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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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1 Answer

up vote 2 down vote accepted

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.

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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
    
oh, i see it now! –  Dylan Jan 22 '13 at 6:39
    
$\frac{r}{R} = e^{\frac{2 \pi \epsilon_0\Delta V}{\lambda}} \rightarrow r = Re^{\frac{2 \pi \epsilon_0 \Delta V}{\lambda}}$ –  Dylan Jan 22 '13 at 6:42
    
Yes, you got it. –  André Nicolas Jan 22 '13 at 6:43
    
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
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