# Pre Calculus

Determine an approximate solution to the inequality 2^x=x^2.

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1. You typed an equation, not an inequality

If you meant 2^x > x^2 , follow my method in your previous post, or

let's look at 2^x = x^2
obviously one trivial solution is x = 2
what about x = 4 ?
LS = 2^4 = 16
RS = 4^2 = 16 , yup!

the graph shows another solution to the left of the origin.
Wolfram shows this:
http://www.wolframalpha.com/input/?i=+2%5Ex+%3D+x%5E2

x = -.766665

so x = -.766665 , 2, 4

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posted by Reiny

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