Topics

Solving exponential equations

Covers AS91261

What makes an equation exponential

An exponential equation has the unknown in the power:

$3^x = 81 \qquad 5^{2x-1} = 100 \qquad 2000(1.06)^t = 5000$

There are two ways to solve one, and choosing correctly saves a great deal of work.

Method 1 — matching the bases

Use when both sides can be written as powers of the same number.
- Rewrite both sides with a common base.
- Equate the exponents.
- Solve the resulting equation.
Worked through — solve $8^{x} = 32$ :
- Both sides are powers of 2: $\left(2^3\right)^x = 2^5$
- So $2^{3x} = 2^5$
- Equate: $3x = 5$ , giving $x = \dfrac{5}{3}$
This method gives an exact answer with no calculator, so use it whenever the bases can be matched.
The bases to recognise: powers of 2 (2, 4, 8, 16, 32, 64), powers of 3 (3, 9, 27, 81), powers of 5 (5, 25, 125).
A fraction is a negative power: $\dfrac{1}{9} = 3^{-2}$ .

Method 2 — taking logarithms

Use when the bases cannot be matched, which is most real contexts.
- Isolate the exponential term so it is alone on one side.
- Take logs of both sides — any base works; $\log$ or $\ln$ on your calculator.
- Bring the exponent down using $\log\left(M^n\right) = n\log M$ .
- Solve the resulting linear equation.
Worked through — solve $5^x = 40$ :
- Take logs: $\log\left(5^x\right) = \log 40$
- Bring down: $x\log 5 = \log 40$
- Solve: $x = \dfrac{\log 40}{\log 5} = 2.2920...$
Note that $\dfrac{\log 40}{\log 5}$ is not $\log 8$ . Dividing two logs is not the log of the quotient.

Isolating first

The exponential must be alone before you take logs.
For $3 \times 2^x + 5 = 53$ $3 \times 2^{x} + 5 = 53$ :
- Subtract 5: $3 \times 2^x = 48$
- Divide by 3: $2^x = 16$
- Now solve: $x = 4$ (by matching bases, since $16 = 2^4$ )
Taking logs of $3 \times 2^x + 5$ directly achieves nothing, because there is no law for the log of a sum.

Exponents that are expressions

The exponent often contains more than just $x$ $x$ :
- $2^{3x-1} = 50$ gives $(3x-1)\log 2 = \log 50$ , so $3x - 1 = \dfrac{\log 50}{\log 2}$ , and then solve for $x$ .
Keep the whole exponent in brackets when you bring it down. Dropping the brackets is the standard error here.

Hidden quadratics

An equation with two exponential terms whose exponents differ by a factor of 2 is a quadratic in disguise:

$2^{2x} - 5\left(2^x\right) + 4 = 0$

Since $2^{2x} = \left(2^x\right)^2$ $2^{2 x} = (2^{x})^{2}$ , substitute $u = 2^x$ $u = 2^{x}$ :
- $u^2 - 5u + 4 = 0$ , so $(u-1)(u-4) = 0$ , giving $u = 1$ or $u = 4$
- Then $2^x = 1$ gives $x = 0$ , and $2^x = 4$ gives $x = 2$
Reject any negative or zero value of $u$ , because $2^x$ is always strictly positive.

Checking your answer

Substitute back. Exponential answers are often decimals, so a quick check catches sign errors and misplaced brackets.
Sense-check the size. If $3^x = 200$ , then $x$ must be between 4 and 5, since $3^4 = 81$ and $3^5 = 243$ . An answer of 66 is obviously wrong.

Tips and Tricks

Try matching bases first. If both sides are powers of the same number, you get an exact answer in two lines and no calculator is needed. Reaching straight for logs on $8^x = 32$ turns an exact answer into a decimal.

Isolate the exponential before taking logs. There is no law for $\log(A + B)$ , so any equation with something added to the exponential term must be rearranged first.

Bracket the whole exponent when you bring it down. $\log\left(2^{3x-1}\right) = (3x-1)\log 2$ , not $3x - 1\log 2$ .

$\dfrac{\log a}{\log b}$ is not $\log\left(\dfrac{a}{b}\right)$ . The quotient law needs the division inside a single log.

An exponential is never zero or negative. In a hidden-quadratic question, any value of $u = a^x$ that comes out $\leq 0$ must be rejected, with the reason stated.

Sense-check the size of your answer against nearby whole powers. It takes five seconds and catches calculator-entry errors.

Worked ExampleTwo methods, and knowing which to use

Solve (a) $9^{x+1} = 27^{x-1}$ exactly, and (b) $700(1.045)^t = 1200$ , giving $t$ to 3 significant figures.

Part (a), Step 1 — Look for a common base

Both 9 and 27 are powers of 3:

$9 = 3^2 \qquad 27 = 3^3$

Since the bases can be matched, this method gives an exact answer with no calculator.

