Question

How do you solve `16^x = x^2` ?

Answer 1

Took it up to a point beyond which I did not know how to mathematically resolve.

Explanation:

Given: `16^x =x^2`

To get rid of the `x` on the right side take logs to base `x`

`xlog_x(16)=2`

`log_x(16)=2/x`

`=> x^(2/x)=16`

`=> (x^(1/x))^2 =4^2`

`=> x^(1/x)=4`
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You could do:

`1/x ln(x)=ln(4)`

But I have idea where this would take you!

Answer 2

`x = -1/2`

Explanation:

The graph of `16^x` intersects the graph of `x^2` once, at `(-1/2, 1/4)`

`16^(-1/2) = 1/sqrt(16) = 1/4 = (-1/2)^2`

graph{(y-x^2)(y-16^x) = 0 [-5.313, 4.687, -0.7, 4.3]}