Solve exponential equations by taking logarithms and logarithmic equations by converting to index form, rejecting invalid solutions

What this dot point is asking

SEAB wants you to solve equations where the unknown sits in an exponent (such as $5^x = 20$) or inside a logarithm (such as $\ln(2x + 1) = 3$). The two cases are mirror images: take logarithms to bring an exponent down, or rewrite a logarithm in index form to free the unknown. Then solve and check that every answer is valid.

The answer

Solving exponential equations

When the unknown is an exponent and you cannot match the bases, take the logarithm of both sides and use the power law to bring the exponent down:

Any base of logarithm works; $\log$ (base $10$) or $\ln$ (base $e$) suit the calculator.

The natural exponential and logarithm

The number $e \approx 2.718$ gives the natural exponential $e^x$ and its inverse the natural logarithm $\ln x = \log_e x$. They undo each other:

So $e^x = b$ is solved by $x = \ln b$.

Solving logarithmic equations

When the unknown is inside a logarithm, first combine into a single log using the log laws if needed, then convert to index form:

This turns the logarithmic equation into an ordinary algebraic one.

Always check validity

The argument of a logarithm must be positive, so after solving, reject any value that would make the inside of a log zero or negative. This step is where many marks are lost.

Hidden quadratics in exponentials

Some equations mix two powers of the same base, such as $e^{2x}$ with $e^{x}$. Since $e^{2x} = (e^x)^2$, substituting $u = e^x$ turns the equation into a quadratic in $u$; solve it, then revert with $x = \ln u$, discarding any non-positive $u$ because an exponential is never negative.

Logarithmic equations with logs on both sides

If both sides are single logarithms of the same base, such as $\log_a P = \log_a Q$, then the arguments are equal: $P = Q$. Combine each side into one logarithm first if needed, then equate the arguments and solve, again checking that every argument stays positive.

Examples in context

Example 1. Half-life and decay. Radioactive decay $N = N_0 e^{-kt}$ asks for the time when $N$ drops to half: taking the natural logarithm of $\tfrac{1}{2} = e^{-kt}$ gives $t = \dfrac{\ln 2}{k}$, the standard half-life result.

Example 2. Time to reach a target. A savings balance growing as $A = 1000(1.05)^n$ reaching $1500$ requires solving $1.05^n = 1.5$; taking logs gives the number of years, a direct financial use of the method.

Try this

Q1. Solve $2^{x} = 50$, to three significant figures. [3 marks]

• Cue. $x = \dfrac{\log 50}{\log 2} = 5.64$.

Q2. Solve $e^{x} = 7$, to three significant figures. [2 marks]

• Cue. $x = \ln 7 = 1.95$.

Q3. Solve $\log_2 (x - 1) = 4$. [2 marks]

• Cue. $x - 1 = 2^4 = 16$, so $x = 17$.