Part (a), Step 2 — Rewrite both sides

$\left(3^2\right)^{x+1} = \left(3^3\right)^{x-1}$

Apply the power-of-a-power law, multiplying the indices. Keep the brackets so the multiplication distributes:

$3^{2(x+1)} = 3^{3(x-1)}$

Part (a), Step 3 — Equate the exponents

If two powers of the same base are equal, their exponents must be equal:

$2(x+1) = 3(x-1)$

Part (a), Step 4 — Solve

$2x + 2 = 3x - 3$

$2 + 3 = 3x - 2x$

$x = 5$

Part (a), Step 5 — Check

$9^{6} = 531441 \qquad 27^{4} = 531441 \;\checkmark$

Part (b), Step 1 — Recognise that bases cannot be matched

1200, 700 and 1.045 are not powers of a common number, so logarithms are needed.

Part (b), Step 2 — Isolate the exponential

Divide both sides by 700 so the exponential stands alone:

$(1.045)^t = \frac{1200}{700} = 1.714285...$

Part (b), Step 3 — Take logs of both sides

$\log\left((1.045)^t\right) = \log(1.714285...)$

Part (b), Step 4 — Bring the exponent down

Using $\log\left(M^n\right) = n\log M$ :

$t\log(1.045) = \log(1.714285...)$

Part (b), Step 5 — Solve for $t$

$t = \frac{\log(1.714285...)}{\log(1.045)} = \frac{0.2341...}{0.0191...}$

$t = 12.2725...$

$t \approx 12.3$

Part (b), Step 6 — Check

$700(1.045)^{12.2725} = 700 \times 1.7143 = 1200 \;\checkmark$

Sense-check: at about 4.5% growth, a value should take roughly 16 years to double. Growing by a factor of 1.71 — less than doubling — in 12.3 years is consistent ✓

Solving exponential equations

Covers AS91261

What makes an equation exponential

An exponential equation has the unknown in the power:

$3^x = 81 \qquad 5^{2x-1} = 100 \qquad 2000(1.06)^t = 5000$

There are two ways to solve one, and choosing correctly saves a great deal of work.

Method 1 — matching the bases

Use when both sides can be written as powers of the same number.
- Rewrite both sides with a common base.
- Equate the exponents.
- Solve the resulting equation.
Worked through — solve $8^{x} = 32$ :
- Both sides are powers of 2: $\left(2^3\right)^x = 2^5$
- So $2^{3x} = 2^5$
- Equate: $3x = 5$ , giving $x = \dfrac{5}{3}$
This method gives an exact answer with no calculator, so use it whenever the bases can be matched.
The bases to recognise: powers of 2 (2, 4, 8, 16, 32, 64), powers of 3 (3, 9, 27, 81), powers of 5 (5, 25, 125).
A fraction is a negative power: $\dfrac{1}{9} = 3^{-2}$ .

Method 2 — taking logarithms

Use when the bases cannot be matched, which is most real contexts.
- Isolate the exponential term so it is alone on one side.
- Take logs of both sides — any base works; $\log$ or $\ln$ on your calculator.
- Bring the exponent down using $\log\left(M^n\right) = n\log M$ .
- Solve the resulting linear equation.
Worked through — solve $5^x = 40$ :
- Take logs: $\log\left(5^x\right) = \log 40$
- Bring down: $x\log 5 = \log 40$
- Solve: $x = \dfrac{\log 40}{\log 5} = 2.2920...$
Note that $\dfrac{\log 40}{\log 5}$ is not $\log 8$ . Dividing two logs is not the log of the quotient.

Isolating first

The exponential must be alone before you take logs.
For $3 \times 2^x + 5 = 53$ $3 \times 2^{x} + 5 = 53$ :
- Subtract 5: $3 \times 2^x = 48$
- Divide by 3: $2^x = 16$
- Now solve: $x = 4$ (by matching bases, since $16 = 2^4$ )
Taking logs of $3 \times 2^x + 5$ directly achieves nothing, because there is no law for the log of a sum.

Exponents that are expressions

The exponent often contains more than just $x$ $x$ :
- $2^{3x-1} = 50$ gives $(3x-1)\log 2 = \log 50$ , so $3x - 1 = \dfrac{\log 50}{\log 2}$ , and then solve for $x$ .
Keep the whole exponent in brackets when you bring it down. Dropping the brackets is the standard error here.

Hidden quadratics

An equation with two exponential terms whose exponents differ by a factor of 2 is a quadratic in disguise:

$2^{2x} - 5\left(2^x\right) + 4 = 0$

Since $2^{2x} = \left(2^x\right)^2$ $2^{2 x} = (2^{x})^{2}$ , substitute $u = 2^x$ $u = 2^{x}$ :
- $u^2 - 5u + 4 = 0$ , so $(u-1)(u-4) = 0$ , giving $u = 1$ or $u = 4$
- Then $2^x = 1$ gives $x = 0$ , and $2^x = 4$ gives $x = 2$
Reject any negative or zero value of $u$ , because $2^x$ is always strictly positive.

