Exponential Equations
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 maths page covers changing the base, fractions as negative powers, isolating the power, factorising sums of powers, and equations whose index is a quadratic or that have two unknowns.
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 are powers too: \(\dfrac{1}{9} = 3^{-2}\) and \(\dfrac{1}{8} = 2^{-3}\). So an answer can be negative, and changing the base can make it a fraction: \(4^x = 8\) gives \(x = \dfrac{3}{2}\).
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 |
|---|---|---|---|
| \(4, 8, 16, 32, 64\) | \(2^2, 2^3, 2^4, 2^5, 2^6\) | \(\dfrac{1}{8}\) | \(2^{-3}\) |
| \(9, 27, 81, 243\) | \(3^2, 3^3, 3^4, 3^5\) | \(\dfrac{1}{27}\) | \(3^{-3}\) |
| \(25, 125, 625\) | \(5^2, 5^3, 5^4\) | \(\dfrac{1}{25}\) | \(5^{-2}\) |
| \(1\) | \(a^0\) | \(0.01\) | \(10^{-2}\) |
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 and fraction 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 \(5^x = 5^3\).
- 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) \(3 \times 2^{x} = 96\)
(ii) \(7^{2x - 6} = 1\)
(iii) \(0.2 \times 5^{x} = 125\)
Equating indices needs a single power on its own, so divide out any number multiplying the power first.
(i) Divide both sides by \(3\):
| \(2^{x}\) | \(=\) | \(32 = 2^5\) |
| \(x\) | \(=\) | \(\mathbf{5}\) |
(ii) \(1\) is a power of every base: \(1 = 7^0\).
| \(7^{2x - 6}\) | \(=\) | \(7^{0}\) |
| \(2x - 6\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\mathbf{3}\) |
(iii) Divide by \(0.2\) (the same as multiplying by \(5\)):
| \(5^{x}\) | \(=\) | \(625 = 5^4\) |
| \(x\) | \(=\) | \(\mathbf{4}\) |
Check (i): \(3 \times 2^5 = 3 \times 32 = 96\) ✓
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 are part of the Year 10A course.
What if a number multiplies the power?
Divide both sides by that number first, so the power is on its own. For example, \(3 \times 2^x = 24\) gives \(2^x = 8 = 2^3\), so \(x = 3\). Do not multiply the \(3\) into the base.
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 \(x = 2\).