Turning curves into straight lines

A curve is hard to read precisely. Is it $y = 2x^2$ or $y = 2x^{2.1}$? You can't tell by eye.

A straight line is easy: you can measure its gradient and intercept.

So the strategy is:

1. take logs of the equation;
2. plot the log quantities, which gives a straight line;
3. read the constants off that line.

Take $\ln$ of both sides of $y = kx^n$:

$\ln y = \ln(kx^n)$

Split the product with the product law:

$\ln y = \ln k + \ln(x^n)$

Then bring the power down with the power law:

$\ln y = \ln k + n\ln x$

Now compare with a straight line, $Y = mX + c$:

$\underbrace{\ln y}_{Y} = \underbrace{n}_{m}\,\underbrace{\ln x}_{X} + \underbrace{\ln k}_{c}$

So:

• plot $\ln y$ (up) against $\ln x$ (across);
• the gradient is $n$;
• the intercept is $\ln k$, so $k = e^{\text{intercept}}$.

Take $\ln$ of both sides and use the same two laws:

$\ln y = \ln k + x\ln a$

Compare with $Y = mX + c$. This time the thing across is plain $x$, not $\ln x$:

$\underbrace{\ln y}_{Y} = \underbrace{\ln a}_{m}\,\underbrace{x}_{X} + \underbrace{\ln k}_{c}$

So:

• plot $\ln y$ against $x$;
• the gradient is $\ln a$, so $a = e^{\text{gradient}}$;
• the intercept is $\ln k$, so $k = e^{\text{intercept}}$.

The same idea handles $y = Ae^{bx}$. Taking $\ln$ gives $\ln y = \ln A + bx$, so the gradient is $b$ and the intercept is $\ln A$.

1. Choosing the axes. Look at where the unknown sits:

• unknown in a power of $x$ (like $x^n$): plot $\ln y$ against $\ln x$;
• unknown as the base of a power with $x$ up top (like $a^x$): plot $\ln y$ against $x$.

Get this backwards and the gradient means nothing.

2. Turning the intercept back into a constant. The intercept is $\ln k$, not $k$. So the last line of your answer must be $k = e^{\text{intercept}}$.