Composite Functions Using Logarithms

Date: 3/10/96 at 21:33:41
From: Anonymous
Subject: Precal

I have the following precal question and am having a bit
of difficulty with it.  Any help (even an idea about how to
approach it) would be helpful.  Thanks so much.

Suppose f and g are functions defined by f(x) = x+2 and 
g(x) = x. Find all x > -2  for which 

                g(x) * logbase3 f(x)
               3                       = f(x)

In case you can't read that, it says 

3 to the exponent (g(x) * the log f(x) base3) equals f(x).  

   Thanks.

Date: 3/21/96 at 13:32:36
From: Doctor Aaron
Subject: Re: Precal

Hello,

Good question.  

A good place to start is to see that you have an expression that
is close to the form:

  logbasea (f(x))
a                 .  

If this doesn't look useful, think about what the log function 
does. It is essentially the inverse of the exponential 
function (provided the exponential and the log have the same 
base).  

Composing a function with its inverse gives us what is called the 
identity function - it gives us back whatever we give it to 
operate on. That is, if we raise a number to a power, and then 
take the log (of the appropriate base) we get back the power to 
which we raised our number.  

Similarly, if we take the log of a number, then raise the base to 
the power (log(base)), we get our number back.  

so then   

 logbasea (f(x))
a                 = f(x).  

You may have noticed a problem.  You have g(x) in front of the 
log, so we can't simplify as nicely as we would like to.  We can 
change this by using the relation that a*log(b) = log(b^a) where 
^ is an exponent.

This relation makes sense, because the log just asks us how many 
times we have to multiply a number by itself to get another 
number. (Actually we can't multiply something by itself a 
fractional number of times, but log is just the extension of the 
above definition to the real numbers)

Well, if we have to muliply (base) by itself n times to get b, 
we'll have to multiply (base) by itself a*n times to get b^a.

Now rewrite  g(x) * logbase3 f(x) as logbase3 (f(x)^g(x)).

You should be able to use the above ideas to make some headway.
If you run into some more trouble, consult a teacher or write 
back.

-Doctor Aaron,  The Math Forum