How to seperate out a variable from a log

Question

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.

We have $20\ln(1+r/4)=\ln((1+r/4)^{20})=4/3$. So $(1+r/4)^{20}=4/3$ and therefore $1+r/4=(4/3)^{1/20}$. (There are other ways.) — André Nicolas, Oct 23 '14 at 03:29
@AndréNicolas Thanks! This worked along with Andiano. Sorry for no upvote, don't have 15 rep. — Micheal, Oct 23 '14 at 03:39

score 0 · Accepted Answer · edited Apr 13 '17 at 12:21

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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*}

edited Apr 13 '17 at 12:21

Community

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answered Oct 23 '14 at 03:35

Adriano

41,576

I have to say I feel a little silly for not thinking of that. Been a while since I've done much math. Thank! Sorry for no upvote along with the answer marked, says I need 15 rep for it =( – Micheal Oct 23 '14 at 03:39
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I do not see the $\ln(1+\frac{r}{4})=\ln(\frac{5r}{4})$. – André Nicolas Oct 23 '14 at 03:54
@AndréNicolas Me too either – Oct 23 '14 at 03:57

How to seperate out a variable from a log

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