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...
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.