# Trying to convert a nasty logarithm into an exponential

I have the following equation that I must express in terms of $r$:

$$\Delta V = \frac{\lambda}{2 \pi \epsilon_0} \ln(\frac{r}{R})$$

This is a pretty tough one. I am not sure how to get the r out of the logarithm. I have tried to express the logarithm as $\ln(r) - \ln(R)$, but I am not sure where to go from there, or where the coefficient should go. I know that $\Delta V$ goes up in the exponent...

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First solve for $\ln\left(\frac{r}{R}\right)$. We get $$\ln\left(\frac{r}{R}\right)=\frac{2\pi\epsilon_0 \Delta V}{\lambda}.$$ Now take the exponential of both sides, and it's nearly over.
Remark: Under other circumstances, your observation that $\ln\left(\frac{r}{R}\right)=\ln r-\ln R$ would be useful. Here it can be used, but there are quicker ways.
Inside the natural log function would be on the left hand side, and the coefficient of the messy lambda business on the right would be $e$, but I am not sure what to do about the fact that the expression inside the natural log function is a ratio. –  Dylan Jan 22 '13 at 6:38
$\frac{r}{R} = e^{\frac{2 \pi \epsilon_0\Delta V}{\lambda}} \rightarrow r = Re^{\frac{2 \pi \epsilon_0 \Delta V}{\lambda}}$ –  Dylan Jan 22 '13 at 6:42
Oh, I see what you mean. For some reason I was confused about the ratio inside the natural logarithm, i.e. $\frac{r}{R}$. But I was just over-thinking it, perhaps. Thanks again, Mr. Nicolas. –  Dylan Jan 22 '13 at 6:47