Solving Exponential Equations – OCR FSMQ Additional Maths

1 The Basic Method

EL6: solve equations of the form $a^x = b$ for $a > 0$. The obstacle is that $x$ is in the index, where none of your usual moves can reach it. Logarithms bring it down.

The three-step method

1. Take logarithms of both sides
2. Use the power law to bring the index down: $\log\left(a^x\right) = x\log a$
3. Divide to isolate $x$

The result

$a^x = b \;\Longrightarrow\; x = \dfrac{\log b}{\log a}$

Any base works, provided you use the same one on both sides. Base $10$ ($\log$) and base $\mathrm{e}$ ($\ln$) both give identical final answers — use whichever button you prefer.

Worked Example 1 — The standard case

Solve $5^x = 300$, to $3$ significant figures.

①$\log\left(5^x\right) = \log 300$

②$x\log 5 = \log 300$

③$x = \dfrac{\log 300}{\log 5} = \dfrac{2.477121}{0.698970}$

④$x = 3.54$ (3 s.f.)

⑤Check: $5^{3.54} = 298$ — close to $300$, the small difference being the rounding ✓

⑥Sanity check without a calculator: $5^3 = 125$ and $5^4 = 625$, so $x$ must be between $3$ and $4$ ✓

The bracketing check is worth doing every time. It catches the commonest error — accidentally computing $\tfrac{\log 5}{\log 300}$ instead.

Worked Example 2 — The specification's own example

Use logarithms to solve $3^{4x+1} = 100$, giving your answer correct to $3$ significant figures.

①$\log\left(3^{4x+1}\right) = \log 100$

②$(4x+1)\log 3 = \log 100$

③$4x + 1 = \dfrac{\log 100}{\log 3} = \dfrac{2}{0.477121} = 4.19180$

④$4x = 3.19180$

⑤$x = 0.797949\ldots = 0.798$ (3 s.f.)

This is the specification's worked exemplar for "detailed reasoning". It states that the learner must show the logarithm line, the bracketed line, and an intermediate evaluation such as $4x + 1 = 4.19\ldots$. Using a calculator's solver to skip any of these "would not result in a complete analytical method".

Worked Example 3 — A decay equation

Solve $0.8^x = 0.2$.

①$x\log 0.8 = \log 0.2$

②$x = \dfrac{\log 0.2}{\log 0.8} = \dfrac{-0.698970}{-0.096910}$

③$x = 7.21$ (3 s.f.)

④Both logarithms are negative (because both numbers are less than $1$), and the negatives cancel to give a positive $x$.

A positive answer is what you should expect: $0.8^x$ shrinks as $x$ grows, so it takes a positive number of steps to fall to $0.2$.

2 Harder Forms

Worked Example 4 — A coefficient in front

Solve $7 \times 2^x = 500$.

①Divide by the coefficient first: $2^x = \dfrac{500}{7} = 71.42857$

②$x\log 2 = \log 71.42857$

③$x = \dfrac{1.853872}{0.301030} = 6.16$ (3 s.f.)

Do not take logarithms before dividing. $\log(7 \times 2^x)$ is $\log 7 + x\log 2$, which works but is more error-prone. Isolate the power first.

Worked Example 5 — Unknown in both indices

Solve $3^{x+1} = 5^x$.

①Take logarithms: $(x+1)\log 3 = x\log 5$

②Expand: $x\log 3 + \log 3 = x\log 5$

③Collect the $x$ terms: $\log 3 = x\log 5 - x\log 3 = x\left(\log 5 - \log 3\right)$

④$x = \dfrac{\log 3}{\log 5 - \log 3} = \dfrac{0.477121}{0.698970 - 0.477121} = \dfrac{0.477121}{0.221849}$

⑤$x = 2.15$ (3 s.f.)

⑥Check: $3^{3.15} = 31.8$ and $5^{2.15} = 31.8$ ✓

Treat $\log 3$ and $\log 5$ as ordinary numbers and the algebra is just "collect like terms". You can also write the denominator as $\log\tfrac53$, using the quotient law.

Worked Example 6 — A hidden quadratic

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

①Notice $2^{2x} = \left(2^x\right)^2$. Substitute $y = 2^x$:

②$y^2 - 6y + 8 = 0$

③$(y-2)(y-4) = 0$, so $y = 2$ or $y = 4$.

④$2^x = 2$ gives $x = 1$.

⑤$2^x = 4$ gives $x = 2$.

⑥$x = 1$ or $x = 2$.

Spot the pattern: an equation containing both $a^{2x}$ and $a^x$ is a quadratic in disguise. Substituting $y = a^x$ makes it obvious.

Worked Example 7 — A hidden quadratic with a rejection

Solve $3^{2x} - 3^x - 6 = 0$.

①Let $y = 3^x$: $y^2 - y - 6 = 0$

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

③Reject $y = -2$: $3^x$ is always positive, so it can never equal $-2$.

④$3^x = 3$ gives $x = 1$.

⑤Check: $3^2 - 3 - 6 = 9 - 9 = 0$ ✓

Always check whether a value of $y$ is attainable. Since $y = a^x > 0$ for any real $x$, a negative or zero value of $y$ must be rejected — and you should say why.

3 Exponential Inequalities

The sign rule

If $a > 1$, the function $a^x$ is increasing, so the inequality is preserved
If $0 < a < 1$, it is decreasing, so the inequality reverses

Worked Example 8 — An increasing base

Solve $2^x > 1000$.

①$x\log 2 > \log 1000 = 3$

②$\log 2 > 0$, so dividing does not flip the sign:

③$x > \dfrac{3}{0.301030} = 9.97$

④Check: $2^9 = 512 < 1000$ and $2^{10} = 1024 > 1000$ ✓ — so for whole numbers, $x \geqslant 10$.

