How do you solve $27^(4x)=9^(x+1 )$ ?

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Answer 1 · 2015-08-17T13:03:51

$color(blue)(x=1/5$

$27^(4x)=9^(x+1)$

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

So our expression can be re written as

$3^(3(4x))=3^(2(x+1)$

$3^(12x)=3^(2x+2)$

As the bases are equal we can equate the exponents.

$12x=2x+2$

$12x-2x=2$

$10x=2$

$color(blue)(x=1/5$

How do you solve 27^(4x)=9^(x+1 ) 27^(4x)=9^(x+1 ) ?