + 4) + \log_3(x - 4) = 2$.

▶ Show solution

$\log_3\left((x+4)(x-4)\right) = 2 \Rightarrow x^2 - 16 = 9 \Rightarrow x^2 = 25$.

$x = 5$ or $x = -5$. But $x = -5$ gives $\log_3(-1)$, undefined.

$x = 5$. Check: $\log_3 9 + \log_3 1 = 2 + 0 = 2$ ✓

Question 7

Solve $2\ln x = \ln(x + 6)$.

▶ Show solution

$\ln x^2 = \ln(x+6) \Rightarrow x^2 = x + 6 \Rightarrow x^2 - x - 6 = 0$.

$(x - 3)(x + 2) = 0$. Reject $x = -2$ since $\ln(-2)$ is undefined.

$x = 3$.

Question 8

£$2000$ is invested at $5\%$ compound interest per year. Find the number of complete years after which the investment is first worth more than £$4000$.

▶ Show solution

Need $2000 \times 1.05^n > 4000$, i.e. $1.05^n > 2$.

$n\ln 1.05 > \ln 2$. Since $\ln 1.05 > 0$ the sign stays: $n > \dfrac{\ln 2}{\ln 1.05} = 14.2\ldots$

So after $15$ years. (Check: $1.05^{14} \approx 1.980$, $1.05^{15} \approx 2.079$ ✓)

Question 9

In this question you must show detailed reasoning. Solve $2^{2x+1} - 9 \times 2^x + 4 = 0$.

▶ Show solution

$2^{2x+1} = 2 \times \left(2^x\right)^2$. Let $y = 2^x$: $2y^2 - 9y + 4 = 0$.

$(2y - 1)(y - 4) = 0$, so $y = \tfrac12$ or $y = 4$.

$2^x = \tfrac12 = 2^{-1} \Rightarrow x = -1$; $2^x = 4 = 2^2 \Rightarrow x = 2$.

Question 10

Two investments are worth £$A$ and £$B$ after $t$ years, where $A = 1000 \times 1.04^t$ and $B = 800 \times 1.06^t$.

(a) State the initial value of each investment and its annual percentage growth. (b) Show that $B > A$ when $\left(\dfrac{53}{52}\right)^t > 1.25$. (c) Hence find the number of complete years after which $B$ is first worth more than $A$. (d) Give one reason why the model may not be realistic over a long period.

▶ Show solution

(a) $A$: £$1000$, growing by $4\%$ a year. $B$: £$800$, growing by $6\%$ a year.

(b) $800 \times 1.06^t > 1000 \times 1.04^t \Rightarrow \dfrac{1.06^t}{1.04^t} > \dfrac{1000}{800} \Rightarrow \left(\dfrac{1.06}{1.04}\right)^t > 1.25$, and $\dfrac{1.06}{1.04} = \dfrac{106}{104} = \dfrac{53}{52}$.

(c) $t\ln\dfrac{53}{52} > \ln 1.25$. As $\ln\frac{53}{52} > 0$: $t > \dfrac{\ln 1.25}{\ln (53/52)} = 11.7\ldots$

So after $12$ complete years. (At $t = 11$: $A \approx 1539$, $B \approx 1519$. At $t = 12$: $A \approx 1601$, $B \approx 1610$ ✓)

(d) Interest rates are unlikely to stay fixed for many years; money may be withdrawn or added.