# The relation between an exponential function and a logarithmic function

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.)?

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They are inverse in the sense that $\log_a(a^b) = b$. For $a = e$, you have the natural logarithm, $\ln(e^b) = b$. –  Pål GD Jan 15 '13 at 13:59

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 $\Bbb R$. This means 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 $f^{-1}(x)=\log_a(x)$

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Just to generalize:

A function $g: Y \to X$ is an inverse function of $\;f: X \to Y\;$ (and vice versa) if and only if $$\;g \circ f = id_X\; \text { and}\;\,f\circ g = id_Y:\;$$ that is, if and only if $\;g(f(x)) = x\;$ for all $x \in X$ and $\;f(g(y)) = y\,$ for all $\,y \in Y$.

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Actually the logarithm function is defined as the inverse of exponent function. It's not a property of these functions, it's how the logarithm is introduced.

If $a^b=c$ then the power by which you raise a to obtain c is the logarithm: $b=log_a c$.

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That's the case in some developments. There are also texts that start by defining $\log x = \int_1^x \frac 1x\,dx$ and then define $\exp(x)$ as the alternative of the logarithm. Yet other developments define the logarithm and the exponential separately and then have to prove explicitly that they are inverses. –  Henning Makholm Jan 15 '13 at 15:10