How do you solve #2^x = x^2# ?

Question

#L1: y= x^2#
#L2: y= x^2#
Find the intersection.

#2^x =x^2#

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Answer 1 · 2018-03-03T18:27:29

Real solutions:

#x=2#, #x=4# or #x = -2/ln 2 W(ln 2/2) ~~ -0.766665#

Given:

#2^x = x^2#

This has two rational solutions #x=2# and #x=4#, since:

#2^color(blue)(2) = 4 = color(blue)(2)^2#

#2^color(blue)(4) = 16 = color(blue)(4)^2#

Let us look at the graphs of #x^2# and #2^x# to see if we should expect more solutions:

graph{(y-x^2)(y-2^x) = 0 [-3.353, 6.65, -4, 20]}

It seems there is a third point of intersection somewhere around #x = -3/4#

Noting that this is negative, take the negative square root of #x^2# to find:

#x = -2^(x/2) = -e^(ln 2/2 x)#

Multiply both ends by #ln 2/2 e^(-ln 2/2 x)# to get:

#(-ln 2/2x) e^(-ln 2/2 x) = ln 2/2#

This is in the form:

#ze^z = f(z)#

So we can use the Lambert W function to find:

#-ln 2/2x = W(ln 2/2)#

So:

#x = -2/ln 2 W(ln 2/2) ~~ -0.766665#