The exponential function and the number e, the natural logarithm, the laws of logarithms, solving exponential equations, and using logarithms to linearise and model real data.

What this dot point is asking

Edexcel wants you to understand the exponential function $e^x$ and its special property, use the natural logarithm $\ln x$ as the inverse of $e^x$, apply the laws of logarithms, solve equations of the form $a^x = b$, and use logarithms to turn a relationship of the form $y = ax^n$ or $y = kb^x$ into a straight line so you can model real data.

The answer

The exponential function and e

The function $y = e^x$ grows in proportion to its own value, so its gradient at any point equals its height: $\dfrac{d}{dx}(e^x) = e^x$. This makes $e \approx 2.718$ the natural base for growth and decay models such as $N = N_0 e^{kt}$, where $k > 0$ gives growth and $k < 0$ gives decay. The curve $y = e^x$ passes through $(0, 1)$, never touches the $x$-axis, and rises without bound, while its reflection $y = e^{-x}$ decays towards zero. The constant $N_0$ is the value at $t = 0$, since $e^0 = 1$, and these two shapes underpin every growth-and-decay question on the paper.

Logarithms

Two special values follow straight from the definition: $\log_a 1 = 0$ because $a^0 = 1$, and $\log_a a = 1$ because $a^1 = a$. The power law is the workhorse, because it turns an exponent into a multiplier you can solve for.

Solving exponential equations

Linearising data

If $y = kx^n$, then $\log y = \log k + n\log x$, so plotting $\log y$ against $\log x$ gives a straight line of gradient $n$ and intercept $\log k$. If $y = kb^x$, then $\log y = \log k + x\log b$, so plotting $\log y$ against $x$ gives a straight line.

Examples in context

Try this

Q1. Solve $5^{2x} = 100$, giving your answer to three significant figures. [3 marks]

• Cue. $2x\ln 5 = \ln 100$, so $x = \dfrac{\ln 100}{2\ln 5} \approx 1.43$.

Q2. The model $N = 50e^{0.2t}$ gives a population at time $t$ years. Find $t$ when $N = 200$. [3 marks]

• Cue. $4 = e^{0.2t}$, so $0.2t = \ln 4$ and $t = \dfrac{\ln 4}{0.2} \approx 6.93$ years.