EL6 - Solving exponential equations

Find an unknown exponent by recognising a common base or using logarithms. You will solve equations with linear expressions in their exponents, including equations with the unknown on both sides, and show the analytical steps needed for detailed reasoning.

Equating exponents with a common base

How many doublings multiply a quantity by exactly $10$ ? Three give $8$ ; four give $16$ . The equation $2^x=10$ has an answer between $3$ and $4$ , and logarithms will locate it. First, take the shortcut when the numbers already share a base.

You already know

$\log_a b$ is the exponent in $a^x=b$ , with $a>0$ , $a\ne1$ , $b>0$ . The index law $(a^r)^s=a^{rs}$ combines powers; the logarithm law $\log_{10}(a^r)=r\log_{10}a$ brings an exponent down.

For a fixed positive base other than $1$ , the exponential function is strictly rising or strictly falling. Equal outputs must therefore have equal exponents. This is why rewriting $4$ and $8$ as powers of $2$ is useful.

Equal powers of a common base

a^u=a^v\quad\Longleftrightarrow\quad u=v

This requires $a>0$ and $a\ne1$ . If $a=1$ , every exponent gives $1$ , so equality tells you nothing about the exponents.
Not on the formulae sheet: learn it.

Check feasibility before solving. For $a>0$ , $a^x=b$ has no real solution when $b\le0$ . If $a=1$ , it has every real $x$ as a solution when $b=1$ , and no solution otherwise. If $a>0$ , $a\ne1$ , $b>0$ , it has exactly one solution.

Worked example 1: common-base powers

Question. Solve $4^{x+1}=8$ .

Plan. Write both sides as powers of $2$ , then equate the exponents.

\begin{aligned} (2^2)^{x+1}&=2^3 && \text{common base}\\ 2(x+1)&=3 && \text{equate exponents}\\ x+1&=\frac32 && \text{divide by }2\\ x&=\frac12 && \text{subtract }1 \end{aligned}

Why not equate $x+1$ to $3$ ? The original base $4$ is $2^2$ , so the left exponent doubles when the base becomes $2$ .

Answer. $x=\frac12$ .

Check. $4^{1.5}=(\sqrt4)^3=8$ .

Change only the right-hand side to its reciprocal. The corresponding exponent changes sign.

Worked example 2: a reciprocal on the right

Question. Solve $4^{x+1}=\frac18$ .

Plan. Use the same common base and express the reciprocal as a negative power.

\begin{aligned} 2^{2(x+1)}&=2^{-3} && \text{write both sides in base }2\\ 2(x+1)&=-3 && \text{equate exponents}\\ x+1&=-\frac32 && \text{divide by }2\\ x&=-\frac52 && \text{subtract }1 \end{aligned}

Why is the negative answer reasonable? It makes the complete exponent $x+1=-3/2$ , so the output is a positive reciprocal.

Answer. $x=-\frac52$ .

Check. $4^{-3/2}=1/4^{3/2}=1/8$ .

Common-base working keeps exact answers exact and often avoids a calculator entirely.

Watch out

Wrong: $(2^2)^{x+1}=2^{2x+1}$ . The multiplier $2$ acts on the whole exponent. Right: $2^{2(x+1)}=2^{2x+2}$ .

Keep a reciprocal target but change the common base.

Using logarithms when no common base is available

A logarithm moves an unknown exponent into an ordinary equation, even when the numbers have no convenient common base.

Take the logarithm of both complete sides. Base $10$ is convenient on a calculator, and using the same base throughout is essential.

Solving a positive-base exponential

a^x=b\quad\Longrightarrow\quad x\log_{10}a=\log_{10}b \quad\Longrightarrow\quad x=\frac{\log_{10}b}{\log_{10}a}

Here $a>0$ , $a\ne1$ , $b>0$ . The denominator is non-zero because $\log_{10}a=0$ only when $a=1$ .
Not on the formulae sheet: learn it.

For a more complicated exponent, bring that entire expression down before rearranging it.

Worked example 3: an unknown exponent

Question. Solve $2^x=7$ , giving your answer correct to $3$ significant figures.

Plan. Take base-ten logarithms and divide by $\log_{10}2$ .

\begin{aligned} \log_{10}(2^x)&=\log_{10}7 && \text{take logs}\\ x\log_{10}2&=\log_{10}7 && \text{power law}\\ x&=\frac{\log_{10}7}{\log_{10}2}=2.80735\ldots && \text{divide, then evaluate} \end{aligned}

Why divide by a logarithm? After applying the power law, $\log_{10}2$ is an ordinary non-zero coefficient of $x$ .

Answer. $x=2.81$ to $3$ significant figures.

Check. $2^2=4<7<8=2^3$ , and the unrounded result gives $2^x=7$ .

Change only the exponent from $x$ to $3x-1$ . Keep the whole expression together until after taking logarithms.

Worked example 4: a linear expression in the exponent

Question. Solve $2^{3x-1}=7$ , giving your answer correct to $3$ significant figures.

Plan. Take logs, find the complete exponent and then solve for $x$ .

\begin{aligned} (3x-1)\log_{10}2&=\log_{10}7 && \text{logarithms and power law}\\ 3x-1&=\frac{\log_{10}7}{\log_{10}2}=2.80735\ldots && \text{divide by the coefficient}\\ x&=\frac{1+\log_{10}7/\log_{10}2}{3} && \text{rearrange before rounding}\\ &=1.26911\ldots && \text{evaluate} \end{aligned}

Why retain the full calculator value? Rounding $3x-1$ early can change the final significant figures.

Answer. $x=1.27$ to $3$ significant figures.

Check. Using the unrounded value, $3x-1=2.80735\ldots$ , the exponent from the previous example.

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Numerical methods

EL6 - Solving exponential equations

Equating exponents with a common base

Using logarithms when no common base is available

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