I'm using a publicly available textbook to revise some maths and learn a bit of basic astrophysics. The section on logarithms discusses power laws and log log graphs.

It uses the following generalised example of $y = ax^k$ can be plotted as $\log y = \log a + k \log x$. This results in a straight line graph where the gradient is equal to $k$ and the intercept gives the value of $\log a$. But how can there be an intercept? When $x = 0$ isn't $k \log x$ undefined?

At the moment I'm just pretending that it's a trick that we play so that we can use a useful graph to infer the result rather than a precise calculation? i.e. $\log x$ is effectively $0$ at the origin rather than undefined.

Textbook extract

  • 1
    $\begingroup$ If you read the given section closely, the graph is between $\log y$ and $\log x$. The intercept is defined by putting one of the axis as 0. In this case, the "$y$-intercept" is actually the $\log y$ intercept and is found out by putting $\log x$ (the other axis) as zero, not $x =0$. $\endgroup$ May 11 at 17:35
  • $\begingroup$ That took me a minute and a few reads to flip my perspective round but I see it now. Thank you Adam and Joe for both taking the time to explain it, and to whoever fixed my formatting for me! $\endgroup$
    – HighWater
    May 12 at 0:22

As Adam mentions in the comments, you are plotting $\log y$ against $\log x$, and so the intercepts actually occur when $\log y=0$ and when $\log x = 0$. For ease of notation, let $X=\log x$ and $Y=\log y$. The equation $$ \log y = k\log x + \log a $$ becomes $$ Y = kX + \log a \, , $$ which you can recognise as the graph of a linear equation. The $Y-$intercept (what is called the 'y-intercept' in the testbook) occurs when $X=0$, so $\log x = 0$. This equation has the solution $x=1$, regardless of the base of the logarithm used.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.