I have been told multiple times that the logarithmic function is the inverse of the exponential function and vice versa. My question is; what are the implications of this? How can we see that they're the inverse of each other in basic math (so their graphed functions, derivatives, etc.)?
[Math] The relation between an exponential function and a logarithmic function
exponential functionfunctionsinverselogarithms
Best Answer
It means that $e^x$ is a bijection from $\Bbb R$ onto $(0, \infty)$, and that $\ln(x)$ is a bijection of $(0,\infty)$ onto $\Bbb R$. In other words, both functions pair up points in their domain and range: roughly speaking, "nothing is left out", and "no two pairs overlap".
The inverse properties say that: $e^{\ln(x)}=x$ for all $x\in (0,\infty)$ and $\ln(e^x)=x$ for all $x\in \Bbb R$. These properties are useful for solving equations, for one thing. If you see $e^x=4$, then by applying $\ln$ to both sides:
$$ \ln(e^x)=\ln(4) $$ and by that cancellation, the left hand side is just $x$, so it is now solved for $x$.
Graphically you can check that they are inverses. If you graph both $e^x$ and $\ln(x)$ on the same axes, then you will observe that they are reflections of each other across the line $y=x$. This is the case for all pairs of mutually inverse functions.
That graphical reflection translates, in symbols, to: if $(x_0,y_0)$ is on the graph of $e^x$, that means $e^x$ sends $x_0$ to $y_0$. Since the functions are inverses, that means $\ln(x)$ sends $y_0$ to $x_0$. Therefore, $(y_0,x_0)$ is on the graph of $\ln(x)$. The act of switching the coordinates of the ordered pairs produces the reflection across $y=x$.
There is one relationship you can derive about the derivatives. Since $f(f^{-1}(x))=x$, differentiating both sides (using the chain rule on the left) says that $f'(f^{-1}(x))\cdot (f^{-1})'(x)=1$. Solving for $(f^{-1})'(x)$ you get that:
$$ (f^{-1})'(x)=\frac{1}{f'(f^{-1}(x))} $$
Using $f(x)=e^x$ and knowing that $f'=f$ for this function, you get:
$$ \frac{d\ln(x)}{dx}=\frac{1}{e^{\ln(x)}}=\frac{1}{x} $$
There are a lot of details and rigor I have not mentioned, but I hope this gives you a bit of a flavor of what is going on. Don't use this as an excuse not to look at your text!
PS: I interpreted "the exponential" to mean $f(x)=e^x$, but of course everything above (with some care, especially in the case of the derivative) can be changed over to the case of $f(x)=a^x$ and $f^{-1}(x)=\log_a(x)$