2.10 Logarithmic scaling and linearizing dataIB Maths: Applications and Interpretation HL: Revision notes

Logarithms turn multiplying into adding, so they squeeze very large or very small numbers into a manageable scale. On a $\log_{10}$ scale, increasing the value by 1 multiplies the original quantity by 10; an increase of 3 multiplies it by 1000. Example: for $N=3\,000\,000$, $\log_{10}N=\log_{10}3+6=6.48$. To go back, use $N=10^{4.5}=3.16\times10^{4}$. The laws you need are $\log(xy)=\log x+\log y$, $\log\frac xy=\log x-\log y$ and $\log(x^m)=m\log x$, with base 10 or $e$ (written $\ln$).

If $y=ab^{x}$, take logarithms: $\ln y=\ln a+x\ln b.$ A plot of $\ln y$ against $x$ is a straight line with gradient $\ln b$ and vertical intercept $\ln a$. This is a semi-log relationship (one variable logged). Example: $\ln N=4.2+0.35t$ gives $a=e^{4.2}=66.7$ and $b=e^{0.35}=1.42$. Use your GDC to find the regression line of $\ln y$ on $x$, then undo the logarithm to get $a$ and $b$.

If $y=kx^{n}$, take logarithms: $\ln y=\ln k+n\ln x.$ A plot of $\ln y$ against $\ln x$ is a straight line with gradient $n$ and vertical intercept $\ln k$. This is a log-log relationship (both variables logged). Example: for planets, $\log_{10}T=1.5\log_{10}d-0.70$ gives $n=1.5$ and $k=10^{-0.70}=0.200$, so $T=0.200d^{1.5}$. Here the gradient is the exponent itself. Any base works, as long as you use the same base to undo it.

You will not be asked to draw these graphs, but you must interpret them.

• Straight line on a semi-log graph ($\ln y$ against $x$): exponential model.
• Straight line on a log-log graph: power model.
• Not straight on either: try another model. Use Pearson's product-moment correlation coefficient $r$ for the transformed data: a value close to $\pm1$ supports the model; a value close to 0 means the model is a poor fit. A log scale is a sensible choice when the data cover a wide range, or when the focus is the rate of growth (constant ratio) rather than the absolute change, since equal growth rates become equal steps.

Method: 1. Choose the transformation. 2. Find the line of best fit by GDC. 3. Compare with $\ln y=\ln a+x\ln b$ or $\ln y=\ln k+n\ln x$. 4. Undo the logarithms. Example: cases $C$ against days $t$ gives $\ln C=2.09+0.526t$ with $r=1.00$, so $C=8.05\times1.69^{t}$. At $t=12$, $C\approx4440$; $C=10\,000$ when $t=13.5$. Always comment: $r$ close to 1 supports the model, but predictions outside the data range are extrapolation, and real growth cannot continue forever, for example because a population is finite.