# how to see the logarithm as the inverse function of the exponential?

I saw here in math.stackexchange some proofs of how the log and exp functions are related to each other, but I want to get an intuition for that.

In layman terms, how would you explain the connection between these 2 functions?

-
When logarithms are first introduced, they are usually (as far as I know) introduced exactly as inverting exponentiation (i.e., $\log_b x$ is defined to be the number you have to raise $b$ to to get $x$), so I'm not sure what you're asking for. How were logarithms introduced to you without connection to exponentiation? –  Steven Taschuk Jun 17 at 16:19
The other common way is to introduce log as the integral of $1/x$, and define the exponential function as the inverse of that. I’m not aware of any presentation that introduces the functions separately and then shows that there is a relationship between them. –  Lubin Jun 17 at 17:56
Is that definition really used for students' first encounter with logarithms? I'd expect it to be presented only to people who already know what logarithms are, as part of an explicit project to put that existing knowledge on firm theoretical foundations. It's a very nice definition for that purpose, but this is renovation, not the first time the house is built. –  Steven Taschuk Jun 17 at 20:52
@Lubin: I didn't know there are schools that teach integration before logarithms. –  Mehrdad Jun 17 at 21:03
For a really good introduction--both intuitive and rigorous--see Chap. 18 in Spivak's Calculus (4th ed.). (You may also want to check out the rest of the book if you're interested in an awesome treatment of calculus). –  Coffee_Table Jun 17 at 23:12

You want to understand that exponential and logarithms are inverse of each other:

(i)$f(x) = y$, if we want to find the inverse, we express $x$ in terms of $y$. I will so you graphically.. (This is equally applicable for other exponents)

Which is one way of seeing that both are inverse of each other.

(ii) $a > 0$ and $x > 0$, condition $a \ne 1$, the logarithm base $a$ of $x$,

we write $\log_a(x)$, which is the exponent to which $a$ to be raised to obtain $y$. $\log_a(x) = y$, which is same as $a^y = x$. The functions $\log_a(x)$ and $a^x$ are clearly inverses of each other. The domain of logarithm base $a$ is all positive numbers and range is all real numbers.

Using the fact that the domain and range of any invertible function are just the range and domain of its inverse.

Here, the domain $\rightarrow$ logarithm base $a$ function is the range $\rightarrow$ $a^x$ function (remember for all positive no.)

and the range $\rightarrow$ logarithm base $a$ function is the domain $\rightarrow$ $a^x$ function (here it is for all numbers).

-
+1. Of course, this works best if you've already introduced this geometric property of the graphs of inverse functions, and connected it to the algebraic definition somehow (e.g. by noting that, if $y = f(x)$, then $x = f^{-1}(y)$). –  Ilmari Karonen Jun 17 at 20:23

A simple way to state how they're related would be that one function will undo the other.

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

-
+1 for "undo" rather than "cancel". –  Jp McCarthy Jun 17 at 12:04
but I can't still see the connection that does the actual "undoing"... –  Draconar Jun 17 at 12:05
@Draconar It's a bit like how subtracting 5 will undo adding 5 and how dividing by 5 will undo multiplying by 5, and how taking the square root will undo squaring. –  Darksonn Jun 17 at 12:06
@Draconar Yes, they undo each other by definition, and all their other properties are a consequence of that definition. –  JoeyBF Jun 17 at 12:08

For a positive real number $\;a\neq 1\;$, we have by definition

$$a^y=x\iff \log_ax=y\;,\;\;x>0$$

In words: the logarithm in base $\;a\;$ of a positive number $\;x\;$ is the exponent to which the base must be raised in order to get $\;x\;$ .

Thus, for example, $\;\log_381=4\;$ since the base $\;3\;$ must be raised to the fourth power in order to get $\;81\;$.

Once one swallows and assimilates this, the tight relation between logarithms and exponentials becomes much clearer. For example:

$$\begin{cases}&\log_3x=y&\iff 3^y=x\\{}\\ &3^y=x&\iff \log_3x=y\end{cases}$$

and from this is now clear that $\;f(x):=\log_ax\;,\;\;g(x):=a^x\;$ are inverse functions to each other, meaning:

$$f\circ g(x)=f(g(x))=\log_a(a^x)=x=3^{\log_ax}=g(f(x))=g\circ f(x)$$

-

If purely for a layman discussion I would go with integers and the $log_{10}$ route.

