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### 2^x = x^2

```
Date: 02/13/2002 at 23:26:33
From: Jeremy Kelly
Subject: Exponents

Recently I was trying to do some random math problems to get a little
smarter when I came across something that I couldn't do.

2^x = x^2

I noticed first that there were 2 roots (2 and 4) but there was one
that I didn't know of that was negative. Thinking that this was too
obvious, I switched the problem to 5^x = x^2 and tried to solve. I
took the ln of both sides and tried from there, but I didn't really
know where to go. Using some big 'o' and such, I deduced that x had to
be negative and there was only one Real value for x that worked, (and
graphing proved that), but I didn't know how to find it without
graphing it. If you could help that would be great.

Jeremy Kelly
```

```
Date: 02/14/2002 at 09:24:29
From: Doctor Rick
Subject: Re: Exponents

Hi, Jeremy.

I don't know of a way to solve your problem analytically, but I know a
method by which we can find successively better approximations to the
solution. It's called the Newton-Raphson method, and it can be
understood from graphical considerations. Here is an explanation:

Cow Grazing Half the Circle: Newton-Raphson Method
http://mathforum.org/dr.math/problems/julus1.18.98.html

2^x = x^2 for what x?

into a function:

f(x) = 2^x - x^2

Now the question is, for what x is f(x) = 0; or, what are the roots of
f(x)? The Newton-Raphson method starts with some first guess, x[0],
and finds the next guess, x[1], by a formula. Then, using this guess,
we apply the same formula to find a new guess, x[2]. We continue until
we're as close as we wish. The formula is

x[i+1] = x[i] - f(x[i])/f'(x[i])

We need f'(x), the derivative of f(x). It is

f'(x) = 2^x * ln(2) - 2x

Thus the formula for our problem is

x[i+1] = x[i] - (2^x[i]-x^2)/(2^x[i]*ln(2)-2x)

You can set this up in a spreadsheet. Then try different first guesses
x[0]. You'll find that the algorithm zeroes in on one of the three
roots, depending on the starting value. If I start with x[0] = 0, I
get the root:

x = -0.766664696

after 5 iterations. You can verify:

2^-0.766664696   = 0.587774756
(-0.766664696)^2 = 0.587774756

x[0] = 3, I get the root x = 4. You have observed that there are three
roots.

I'll leave the variant with 5^x, and others, up to you.

- Doctor Rick, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Functions

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