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### A Logarithmic Equation

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Date: 08/06/98 at 10:46:14
From: Morgan Stratton
Subject: Solution to log/exp algebra problem

I had this problem solved before but lost the answer. Here's the
problem:

A*2^(x/y) = B

Solve for x using only log (base 10) elements.

I reviewed some textbooks and believe I had it solved for log(base 2),
x = (log2B-log2A)/y, but I need it in standard log form.

Thanks.
```

```
Date: 08/06/98 at 11:30:38
From: Doctor Mike
Subject: Re: Solution to log/exp algebra problem

Hi Morgan,

Some really handy log relationships are log(a*b) = log(a) + log(b) and
log(a/b) = log(a) - log(b). These hold whether the logs are base 10,
base e, base 2 or whatever.

I am a little confused because you speak both of base 2 and base 10
logs in your first paragraph, so I'll just go with 2 because it fits.
By that, I mean that there is a power of 2 in your problem equation.

The original is A*2^(x/y) = B. I suppose A and B are constants and
x and y are two variables related by this equation. I'm going to show
a series of steps where I do the same thing to both sides of the
equation. The final version will be what you want. Okay?

Take the log (base 2) of both sides. The right side will then become
log(B). The left side is the log of a product, so using one of the
handy relations I gave above, you convert that to the sum of the logs
of the two things that were multiplied together, giving:

log(A) + log( 2^(x/y) ) = log(B)

The next step is a little subtle, but it is really important for you to
understand it. As I have said, I am working with base 2 logs. This
means that when you raise 2 to the power log(s), you get s. Now, what
power do you have to raise 2 to in order to get 2^(x/y)? Well, that is
a no brainer! It's right there under our noses. 2^(x/y) is already in
the form of 2 raised to a power, so you have to raise 2 to the power:

(x/y)

to get 2^(x/y). So, the net of all this is that log(2^(x/y)) is x/y.
The equation then reduces to:

log(A) + x/y  =  log(B)

Now, to clear the denominator multiply both sides by y to get:

y*log(A) + x  =  y*log(B)

You want to solve for x, so to get x by itself, subtract y*log(A)
from both sides of the equation. This gives:

x  =  y*log(B) - y*log(A)

Then, by the Distributive Law you get:

x  =  y*(log(B) - log(A))   =   (log(B) - log(A))*y

I'm going into all this detail because somewhere you missed something
and wound up with x being equal to (log(B) - log(A)) divided by y.

Anyway, we are finally near the light at the end of the tunnel. By
the other handy relation I told you about, log(B) - log(A) = log(B/A).
So the final and simplest version is:

x  = y*log(B/A)

Let's check that answer. Does it satisfy A*2^(x/y) = B? Just plug
that final expression for x into the original equation, and you will
find that the left side really does simplify to just "B".

I hope this helps.

- Doctor Mike, The Math Forum
Check out our web site! http://mathforum.org/dr.math/
```
Associated Topics:
High School Logs
Middle School Logarithms

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