The number 981 is 3 digits so $log_{10}981$ is close to 3. If you do $10^3$, or $exp(3)$ you get a number close to 981; $exp(3)$ is of the same magnitude as $981$.

In this context the usefulness, in my opinion, would be to use $log$ to describe a magnitude, and also an intuitive way to understand exponential growth. In each step of exponential growth you add one* more digit.

*or whatever the growth rate may be.

-

Consider first for any positive fixed $x$ : $$\tag{1} y_t:=\int_1^x z^{t-1}\,dz=\frac {x^t-1}t$$ then (see too this discussion) : $$\tag{2}y_0=\lim_{t\to 0}\int_1^x z^{t-1}\,dz=\lim_{t\to 0}\frac {e^{t\log(x)}-1}t=\log(x)$$ But $(1)$ may be reversed to get : $$\tag{3}x=\left(1+t\;y_t\right)^{1/t}$$ and again as $\,t\to 0$ we get the usual $$\tag{4}x=e^{\,y_{\,0}}$$ that is the wished converse of $(2)$.

Hoping this will give some intuition too,

-
It may be better to use $\displaystyle y_0=\lim_{t\to 0}\int_1^x z^{t-1}\,dz=\int_1^x \frac{dz}z=\log(x)$ for $(2)$. –  Raymond Manzoni Jun 17 at 23:47

Specifically, if $a=e$, then $\ln b =c$ if and only if $e^c=b$.

-

For the layman it might help to know that logarithms were used as a tool for multiplication back when ships sailed by celestial navigation. The sum of the logarithms of two numbers is the logarithm of their product. Extensive log tables were published and used by rote. This made a nice extra topic for 6 grade math in 1958. Our teacher gave no hint of a link to integrals, but the relation to exponents was taught.

As a physical correlate, the slide rule uses one axis graphs of an exponential function to convert multiplication to addition of lengths, a graphic log table.

-

You could point out their opposite algebraic properties.

For logarithms you have the product rule:

$$\log(a b) = \log(a) + \log(b)$$

So the logarithm of a product is the sum of the logarithms of its factors.

For exponential functions you have the opposite relationship:

$$\exp(a + b) = \exp(a) \exp(b)$$

So the exponent of a sum is the product of the exponents of its summands.

If we chose the $\log$ function accordingly so that $\log(e)=1$ and note that $\exp(x)=e^x$ it's easy to see that these properties imply the inverse function relationship, since the two rules give us

$$\log(\exp(x)) = x \log(e) = x$$

(At least that should be clear when $x$ is an integer, but it also works with rationals. For real numbers you probably have to show cauchy convergence but I guess that would not be necessary for a layman.) Note that this doesn't imply that $e$ is Euler's number, but I think this doesn't matter for your question, since the inverse relationship is valid for any base. Probably more important is that there are existing two such functions for any $e\not=0$.

-

By definition. That's how I learned it in high school.

The teacher started class with something like $2^y=x$ and asked us to solve for $y$. Nobody could do it. Then she said "define this thing called a logarithm such that $\operatorname{log}_{base}(\mathrm{answer})=\mathrm{exponent}$" and gave us the mnemonic "b ase a nswer e xponent" to remember it. I still use "bax" once in a while.

All the other properties came out of "well, it turns out that when you define a log this way you get such and such properties" like the multiplication rules, which we then spent the next few classes proving.

-

A simple way to show this would be that they are inverses of each other. In fact, that is how the $\exp$ function is defined:

$$f(x) = \ln(x)$$ $$f^{-1}(x) = \exp(x) \equiv e^x$$

Because of how inverse functions are related ($f(f^{-1}(x)) \equiv x$ and $f^{-1}(f(x)) \equiv x$), the following identities appear:

$$\log(\exp(x)) \equiv \ln(e^x) \equiv x$$ $$\exp(\ln(x)) \equiv e^{\ln(x)} \equiv x$$

Side-note: In some cases, $\log(x)$ is used to mean the natural, or base-$e$, logarithm which is sometimes written as $\ln(x)$.

-