The exponential function and $\mathrm{e}^x$, logarithms and their laws, solving equations with logs, and fitting exponential models with a log-linear graph.

What this dot point is asking

WJEC wants you to know the shape and key features of exponential functions including $\mathrm{e}^x$, to use the laws of logarithms to manipulate and simplify expressions, to solve equations where the unknown is in an exponent, and to fit an exponential model to data by plotting a log-linear graph and reading off the constants. Logarithms unlock any equation with the variable in the power, so this topic is heavily examined.

The answer

Exponential functions

An exponential function $y = a^x$ (with $a > 0$) passes through $(0, 1)$, is always positive, and increases for $a > 1$ or decreases for $0 < a < 1$. The special base $\mathrm{e} \approx 2.71828$ gives the natural exponential $\mathrm{e}^x$, which has the property that its gradient equals its value, central to the calculus that follows.

Logarithms and their laws

A logarithm answers the question "to what power must the base be raised?".

Solving exponential equations

Modelling with log-linear graphs

A relationship of the form $y = k a^x$ becomes linear when you take logarithms: $\log y = \log k + x\log a$. Plotting $\log y$ (vertical) against $x$ (horizontal) gives a straight line with gradient $\log a$ and vertical intercept $\log k$, so you can recover the constants from a fitted line.

Examples in context

Example 1. Carbon dating. A sample decays as $N = N_0\,\mathrm{e}^{-0.000121t}$ with $t$ in years. If $N = 0.25 N_0$, then $\mathrm{e}^{-0.000121t} = 0.25$, so $-0.000121t = \ln 0.25 = -1.386$, giving $t \approx 11\,500$ years. The natural log turns the decay model into a single division.

Example 2. Reading a log graph. Data plotted as $\log y$ against $x$ give a line of gradient $0.30$ and intercept $0.70$. Then $\log a = 0.30$ so $a = 10^{0.30} \approx 2.0$, and $\log k = 0.70$ so $k = 10^{0.70} \approx 5.0$. The model is $y \approx 5.0 \times 2.0^x$. The straight-line fit delivers both constants.

Try this

Q1. Write $2\log x - \log 3$ as a single logarithm. [2 marks]

• Cue. $2\log x = \log x^2$, then subtract: $\log\dfrac{x^2}{3}$.

Q2. Solve $\ln(2x + 1) = 3$, giving $x$ to three significant figures. [3 marks]

• Cue. $2x + 1 = \mathrm{e}^3 = 20.09$, so $x = 9.54$.

Q3. A quantity satisfies $y = 4 \times 3^x$. Find $x$ when $y = 36$. [2 marks]

• Cue. $3^x = 9$, so $x = 2$ directly.