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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/
```
Associated Topics:
High School Basic Algebra
High School Exponents

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