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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 
log_b(a) are related. I'll start with the base-conversion formula, 
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)

If we had thought of that to start with, the problem would have been 
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

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