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

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Answer 1 · 2015-07-08T08:39:26

$x = 2$ $color(blue)(x=2$

$9^{2 x - 1} = 3^{2 . (2 x - 1)} = 3^{4 x - 2}$ $9^(2x-1) = 3^(2.(2x-1)) = 3^(4x-2)$

$27^{x} = 3^{3 x}$ $27^x = 3^(3x)$

Upon equating:

$3^{4 x - 2} = 3^{3 x}$ $3^(4x-2) = 3^(3x)$
Since the base is equal , we can equate the powers:

$4 x - 2 = 3 x$ $4x-2 =3x$
$x = 2$ $color(blue)(x=2$

How do you solve 9^(2x-1)=27^x92x−1=27x9^(2x-1)=27^x?