Pure 6.5 - Solving exponential equations | Edexcel A-Level Mathematics

Find an unknown exponent using logarithms, and choose a common-base shortcut when one is available. You will solve equations with linear expressions in the exponent, keep exact answers exact and check that a real solution is possible.

What the equation means

What power of $2$ gives $5$ ? Since $2^2=4$ and $2^3=8$ , the answer sits between $2$ and $3$ . A logarithm gives that awkward-looking exponent an exact name: $\log_2 5$ .

You already know

Index laws give $(a^u)^v=a^{uv}$ . Logarithms undo powers, and the power law gives $\ln(a^u)=u\ln a$ for $a>0$ .

A horizontal line at height $5$ meets the increasing curve once. The intersection tells us both that the solution exists and that there is only one.

[DIAGRAM: asset_name: A logarithm locates an exponent; asset_slug: p1_6_5_solve_equations_of_the_form_a_x_b__diagram_01; caption: The intersection of the exponential curve and the horizontal line gives the required exponent; description: The graph of y=2^x meets y=5 at x=log base 2 of 5, between 2 and 3. A vertical guide links the intersection to its horizontal coordinate.]

Logarithm

For $a>0$ , $a\ne1$ and $b>0$ , the logarithm $\log_a b$ is the unique real exponent $x$ satisfying $a^x=b$ . For example, $\log_2 8=3$ . There is no real $\log_2(-8)$ , because $2^x$ is always positive.

The same uniqueness holds for $0<a<1$ : the curve decreases, but still reaches every positive output exactly once. The base $1$ is different: $1^x=1$ for every real $x$ , so $1^x=1$ has infinitely many solutions and $1^x=5$ has none. Negative bases are outside this real-exponential method because they do not define a real value for every real exponent.

Here is a small surprise: $\log_2 5$ is not even a fraction. If it were $p/q$ for positive integers $p,q$ , then $2^p=5^q$ . The left side is even and the right side is odd, an impossibility. An exact logarithm is a complete answer, even when no fraction will do.

Key idea

A positive output gives exactly one real exponent when the positive base differs from $1$ . A zero or negative output gives no real solution.

Check the output before reaching for a calculator.

Using change of base

To calculate the exponent in $2^x=5$ , take the same logarithm of both sides. Natural logarithms are convenient, but base $10$ works too.

Change of base

For $a>0$ , $a\ne1$ , $b>0$ , and any logarithm base $c>0$ , $c\ne1$ ,

\log_a b=\frac{\log_c b}{\log_c a}

In particular, $\log_a b=\dfrac{\ln b}{\ln a}$ .

In the formula booklet.

You can see why the quotient appears: if $a^x=b$ , the power law gives $x\ln a=\ln b$ . Since $a\ne1$ , $\ln a\ne0$ , so division is legitimate.

Worked example 1: an exact exponent

Question. Solve $2^x=5$ , giving an exact answer and a value to $3$ significant figures.

Plan. Take natural logarithms, bring down the exponent and divide by the logarithm of the base.

\begin{aligned} \ln(2^x)&=\ln5 &&\text{same operation on both sides}\\ x\ln2&=\ln5 &&\text{power law}\\ x&=\frac{\ln5}{\ln2} &&\text{divide by }\ln2 \end{aligned}

Why this step? Taking a logarithm turns the unknown power into a multiplier, which ordinary algebra can isolate.

Answer. $x=\dfrac{\ln5}{\ln2}=2.32$ to $3$ significant figures; the logarithmic quotient is exact.

Check. The unrounded value is $2.321928\ldots$ , between $2$ and $3$ , and $2^{\ln5/\ln2}=5$ .

Worked example 2 changes only the base, replacing $2$ by $1/2$ . The required exponent changes sign because a reciprocal base reverses growth.

Worked example 2: a base below one

Question. Solve $\left(\dfrac12\right)^x=5$ , exactly and to $3$ significant figures.

Plan. Use the same logarithmic quotient and keep the negative denominator.

\begin{aligned} x\ln\left(\frac12\right)&=\ln5 &&\text{take logs and use the power law}\\ x&=\frac{\ln5}{\ln(1/2)} &&\text{isolate }x\\ &=-\frac{\ln5}{\ln2} &&\text{reciprocal log law} \end{aligned}

Why this step? $\ln(1/2)=-\ln2$ , so a negative exponent is needed to make a base below $1$ produce an output above $1$ .

Answer. $x=-\dfrac{\ln5}{\ln2}=-2.32$ to $3$ significant figures.

Check. $(1/2)^{-\ln5/\ln2}=2^{\ln5/\ln2}=5$ .

Keep the numerator and denominator in the same base. $\ln b/\log_{10}a$ is not change of base. For another increasing-base equation, $3^x=11$ gives $x=\ln11/\ln3\approx2.18$ .

Watch out

Wrong: $\ln(2^x)=(\ln2)^x$ . Right: $\ln(2^x)=x\ln2$ . The power becomes a multiplier, not a power of the logarithm.

The next equation keeps the reciprocal base from Worked example 2 and changes only the positive output.

Linear expressions in the exponent

In $2^{3x-1}=3$ , the logarithm first reveals $3x-1$ , rather than $x$ . Undo the exponential before solving that linear equation.

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Last Accessed

Pure Mathematics (Papers 1 and 2)

Statistics (Paper 3, Section A)

Mechanics (Paper 3, Section B)

Pure 6.5 - Solving exponential equations

What the equation means

Using change of base

Linear expressions in the exponent

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