Exponential Equations
An exponential equation has the unknown in the index, as in \(125^{x-1}=25\). It can often be solved by writing both sides as powers of the same base and then setting the indices equal. In NSW Year 10 Advanced maths the equations include fractions, surds such as \(7\sqrt{7}\), and quadratic indices (NESA outcome MA5‑IND‑P‑02).
You will solve equations by rewriting numbers as powers of 2, 3 or 5, including negative and fractional indices, and solve harder ones with a sum of powers, two different bases with the same index, and a pair of simultaneous exponential equations. You will show that an equation has no solution. Problems include folding paper, a rare coin doubling in value every 5 years and algae spreading over two dams. It is part of Year 10 Advanced Indices.
Every question with a fully worked solution.
- Exponential Equations - Video - Solving Exponential equations Watch
Theory
To solve an exponential equation such as \(9^{x+1} = 27^{x-1}\), write both sides as powers of the same base, then equate the indices. This Year 10 Advanced maths page covers fractions and surds as powers, factorising sums of powers, quadratic and simultaneous index equations, and growth models.
An exponential equation has the unknown in the index, such as \(2^x = 32\) or \(4^{2x-1} = 8\).
If two powers of the same base are equal, their indices are equal: \(a^p = a^q\) means \(p = q\) (for a base \(a \gt 0\), \(a \ne 1\)). This is true because the graph of \(y = a^x\) never takes the same height twice.
Fractions and roots are powers too: \(\dfrac{1}{9} = 3^{-2}\) and \(\sqrt{2} = 2^{\genfrac{}{}{}{1}{1}{2}}\). So an answer can be negative or a fraction.
The method needs a single power on each side. A sum such as \(2^{x+1} + 2^x\) must be factorised first.
Powers to recognise:
| Write | As | Write | As |
|---|---|---|---|
| \(4, 8, 16, 32, 64\) | \(2^2, 2^3, 2^4, 2^5, 2^6\) | \(\dfrac{1}{8}\) | \(2^{-3}\) |
| \(9, 27, 81, 243\) | \(3^2, 3^3, 3^4, 3^5\) | \(\sqrt{3}\) | \(3^{\genfrac{}{}{}{1}{1}{2}}\) |
| \(25, 125, 625\) | \(5^2, 5^3, 5^4\) | \(\sqrt[3]{25}\) | \(5^{\genfrac{}{}{}{1}{2}{3}}\) |
| \(1\) | \(a^0\) | \(0.01\) | \(10^{-2}\) |
How to solve an exponential equation
- Isolate the power: divide by any number multiplying it (in \(0.5 \times 2^n = 64\), get \(2^n = 128\) first).
- Choose the base: the smallest prime that every number is a power of.
- Rewrite every number, fraction and surd as a power of that base, and simplify each side to a single power.
- Equate the indices and solve. Check by substituting.
Harder types
- Sum of powers: factorise, such as \(2^{x+2} + 2^x = 2^x(4 + 1)\).
- Different bases with the same index: divide: \(10^x = 125 \times 2^x\) gives \(\left(\dfrac{10}{2}\right)^x = 125\), so \(5^x = 5^3\).
- Base \(\sqrt{a}\): \(\left(\sqrt{3}\right)^{x} = 3^{\genfrac{}{}{}{1}{x}{2}}\), so the index becomes a fraction.
- Quadratic index: equate, rearrange to zero and factorise; check both roots.
- Two unknowns: two exponential equations give two linear equations to solve simultaneously.
- No solution: if equating gives a false statement like \(3 = 6\), the equation has no solution.
- Growth models like \(N = 50 \times 2^{\genfrac{}{}{}{1}{t}{4}}\): substitute the target, divide, then equate. Two models are equal when their formulas are set equal.
(i) \(9^{x+1} = 27^{x-1}\)
(ii) \(\left(\dfrac{1}{4}\right)^{x} = 32\).
