Question

How do you solve `16^(3x) = 64^x + 9`?

Answer 1

`x=0.3041`, nearly.

Explanation:

The equation can be reorganized as `(4^(3x))^2-4^(3x)-9=0`.

This is a quadratic in `4^(3x)`. Solving,

`4^(3x)=(4^3)^x= 64^x=(1+-sqrt 37)/2=3.5414, -2.5414`, nearly.

As #64^x>0, the negative root is inadmissible.

So, inversely, `x = log (3.5414)/log 64=0.3041`, nearly. .