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

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Answer 1 · 2016-05-19T05:55:38

$x = 0.3041$ $x=0.3041$ , nearly.

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

This is a quadratic in $4^{3 x}$ $4^(3x)$ . Solving,

$4^{3 x} = {(4^{3})}^{x} = 64^{x} = \frac{1 \pm \sqrt{37}}{2} = 3.5414, - 2.5414$ $4^(3x)=(4^3)^x= 64^x=(1+-sqrt 37)/2=3.5414, -2.5414$ , nearly.

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

So, inversely, $x = \frac{log (3.5414)}{log 64} = 0.3041$ $x = log (3.5414)/log 64=0.3041$ , nearly. .

How do you solve 163x=64x+916^(3x) = 64^x + 9?