(i) \(9\) and \(27\) are both powers of \(3\):
| \((3^2)^{x+1}\) | \(=\) | \((3^3)^{x-1}\) |
| \(2x + 2\) | \(=\) | \(3x - 3\) |
| \(x\) | \(=\) | \(\boldsymbol{5}\) |
(ii) A fraction is a negative power: \(\dfrac{1}{4} = 2^{-2}\) and \(32 = 2^5\).
| \((2^{-2})^{x}\) | \(=\) | \(2^5\) |
| \(-2x\) | \(=\) | \(5\) |
| \(x\) | \(=\) | \(\boldsymbol{-\dfrac{5}{2}}\) |
Check (i): \(9^6 = 3^{12}\) and \(27^4 = 3^{12}\) ✓
(i) \(25^{x} = 5\sqrt{5}\)
(ii) \(8^{x} = \dfrac{1}{4\sqrt{2}}\).
Write each surd as a fractional power: \(\sqrt{5} = 5^{\genfrac{}{}{}{1}{1}{2}}\).
(i)
| \(5^{2x}\) | \(=\) | \(5^1 \times 5^{\genfrac{}{}{}{1}{1}{2}} = 5^{\genfrac{}{}{}{1}{3}{2}}\) |
| \(2x\) | \(=\) | \(\dfrac{3}{2}\) |
| \(x\) | \(=\) | \(\boldsymbol{\dfrac{3}{4}}\) |
(ii)
| \(2^{3x}\) | \(=\) | \(\dfrac{1}{2^2 \times 2^{\genfrac{}{}{}{1}{1}{2}}} = 2^{-\genfrac{}{}{}{1}{5}{2}}\) |
| \(3x\) | \(=\) | \(-\dfrac{5}{2}\) |
| \(x\) | \(=\) | \(\boldsymbol{-\dfrac{5}{6}}\) |
A difference of two powers is not one power, so you cannot equate indices yet. Take out the common factor \(3^x\):
| \(3^{x+2} - 3^{x}\) | \(=\) | \(3^x \times 3^2 - 3^x\) |
| \(\) | \(=\) | \(3^x(9 - 1) = 8 \times 3^x\) |
| \(8 \times 3^x\) | \(=\) | \(72\) |
| \(3^x\) | \(=\) | \(9 = 3^2\) |
| \(x\) | \(=\) | \(\boldsymbol{2}\) |
Check: \(3^4 - 3^2 = 81 - 9 = 72\) ✓
| \(5^{x^2 + 2x}\) | \(=\) | \(5^3\) |
| \(x^2 + 2x - 3\) | \(=\) | \(0\) |
| \((x + 3)(x - 1)\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\boldsymbol{-3} \text{ or } x = \boldsymbol{1}\) |
Both answers work: \((-3)^2 + 2(-3) = 3\) and \(1^2 + 2(1) = 3\). Unlike a length, \(x\) here may be negative, so keep both.
Common pitfalls
Frequently asked questions
How do you solve an exponential equation?
Write both sides as powers of the same base, simplify each side to a single power, then set the indices equal and solve the equation you get.
What if the numbers are not powers of the same base?
Look for a prime that both numbers are built from, such as 2 for 4 and 8, or 3 for 9 and 27. If there is no common base, you need logarithms, which come later in senior maths.
How do you deal with a square root in an exponential equation?
Write it as a power with index \(\genfrac{}{}{}{1}{1}{2}\). For example, \(2\sqrt{2} = 2^{\genfrac{}{}{}{1}{3}{2}}\).
Can the answer to an exponential equation be negative?
Yes. The power itself is always positive, but the index can be negative: \(2^x = \dfrac{1}{8}\) gives \(x = -3\).
Why can't you equate indices when the powers are added?
Because a sum is not a single power. Factorise first: \(3^{x+1} + 3^x = 4 \times 3^x\). If that equals \(36\), then \(3^x = 9\) and \(x = 2\).