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

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).

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 $$e^{\ln(x)}=x$$

There are a couple of things you must know about logarithms to understand that piece of mathematics.

First of all, you need to know that $$\ln{e}=1$$. It follows directly from the fact that $$e^1=e$$.

Secondly, you need to know the fact that you can slide whatever you've got in front of the logarithm up and make it an exponent on the thing that's inside the logarithm: $$x\ln{y}=\ln{y^x}$$.

Also, when you add logarithms, you multiply whatever you have under the logarithm signs together as long as the product is a number that's greater than zero: $$\ln{x}+\ln{y}=\ln{(xy)},\ xy>0$$. (You can only take the logarithm of a positive number)

And the last fact you're going to need is the fact that $$\ln{x}=\ln{y}\implies x=y$$.

\begin{align} \ln{A}&=\ln{P}+rn\cdot 1\\ \ln{A}&=\ln{P}+rn\cdot \ln{e}\\ \ln{A}&=\ln{P}+\ln{e^{rn}}\\ \ln{A}&=\ln{(Pe^{rn})}\\ A&=Pe^{rn} \end{align}