How do we get from $\ln A=\ln P+rn$ to $A=Pe^{rn}$ and similar logarithmic equations?

The basic idea behind all basic algebraic manipulations is that you are trying to isolate some variable or expression from the rest of the equation (that is, you are trying to "solve" for $A$ in this equation by putting it on one side of the equality by itself).

For this particular example (and indeed, most questions involving logarithms), you will have to know that the logarithm is "invertible"; just like multiplying and dividing by the same non-zero number changes nothing, taking a logarithm and then an exponential of a positive number changes nothing.

So, when we see $\ln(A)=\ln(P)+rn$, we can "undo" the logarithm by taking an exponential. However, what we do to one side must also be done to the other, so we are left with the following after recalling our basic rules of exponentiation: $$ A=e^{\ln(A)}=e^{\ln(P)+rn}=e^{\ln(P)}\cdot e^{rn}=Pe^{rn} $$

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

$\begin{align} \ln(A)&=\ln(P)+rn\newline \ln(A)-\ln(P)&=rn\newline \ln(A/P)&=rn\newline A/P&=e^{rn}\newline A&=Pe^{rn} \end{align}$

Hint: Write $$e^{\ln(A)}=e^{\ln(P)+rn}$$ and use that $$e^{\ln(x)}=x$$