I have: $$20\ln(1 + r/4) = \ln(4/3)$$
I'm trying to solve for $r$. Now if it was just $\ln(r/4)$, it would be easy: $\ln(r) - \ln(4)$, but in this case with a $1 + $ in front, I'm a little confused how to get the $r$ out of there.
This is part of a larger more complex homework problem, and I'm stuck at this step. I've tried searching the internet for this, but I only find log properties for multiplication and division.
EDIT: Whoops. Yeah this answer was completely wrong. I can't delete this unfortunately, so I'll copy André Nicolas' solution from the comments in full here: \begin{align*} 20\ln(1 + r/4) &= \ln(4/3) \\ \ln((1 + r/4)^{20}) &= \ln(4/3) \\ (1 + r/4)^{20} &= 4/3 \\ 1 + r/4 &= (4/3)^{1/20} \\ r &= 4((4/3)^{1/20} - 1) \end{align*}