Exponential functions and their graphs, the number e and the natural logarithm, the laws of logarithms, solving exponential and logarithmic equations, and using logarithms to estimate parameters in exponential and power-law models.

What this dot point is asking

OCR wants you to know the shape and behaviour of exponential functions $a^x$ and $e^x$, understand $e$ as the base for which the gradient of $e^x$ equals $e^x$, use the natural logarithm $\ln x$ as the inverse of $e^x$, apply the three laws of logarithms, solve exponential and logarithmic equations, and use logarithms to linearise exponential ($y = ab^x$) and power-law ($y = ax^n$) models so that a straight-line graph reveals the parameters.

The answer

Exponential functions and e

An exponential function $y = a^x$ (with $a > 0$) passes through $(0, 1)$, is always positive, and either grows (if $a > 1$) or decays (if $a < 1$). The special base $e \approx 2.718$ has the property that the curve $y = e^x$ is its own gradient function.

The laws of logarithms

A logarithm answers "what power gives this number": $\log_a b = c$ means $a^c = b$. The three laws turn products into sums and powers into multipliers, which is what lets us solve equations with the unknown in an exponent.

Solving exponential equations

When the unknown is in the exponent, take logs of both sides and bring the power down with the third law.

Solving logarithmic equations

Combine logarithms into a single one with the laws, then rewrite in exponential (index) form. Always check answers, because the logarithm of a negative number or zero is undefined.

Examples in context

Linearising a model

If $y = ab^x$, taking logs gives $\log y = \log a + x\log b$, which is linear in $x$: plotting $\log y$ against $x$ gives a straight line with gradient $\log b$ and intercept $\log a$. If $y = ax^n$, taking logs gives $\log y = \log a + n\log x$, linear in $\log x$ with gradient $n$. Recognising which form a model takes tells you which graph to plot.

Reading a log graph

Try this

Q1. Solve $\ln(2x + 1) = 3$, giving your answer to three significant figures. [2 marks]

• Cue. $2x + 1 = e^3 = 20.09$, so $x = \dfrac{19.09}{2} \approx 9.54$.

Q2. Write $2\log_{10} x - \log_{10} 3$ as a single logarithm. [2 marks]

• Cue. $\log_{10} x^2 - \log_{10} 3 = \log_{10}\dfrac{x^2}{3}$.