Solve for X Algebraically

Date: 07/24/2003 at 21:09:34
From: John
Subject: Exponents

Many people (including myself) have been stumped by this question:

   Solve for x algebraically, (no graphing)

   2^x + 2^(-x) = 5

I tried doing different things, but ended up just going in a circle.

Date: 07/24/2003 at 23:59:07
From: Doctor Ian
Subject: Re: Exponents

Hi John,

Let's begin by writing it this way:

          1   
  2^x  + --- = 5
         2^x

Now we have a fraction, and the easiest way to get rid of a fraction
is to multiply both sides by the denominator:

  (2^x)(2^x) + 1 = 5(2^x)

     (2^x)^2 + 1 = 5*(2^x)

Hmmmm.  This looks a little like a quadratic equation, doesn't it? 
Suppose we make the substitution

  u = 2^x

Then we have

       u^2 + 1 = 5u

  u^2 - 5u + 1 = 0

Now, suppose we solve this and end up with 

  u = [something]

This means that 

  2^x = [something]

which is simpler to deal with. 
  
Does this help?

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