Question: Suppose $x >1$ and $x$ increases by a factor of $5$. By what factor will $y$ increase given $y = \ln(x)$?
My answer:
So the increase would be $$\frac{(\ln(5x) - \ln x)}{\ln x} \times 100\%$$
And then $$\frac{( \ln 5 + \ln x -\ln x )}{\ln x}\times 100 \%$$
But thats clearly wrong because $\ln x$ get cancelled in the numerator but the denominator is $\ln x$
The increase in $y$, is just given by $ \Delta y=\ln(5x)-\ln x=\ln 5$, which is independent of the value of $x$.
The relative increase in $y$ is given by $\frac{\Delta y}y=\ln5/\ln x$, which is not independent of $x$, and can also be expressed as a percentage.