How do you solve $27^{2 x + 1} = 9^{4 x}$ $27^(2x+1) = 9^(4x)$ ?

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Answer 1 · 2016-06-10T23:47:24

$x = \frac{3}{2}$ $x = 3/2$

$27 = 3^{3}$ $27=3^3$ and $9 = 3^{2}$ $9 = 3^2$

then

$27^{2 x + 1} = 9^{4 x} \equiv {(3^{3})}^{2 x + 1} = {(3^{2})}^{4 x}$ $27^(2x+1) = 9^(4x) equiv (3^3)^{2x+1} = (3^2)^{4x}$

or

$3^{3 (2 x + 1)} = 3^{8 x}$ $3^{3(2x+1)} = 3^{8x}$

so

$3 (2 x + 1) = 8 x$ $3(2x+1)=8x$ Solving for $x$ $x$ gives $x = \frac{3}{2}$ $x = 3/2$

How do you solve 272x+1=94x27^(2x+1) = 9^(4x)?