Inverse of Logarithm: Understanding ax

[nobahar]

Hello!
Run into trouble...again.
This concerns the inverse function of a logarithm
If a function maps x on to log_ax, then the inverse maps log_ax on to x.
So, f(x) = log_ax, can be presented as y=log_ax; therefore, x=a^y.
The book states that the inverse is a^x, why is this the inverse?
I tried determining the inverse by using a basic inverse:
f(x)=3x+1
f^-1(x)=(x-1)/3
If x=2
Then, f(x)=3*2+1=7
and f^-1(x)=(7-1)/3=2
If the product of log_ax=y, going the other way involves a^y=x, yet the inverse is, apparently, a^x.
I don't see why this is so...

P.S. Happy voting!

 

[berkeman]

So, f(x) = logax, can be presented as y=logax; therefore, x=ay.

You are correct. I think the book is just redefining "x". Kind of the way you did when you used x in both the function and inverse function definitions...

f(x)=3x+1
f-1(x)=(x-1)/3

It's better to be explicit about a different "x" and "y" for the two different directions...

 

[statdad]

Your work is correct; the difference is mostly a matter of notation.

One old¹ trick for finding inverses of functions is to begin by reversing the roles of $$x$$ and $$y$$ at the start. The complete process would look like this.

$$\begin{align*} f(x) & = \log_a x \\ y & = \log_a x \tag{Now switch variables}\\ x & =\log_a y\\ a^x & = y, \end{align*}$$

so the inverse function has equation

$$y = a^x$$

We are accustomed to writing functions in the $$y =$$ form in early math courses.

Footnote 1: I know this is an old procedure because I learned it when I was in high school: if it has been around that long, it is indeed old.

 

[nobahar]

Thanks for the quick replies! The responses are much appreciated, and even tickled me , not something one would associate with a maths help forum!
So a^x is really a^y 'in disguise', since it is the inverse function the y is now the x?

 

[HallsofIvy]

Don't write "x" and "y" in isolation. That is what is confusing you. The functions f(x)= ln(x) or f(y)= ln(y) or f(z)= ln(z) are all exactly the same function.

If f(x)= ln(x) then f^-1(x)= e^x or f^-1(y)= e^y or f^-1(a)= e^a, etc. it doesn't matter what you call the variable.

 

[nobahar]

Okay, I thought I had it...
If f(x)=y, then f(x)=log_ax, and a^y=x, and so a^f-1(x)=x?
If f(x)=log_ax and the inverse is a^x, and I attempt an equation it doesn't work... x has to change going in the different 'ends', and so x in the new equation (the inverse) is y on the first equation...
I appreciate this may be getting tiresome to explain... but please bear with me!
Thanks!

 

[HallsofIvy]

Sure it does
$$a^{log_a x}= x$$
and
$$log_a(a^x)= x$$

Once again, f(x)= log_a(x), f(y)= log_a(y), f(z)= log_a(z), etc. are all exactly the same function.

 

[statdad]

The notation is getting to you.

$$\begin{align*} f(x) & = \log_a x\\ y & = \log_a x \\ x & = \log_a y\\ a^x & = y \end{align*}$$

This means that the function $$f(x) = \log_a x$$ has as its inverse

$$f^{-1}(x) = a^x$$