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

Aug 17, 2015

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#### Explanation:

${27}^{4 x} = {9}^{x + 1}$

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

So our expression can be re written as

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

${3}^{12 x} = {3}^{2 x + 2}$

As the bases are equal we can equate the exponents.

$12 x = 2 x + 2$

$12 x - 2 x = 2$

$10 x = 2$

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