Indices and exponential equations · Indices, logarithms & surds

In General Maths you solved equations like $2^x = 32$ by writing both sides as powers of the same number and comparing the indices (see indices). Further Maths questions use exactly the same idea, but hide it better: the bases are mixed, the powers are added instead of multiplied, or the equation turns out to be a quadratic. This lesson takes each of those in turn.

Same base, then compare

The one rule that solves every exponential equation:

a^p = a^q \quad\Longrightarrow\quad p = q \qquad (a > 0,\ a \ne 1)

So first write every number as a power of the same prime. Keep brackets round the index when you use $(a^m)^n = a^{mn}$ , so that every term of it gets multiplied:

A power of a powerMultiply the indices: (2³)ˣ⁻¹ = 2³⁽ˣ⁻¹⁾ = 2³ˣ⁻³

For example, to solve $8^{x - 1} = 4^x \times 2^3$ :

Write the left side in base 2: ${8^{x - 1} = (2^3)^{x - 1} = 2^{3x - 3}}$ .
Write the right side in base 2: ${4^x \times 2^3 = 2^{2x} \times 2^3 = 2^{2x + 3}}$ .
Compare the indices: ${3x - 3 = 2x + 3}$ .
Take $2x$ from both sides and add 3: ${x = 6}$ .

When the numbers have two different primes, such as $10 = 2 \times 5$ , collect the powers of each prime. To solve $10^x \times 2^{x - 2} \times 5^{x - 2} = 1$ :

The 2s and 5s with the same index pair up: ${2^{x - 2} \times 5^{x - 2} = 10^{x - 2}}$ .
So the left side is ${10^x \times 10^{x - 2} = 10^{2x - 2}}$ .
Write 1 as a power of 10: ${1 = 10^0}$ .
Compare the indices: ${2x - 2 = 0}$ , so ${x = 1}$ .

More: same base, then compare

Fractional powers with letters

Roots are powers too: $\sqrt[n]{x^m} = x^{\frac mn}$ and $\frac{1}{x^n} = x^{-n}$ . Change every root and every “one over” into a power of $x$ first. Then the laws do the rest.

For example, to simplify $\left(\dfrac{x^{\frac12}}{\sqrt[3]{x}}\right)^{6}$ :

Write the root as a power: ${\sqrt[3]{x} = x^{\frac13}}$ .
Divide by subtracting the indices: ${x^{\frac12} \div x^{\frac13} = x^{\frac12 - \frac13} = x^{\frac16}}$ .
Raise to the power 6 by multiplying: ${\left(x^{\frac16}\right)^6 = x^1 = x}$ .

More: fractional powers with letters

NECO 2023 · Paper 2 · Q2Find the equation of the circle whose centre is $(4, -5)$ and which passes through $(-3, 2)$ .

Powers that are added: take out the smallest

The laws of indices are for multiplying and dividing powers. When powers are added or taken away, as in $2^{x + 3} - 2^{x + 1}$ , no law joins them. Take out the smallest power as a common factor instead:

Take out the smallest powerEvery other term is it times a number

Worked example · WAEC 2019

WAEC 2019 · Paper 2 · Q1

Simplify: $\dfrac{625^{\left(\frac{3x}{4} - 1\right)} + 125^{(x - 1)}}{5^{(3x - 2)}}$ .

Everything as a power of 5
$625 = 5^4$ and $125 = 5^3$ , so
$\begin{aligned} 625^{\frac{3x}{4} - 1} &= 5^{4\left(\frac{3x}{4} - 1\right)} = 5^{3x - 4} \\ 125^{x - 1} &= 5^{3(x - 1)} = 5^{3x - 3} \end{aligned}$
Think first. Write 625 and 125 as powers of 5.
Take out the smallest power on top
- $5^{3x - 4}$ is the smaller power.
- The other term is it times 5: ${5^{3x - 3} = 5^{3x - 4} \times 5}$ .
- Take $5^{3x - 4}$ out: the top is ${5^{3x - 4}(1 + 5) = 6 \times 5^{3x - 4}}$ .
Think first. The top is 5³ˣ⁻⁴ + 5³ˣ⁻³. Which is smaller?
Divide by the bottom
- Divide the powers by subtracting the indices: ${(3x - 4) - (3x - 2) = -2}$ .
- So the fraction is $6 \times 5^{-2}$ .
- ${5^{-2} = \frac{1}{25}}$ , so the answer is $\dfrac{6}{25}$ .
Think first. How does 5³ˣ⁻⁴ compare with 5³ˣ⁻²?

More: taking out the smallest power

WAEC 2022 · Paper 1 · Q4Simplify: $\dfrac{9 \times 3^{n+1} - 3^{n+2}}{3^{n+1} - 3^{n}}$ .

Equations that hide a quadratic

When an equation has both $a^{2x}$ and $a^x$ in it, it is a quadratic in disguise, because $a^{2x} = (a^x)^2$ . Write every power in terms of $a^x$ , put $u = a^x$ , and solve the quadratic in $u$ . Then go back to $x$ .

A quadratic in disguiseStep through

3²ˣ − 10(3ˣ) + 9 = 0

3²ˣ − 10(3ˣ) + 9 = 0step 0 of 4

An equation with 3²ˣ and 3ˣ in it. Neither law of indices joins them, because they are added, not multiplied.

Try the third equation: one value of $u$ is negative, and the graph shows why it gives no $x$ . A power of a positive number is always positive, so $5^x = -1$ has no solution.

Worked example · WAEC 2011

WAEC 2011 · Paper 2 · Q4

Solve $2^{(2y + 2)} - 9(2^y) = -2$ .

Split the index
- Split the index: ${2^{2y + 2} = 2^2 \times 2^{2y}}$ .
- Write $2^{2y}$ as a square: ${2^2 \times 2^{2y} = 4(2^y)^2}$ .
- So the equation is ${4(2^y)^2 - 9(2^y) = -2}$ .
Think first. What does the +2 in 2²ʸ⁺² turn into?
Let u = 2ʸ
- With ${u = 2^y}$ : ${4u^2 - 9u = -2}$ .
- Add 2 to both sides: ${4u^2 - 9u + 2 = 0}$ .
Think first. Write it as a quadratic in u, with 0 on one side.
Solve for u
- Factorise: ${(4u - 1)(u - 2) = 0}$ .
- So ${u = \frac14}$ or ${u = 2}$ .
- Both are positive, so both will give an answer.
Back to y
- ${2^y = \frac14 = 2^{-2}}$ , so ${y = -2}$ .
- ${2^y = 2 = 2^1}$ , so ${y = 1}$ .

If the equation has $a^{3x}$ as well, the same substitution gives a cubic in $u$ . Find one root by trying the factors of the constant term, then divide it out (see polynomials).

More: hidden quadratics and cubics

Your turn

WAEC 2022 · Paper 2 · Q2

(a)
Solve $2^{(2y + 1)} - 5(2^y) + 2 = 0$ .
Separate values with commas, e.g. 3, −2

Worked solution (try it first)

Split the index: $2^{2y + 1} = 2 \times (2^y)^2$ .
Let $x = 2^y$ : $2x^2 - 5x + 2 = 0$ .
Factorise: $(2x - 1)(x - 2) = 0$ , so $x = \frac12$ or $x = 2$ .
Back to $y$ : $2^y = \frac12 = 2^{-1}$ gives $y = -1$ , and $2^y = 2$ gives $y = 1$ .

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