Logarithmic graphsAQA A-Level Maths: Revision notes

Many relationships are curves that are hard to read parameters from. Taking logarithms can turn them into straight lines, because a straight line $Y=mX+c$ has a gradient and intercept you can measure. The two relationships in the specification are the power law $y=ax^n$ and the exponential law $y=kb^x$. Which graph you draw tells you which model you are testing. In this topic $\lg$ means $\log_{10}$, but any base can be used if you are consistent.

Take $\lg$ of both sides and use the laws of logarithms: $\lg y=\lg(ax^n)=\lg a+n\lg x.$ Compare with $Y=mX+c$ where $Y=\lg y$ and $X=\lg x$. A graph of $\lg y$ against $\lg x$ is a straight line with gradient $n$ and vertical intercept $\lg a$. So $a=10^{\text{intercept}}$. If this graph is a straight line the data fit a power law. Example: gradient $1.5$ and intercept $0.6$ give $n=1.5$ and $a=10^{0.6}=3.98$.

Take $\lg$ of both sides: $\lg y=\lg(kb^x)=\lg k+x\lg b.$ A graph of $\lg y$ against $x$ (not against $\lg x$) is a straight line with gradient $\lg b$ and vertical intercept $\lg k$. So $b=10^{\text{gradient}}$ and $k=10^{\text{intercept}}$. Example: a line through $(0,2)$ and $(4,3.2)$ has gradient $\frac{3.2-2}{4}=0.3$, so $b=10^{0.3}=2.00$ and $k=10^2=100$.

You may be given data pairs $(x,y)$, a table of $(\lg x,\lg y)$ values, or two points on a line of best fit. The method is always the same:

1. Decide which graph is straight: $\lg y$ against $\lg x$ for $y=ax^n$, or $\lg y$ against $x$ for $y=kb^x$.
2. Work out the gradient from two points far apart: $m=\frac{Y_2-Y_1}{X_2-X_1}$.
3. Find the intercept by substituting a point into $Y=mX+c$.
4. Undo the logarithm with $10^{(\cdot)}$. Example: line of best fit through $(0.60,1.38)$ and $(1.20,2.28)$ gives $n=\frac{0.90}{0.60}=1.5$ and $\lg a=1.38-1.5(0.60)=0.48$, so $a=3.02$ and $y=3.02x^{1.5}$.

Once $a$ and $n$ (or $k$ and $b$) are known, substitute into the original relationship to estimate $y$, or into the straight-line equation to find $x$. For example with $\lg y=2.1+0.2x$, $y=50\,000$ gives $x=\frac{4.699-2.1}{0.2}=13.0$. Estimates are less reliable outside the range of the data (extrapolation) and depend on how well the line fits. Exponential models also cannot grow without limit in real situations.