Exponential equations are solved by taking logarithms of both sides and applying the power law to bring the unknown exponent down.

What this dot point is asking

When the unknown sits in the exponent, ordinary algebra cannot isolate it. The logarithm is the tool that "undoes" the exponential, letting you bring the exponent down where you can solve for it. This works because of the power law $\log(a^x)=x\log a$: applying a logarithm converts the exponent into a multiplier, turning an exponential equation into a linear one that ordinary algebra can finish.

The core technique

Equations with base $e$

When the base is $e$, taking $\ln$ is especially clean because $\ln e=1$.

Growth-and-decay applications

Most exam questions dress this technique as a real model - doubling time, half-life, or reaching a target value. In every case the model has the form $Q=Q_0 e^{kt}$, where $Q_0$ is the starting amount and $k$ is the continuous growth rate ($k>0$) or decay rate ($k<0$). Reaching a target means setting $Q$ to the target value, isolating the exponential, and taking logs - exactly the technique above. Doubling and half-life are just the special targets $Q=2Q_0$ and $Q=\tfrac12 Q_0$, where the $Q_0$ cancels and the time depends only on $k$.

Equations that reduce to a quadratic

Some exponential equations hide a quadratic. An equation such as $e^{2x}-5e^x+6=0$ becomes a quadratic if you let $u=e^x$, since $e^{2x}=(e^x)^2=u^2$. Then $u^2-5u+6=0$ factorises to $(u-2)(u-3)=0$, giving $u=2$ or $u=3$. Substituting back, $e^x=2$ gives $x=\ln 2$ and $e^x=3$ gives $x=\ln 3$. Watch for any value of $u$ that is zero or negative: since $e^x>0$ always, such a value must be rejected because no real $x$ produces it.

Choosing which base to use

You may take logs in any base, but the natural log is almost always the cleanest choice in SACE because the models are written with $e$ and because $\ln e=1$ removes a step. If a question is set in base $10$ or base $2$, you can still use $\ln$ throughout and the change-of-base ratio falls out automatically: $x=\dfrac{\ln b}{\ln a}$ for $a^x=b$. The answer is identical whichever base you take; only the intermediate arithmetic differs.

Common errors

Why it matters

Solving exponential equations is the payoff of Topic 4 and a guaranteed exam skill, especially in growth-and-decay contexts. It draws together the log laws, change of base, and the exponential models you differentiated earlier, and it recurs whenever a continuous random variable or model must be inverted for a target value.