How do you solve $9^(2x)=27^(x-1)$ ?

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Answer 1 · 2015-08-19T06:10:42

$color(blue)(x=-3$

$9^(2x)=27^(x−1$

We know that
$9=3^2$
$27=3^3$

So,

$3^(2(2x))=3^(3(x−1)$

$3^(4x)=3^(3x−3)$

Now as bases are equal we can equate exponents

$4x=3x-3$
$color(blue)(x=-3$

How do you solve 9^(2x)=27^(x-1) 9^(2x)=27^(x-1)?