TITLE: How do we get from $\ln A=\ln P+rn$ to $A=Pe^{rn}$ and similar logarithmic equations?
QUESTION [12 upvotes]: I've been self-studying from the amazing "Engineering Mathematics" by Stroud and Booth, and am currently learning about algebra, particularly logarithms.
There is a question which I don't understand who they've solved. Namely, I'm supposed to express the following equations without logs:
$$\ln A = \ln P  + rn$$
The solution they provide is:
$$A = Pe^{rn}$$
But I absolutely have no idea how they got to these solutions. (I managed to "decipher" some of the similar ones piece by piece by studying the rules of logarithms).

REPLY [9 votes]: A key intuition behind logarithms is that multiplication translates to addition, i.e.
$\ln(A*B)=\ln(A)+\ln(P)\qquad$ and
$\qquad\ln(A/B)=\ln(A)-\ln(P)$
We can use this to solve your equation
$\begin{align}
\ln(A)&=\ln(P)+rn\newline
\ln(A)-\ln(P)&=rn\newline
\ln(A/P)&=rn\newline
A/P&=e^{rn}\newline
A&=Pe^{rn}
\end{align}$