x^2=16^x <----- let's look at y = x^2 and y = 16^x
https://www.wolframalpha.com/input/?i=x%5E2%3D16%5Ex
It shows they intersect at x = -1/2
check:
if x = -1/2
LS = (-1/2)^2 = 1/4
RS = 16^(-1/2)
= 1/16^(1/2) = 1/√16 = 1/4
so x = -1/2
Solve:
x^2=16^x
2 answers
By using b^2=4ac: x^2 =16^2. 1=4*1*16^x. 1=2^2+4x. So let x=-1/2. 1=1.