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

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Answer 1 · 2016-05-19T05:55:38

$x=0.3041$ , nearly.

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. .

How do you solve 16^(3x) = 64^x + 916^(3x) = 64^x + 9?