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2^x = 8

```Date: 05/27/2003 at 09:34:52
From: Keith
Subject: 2^x = 8

How can I solve for x in the equation 2^x = 8?
```

```
Date: 05/27/2003 at 09:50:58
From: Doctor Ian
Subject: Re: 2^x = 8

Hi Keith,

It depends on what you mean by 'solve'. You can start substituting
values for x, and see which one works:

2^0 = 1
2^1 = 2
2^2 = 4
2^3 = 8        Success!

This doesn't work so well for something like 2^x = 9:

2^0 =  1
2^1 =  2
2^2 =  4
2^3 =  8       Too small
2^4 = 16       Too large

At that point, if you have a calculator with a 'y^x' key, you can
start trying other exponents:

2^3.1 =  8.57
2^?   =  9           It's somewhere between 3.1 and 3.5
2^3.5 = 11.31

Or, you can use logarithms.

By definition, if

x
b  = a

then

log a = x
b

In this case, if

x
2  = 8

then

log 8 = x
2

To find a logarithm, you would typically use a calculator or table of
logarithms (in much the same way that you'd find the sine or cosine of
an angle, and for much the same reason).

The fly in the ointment is that most calculators and log tables assume
that your base (i.e., 'b' in the description above) is either 10 or e.
To deal with other bases (like 2), you need to use the 'change of
base' formula, which is explained here:

Logarithms: Solving for T
http://mathforum.org/library/drmath/view/55561.html

Of course, this is kind of a pain to remember if you don't use it on a
regular basis, which is why I'd normally use the first method I
described, i.e., try exponents until I get a value that I feel is
close enough to 9 for my purposes.

There is a third approach you can use. If you graph the function

f(x) = 2^x - 8

using a graphing calculator, the value of x where the graph crosses
the x-axis is the value that satisfies the original equation. Do you
see why that's true?

or anything else.

- Doctor Ian, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
Middle School Logarithms

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