Logarithmic graphs and modellingEdexcel International A Level Maths: Revision notes

A relationship of the form $y=ax^{n}$ is not linear, but taking logarithms of both sides (base $10$, written $\lg$, or $\ln$) gives a linear equation. Using the laws of logarithms: $\lg y=\lg a+n\lg x.$ 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 $\lg a$. The two logarithm laws used are $\lg(AB)=\lg A+\lg B$ and $\lg(A^{p})=p\lg A$.

For an exponential relationship $y=kb^{x}$, taking logarithms gives $\lg y=\lg k+x\lg b.$ Now $Y=\lg y$ is plotted against $X=x$ (not $\lg x$). The line has gradient $\lg b$ and vertical intercept $\lg k$. Hence $b=10^{\text{gradient}}$ and $k=10^{\text{intercept}}$. Quick test: if the data give a straight line against $\lg x$, the relationship is a power law; if a straight line against $x$ (when $\lg y$ is plotted), it is exponential.

Pick two points on the line (or use two data points converted to logarithms) and find the gradient $=\frac{\Delta Y}{\Delta X}$. Then find the intercept using $c=Y-mX$ with one point. Worked example (power law): the line passes through $(0.2,0.8)$ and $(0.6,1.6)$. Gradient $=\frac{0.8}{0.4}=2$, so $n=2$. Intercept $=0.8-2(0.2)=0.4$, so $a=10^{0.4}=2.51$. Worked example (exponential): the line passes through $(2,3.4)$ and $(6,4.2)$. Gradient $0.2$ gives $b=10^{0.2}=1.58$; intercept $3.4-2(0.2)=3.0$ gives $k=1000$.

Once $a$ and $n$ (or $k$ and $b$) are known you can predict values by substituting into $y=ax^n$ or $y=kb^x$, or by working with the straight-line equation directly: for example $\lg y=0.3x+0.6$ at $x=8$ gives $\lg y=3$, so $y=1000$. To find the input for a given output, take logarithms of the output first: $\lg500=2\lg x+0.4$, so $\lg x=\frac{\lg500-0.4}{2}$ and $x=10^{1.1495}=14.1$. Evaluate a model by commenting on its limits: a simple exponential model of growth predicts unlimited increase, but real systems are limited by resources, so the model is only valid over a restricted range.