Associated Topics || Dr. Math Home || Search Dr. Math

### Antilogs

```
Date: 08/22/2001 at 09:28:06
From: Cedric
Subject: Antilogs

If log sub(a)10 = 0.250, then log sub(10)a equals ?

Breakdown:

log sub(a)10 = 0.250

The solution requires me to the take the base(a) antilogarithm of
both sides. That would be

10 = a^0.250.

I would like to know why is this the final answer.
```

```
Date: 08/22/2001 at 12:29:05
From: Doctor Peterson
Subject: Re: Antilogs

Hi, Cedric.

I'm not sure I understand; what you show is NOT the final answer,
because it doesn't say what log_10(a) is. I'm assuming you understand
how you got that far, but don't know where to go from here. We started
with

log_a(10) = 0.25

and raised a to the power on each side:

10 = a^0.25

You can solve for a by raising both sides to the 4th power:

10^4 = a^(0.25 * 4)

a = 10^4

Now take the base-10 log of both sides:

log_10(a) = log_10(10^4) = 4

This is an interesting result; let's see in general how log_a(b) and
taking "log" without a base as any convenient log, such as base-10 or
natural log:

log_a(b) = log(b)/log(a)
log_b(a) = log(a)/log(b)

It follows automatically that

log_a(b) = 1/log_b(a)

obvious.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/22/2001 at 12:53:54
From: Cedric
Subject: Re: Antilogs

Why and how do you raise a to the power on each side?

Why is the antilog 10 = a^0.25?

That is the question...
```

```
Date: 08/22/2001 at 14:19:39
From: Doctor Peterson
Subject: Re: Antilogs

Hi again!

Okay, I made the wrong assumption: you didn't know how you got from

log_a(10) = 0.25

to

10 = a^0.25

This is because the antilog IS the power; the exponential function is
the inverse of the logarithm:

a^log_a(x) = x  and  log_a(a^x) = x

That is, the functions

f(x) = a^x

and

g(x) = log_a(x)

are inverses of one another; f(g(x)) = x and g(f(x)) = x for any x.

Therefore, we can undo the base-a logarithm by raising a to that
power:

log_a(10) = 0.25

a^(log_a(10)) = a^0.25    raising a to each power

10 = a^0.25               using the inverse (with x=10)

That's all there is to it.

For more on antilogs, go to our search page and enter the keyword
antilog  . Here is one of the answers you'll find:

Anti-Logarithms
http://mathforum.org/dr.math/problems/elisabeth.02.27.01.html

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Logs

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search