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I have a linear equation y = mx + b that I obtained by fitting a line-of-best-fit to a series of data.

How would I get the equivalent formula if I wanted to plot the same line using log-log axes?

In other words, if the new formula is log(y) = z*log(x) + c , what would I have to do to change my original m and b values to turn them into the z and c values?

pfadenhw
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1 Answers1

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If $y=mx+b$, then $\log(y)$ and $\log(x)$ certainly won't be related linearly anymore, so in the equation $\log(y)=z\log(x)+c$, $z$ and $c$ won't be constants.

In fact, if you insist that $z$ and $c$ are constants, then we would have $$y=10^{\log(y)}=10^{z\log(x)+c}=10^{z\log(x)}10^c = rx^z$$ where $r=10^c$, which is not the same kind of graph as $y=mx+b$.

J. Chapman
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  • So, if I wanted a linear equation for the logged values of x and y, the only way to obtain that would be to log all the x and y values and then re-fit a new line of best fit to the now-logged dataset? And that new line would have no predictable relationship with my original formula y=mx + b ? – pfadenhw Oct 21 '23 at 01:26
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    If you truly want a linear equation for the logged values, I don't see any way around redoing the best-fit line. The original $y=mx+b=m(x+b/m)$ would turn into something like $\log(y)=log(x+b/m)+log(m)$ which doesn't really directly relate $\log(y)$ and $\log(x)$ as you're hoping. This would be a linear equation if your axes are labeled $\log(y)$ and $\log(x+b/m)$, but not if they are $\log(y)$ and $\log(x)$. – J. Chapman Oct 21 '23 at 01:48
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    I suppose the ultimate question is whether a line better fits x and y data or if it better fits log(x) and log(y) data (in which case a power law $y=rx^z$ would have fit the original x and y data. – J. Chapman Oct 21 '23 at 01:52