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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 $\ln x$

Emilio Novati
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LJacob
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  • Why must that be wrong? The factor of increase in $y$ will not be independent of $x$. Try some numerical examples. (Please [edit] the question to reformat with mathjax: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Ethan Bolker Dec 13 '18 at 17:26

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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.

Shubham Johri
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