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

1 Answer
May 19, 2016

#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. .