Checking your answer

Substitute back. Exponential answers are often decimals, so a quick check catches sign errors and misplaced brackets.
Sense-check the size. If $3^x = 200$ , then $x$ must be between 4 and 5, since $3^4 = 81$ and $3^5 = 243$ . An answer of 66 is obviously wrong.

Tips and Tricks

Isolate the exponential before taking logs. There is no law for $\log(A + B)$ , so any equation with something added to the exponential term must be rearranged first.

Bracket the whole exponent when you bring it down. $\log\left(2^{3x-1}\right) = (3x-1)\log 2$ , not $3x - 1\log 2$ .

$\dfrac{\log a}{\log b}$ is not $\log\left(\dfrac{a}{b}\right)$ . The quotient law needs the division inside a single log.

An exponential is never zero or negative. In a hidden-quadratic question, any value of $u = a^x$ that comes out $\leq 0$ must be rejected, with the reason stated.

Sense-check the size of your answer against nearby whole powers. It takes five seconds and catches calculator-entry errors.

Worked ExampleTwo methods, and knowing which to use

Solve (a) $9^{x+1} = 27^{x-1}$ exactly, and (b) $700(1.045)^t = 1200$ , giving $t$ to 3 significant figures.

Part (a), Step 1 — Look for a common base

Both 9 and 27 are powers of 3:

$9 = 3^2 \qquad 27 = 3^3$

Since the bases can be matched, this method gives an exact answer with no calculator.

Part (a), Step 2 — Rewrite both sides

$\left(3^2\right)^{x+1} = \left(3^3\right)^{x-1}$

Apply the power-of-a-power law, multiplying the indices. Keep the brackets so the multiplication distributes:

$3^{2(x+1)} = 3^{3(x-1)}$

Part (a), Step 3 — Equate the exponents

If two powers of the same base are equal, their exponents must be equal:

$2(x+1) = 3(x-1)$

Part (a), Step 4 — Solve

$2x + 2 = 3x - 3$

$2 + 3 = 3x - 2x$

$x = 5$

Part (a), Step 5 — Check

$9^{6} = 531441 \qquad 27^{4} = 531441 \;\checkmark$

Part (b), Step 1 — Recognise that bases cannot be matched

1200, 700 and 1.045 are not powers of a common number, so logarithms are needed.

Part (b), Step 2 — Isolate the exponential

Divide both sides by 700 so the exponential stands alone:

$(1.045)^t = \frac{1200}{700} = 1.714285...$

Part (b), Step 3 — Take logs of both sides

$\log\left((1.045)^t\right) = \log(1.714285...)$

Part (b), Step 4 — Bring the exponent down

Using $\log\left(M^n\right) = n\log M$ :

$t\log(1.045) = \log(1.714285...)$

Part (b), Step 5 — Solve for $t$

$t = \frac{\log(1.714285...)}{\log(1.045)} = \frac{0.2341...}{0.0191...}$

$t = 12.2725...$

$t \approx 12.3$

Part (b), Step 6 — Check

$700(1.045)^{12.2725} = 700 \times 1.7143 = 1200 \;\checkmark$

Sense-check: at about 4.5% growth, a value should take roughly 16 years to double. Growing by a factor of 1.71 — less than doubling — in 12.3 years is consistent ✓

Solving exponential equations

What makes an equation exponential

Method 1 — matching the bases

Method 2 — taking logarithms

Isolating first

Exponents that are expressions

Hidden quadratics

Checking your answer

Part (a), Step 1 — Look for a common base

Part (a), Step 2 — Rewrite both sides

Part (a), Step 3 — Equate the exponents

Part (a), Step 4 — Solve

Part (a), Step 5 — Check

Part (b), Step 1 — Recognise that bases cannot be matched

Part (b), Step 2 — Isolate the exponential

Part (b), Step 3 — Take logs of both sides

Part (b), Step 4 — Bring the exponent down

Part (b), Step 5 — Solve for ttt

Part (b), Step 6 — Check

Solving exponential equations

What makes an equation exponential

Method 1 — matching the bases

Method 2 — taking logarithms

Isolating first

Exponents that are expressions

Hidden quadratics

Checking your answer

Part (a), Step 1 — Look for a common base

Part (a), Step 2 — Rewrite both sides

Part (a), Step 3 — Equate the exponents

Part (a), Step 4 — Solve

Part (a), Step 5 — Check

Part (b), Step 1 — Recognise that bases cannot be matched

Part (b), Step 2 — Isolate the exponential

Part (b), Step 3 — Take logs of both sides

Part (b), Step 4 — Bring the exponent down

Part (b), Step 5 — Solve for ttt

Part (b), Step 6 — Check

Part (b), Step 5 — Solve for $t$

Part (b), Step 5 — Solve for $t$