Asked by Hanu
Determine an approximate solution to the inequality 2^x=x^2.
Answers
Answered by
Reiny
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
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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