Worked Example 9 — A decaying base

Solve $0.9^x < 0.1$.

①$x\log 0.9 < \log 0.1$

②$\log 0.9 = -0.045757$ is negative, so dividing by it reverses the inequality:

③$x > \dfrac{-1}{-0.045757} = 21.85$

④Check: $0.9^{22} = 0.0985 < 0.1$ ✓ and $0.9^{21} = 0.1094 > 0.1$ ✓

The direction should make sense: a decaying quantity gets smaller as $x$ increases, so "below $0.1$" must mean "after a certain point", i.e. $x$ greater than something.

4 Quick Reference

The core result

$a^x = b \Rightarrow x = \dfrac{\log b}{\log a}$.

Any base

$\log$ or $\ln$ — same answer.

Isolate first

Divide out any coefficient before taking logs.

Show the steps

Log line, bracket line, intermediate value.

Index with $+1$

Solve for the whole index, then for $x$.

Both sides unknown

Expand and collect the $x$ terms.

$a^{2x}$ present

Substitute $y = a^x$ for a quadratic.

Reject $y \leqslant 0$

$a^x$ is always positive.

Inequality, $a>1$

Direction preserved.

Inequality, $a<1$

Direction reversed.

5 Practice Questions

Question 1

Solve $2^x = 50$, to $3$ s.f.

▶ Show solution

$x = \dfrac{\log 50}{\log 2} = \dfrac{1.698970}{0.301030} = 5.64$ (3 s.f.)

Question 2

Solve $3^x = 200$, to $3$ s.f.

▶ Show solution

$x = \dfrac{\log 200}{\log 3} = \dfrac{2.301030}{0.477121} = 4.82$ (3 s.f.)

Question 3

Solve $0.5^x = 0.02$, to $3$ s.f.

▶ Show solution

$x = \dfrac{\log 0.02}{\log 0.5} = \dfrac{-1.698970}{-0.301030} = 5.64$ (3 s.f.)

Question 4

Solve $4 \times 3^x = 100$, to $3$ s.f.

▶ Show solution

$3^x = 25$

$x = \dfrac{\log 25}{\log 3} = \dfrac{1.397940}{0.477121} = 2.93$ (3 s.f.)

Question 5

Solve $2^{3x} = 40$, to $3$ s.f.

▶ Show solution

$3x\log 2 = \log 40$

$3x = \dfrac{1.602060}{0.301030} = 5.32193$

$x = 1.77$ (3 s.f.)

Question 6

Solve $5^{x-2} = 30$, to $3$ s.f.

▶ Show solution

$(x-2)\log 5 = \log 30$

$x - 2 = \dfrac{1.477121}{0.698970} = 2.11329$

$x = 4.11$ (3 s.f.)

Question 7

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

▶ Show solution

Let $y = 3^x$: $y^2 - 4y + 3 = 0$

$(y-1)(y-3) = 0$, so $y = 1$ or $y = 3$.

$3^x = 1$ gives $x = 0$; $3^x = 3$ gives $x = 1$.

$x = 0$ or $x = 1$.

Question 8

Solve $2^x > 500$.

▶ Show solution

$x > \dfrac{\log 500}{\log 2} = \dfrac{2.698970}{0.301030} = 8.966$

So $x > 8.97$ (3 s.f.), and for whole numbers $x \geqslant 9$.

Check: $2^9 = 512 > 500$ ✓ and $2^8 = 256 < 500$ ✓

Question 9

Solve $2^{x+1} = 3^x$, to $3$ s.f.

▶ Show solution

$(x+1)\log 2 = x\log 3$

$x\log 2 + \log 2 = x\log 3$

$\log 2 = x\left(\log 3 - \log 2\right)$

$x = \dfrac{0.301030}{0.477121 - 0.301030} = \dfrac{0.301030}{0.176091} = 1.71$ (3 s.f.)

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

Question 10

A savings account pays $4\%$ interest per year. £$2500$ is invested.

(a) Write a formula for the value $V$ after $t$ years. (b) Find the value after $10$ years. (c) Find how long it takes to reach £$4000$, to the nearest month. (d) Find how long it takes to double, and show that this does not depend on the amount invested.

▶ Show solution

(a) $V = 2500 \times 1.04^t$

(b) $V = 2500 \times 1.04^{10} = 2500 \times 1.480244 = £3700.61$ (to the nearest penny)

(c) $2500 \times 1.04^t = 4000$

$1.04^t = 1.6$

$t\log 1.04 = \log 1.6$

$t = \dfrac{0.204120}{0.017033} = 11.984$ years

The fractional part is $0.984 \times 12 = 11.8$ months, which rounds to $12$ months — so the value reaches £$4000$ after $\mathbf{12}$ years (to the nearest month).

Check: $2500 \times 1.04^{12} = £4002.58$ ✓ (just over £4000), while $1.04^{11}$ gives £3849 ✓

(d) Doubling means $V = 5000$:

$2500 \times 1.04^t = 5000 \;\Rightarrow\; 1.04^t = 2$

$t = \dfrac{\log 2}{\log 1.04} = \dfrac{0.301030}{0.017033} = 17.67$ years.

Why the initial amount is irrelevant: starting with any amount $P$, doubling requires $P \times 1.04^t = 2P$. The $P$ cancels, leaving $1.04^t = 2$ regardless of its value.

So £$100$ and £$1$ million both take $17.67$ years to double at $4\%$. This constant doubling time is the defining property of exponential growth — and the reason the "rule of 72" works: $\tfrac{72}{4} = 18$ years, a good quick approximation to $17.67$.

Solving Exponential Equations (EL6) · OCR FSMQ Additional Maths · Created with MathJax