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

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Answer 1 · 2015-08-19T06:10:42

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

$9^{2 x} = 27^{x - 1}$ $9^(2x)=27^(x−1$

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

So,

$3^{2 (2 x)} = 3^{3 (x - 1)}$ $3^(2(2x))=3^(3(x−1)$

$3^{4 x} = 3^{3 x - 3}$ $3^(4x)=3^(3x−3)$

Now as bases are equal we can equate exponents

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

How do you solve 92x=27x−1 9^(2x)=27^(x-1)?