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Answer 1 · 2016-10-13T21:20:47

$y = 3 x - 1$ $y = 3x-1$

We will use the following properties of exponents and of logarithms:

With those

$\frac{3^{y}}{27^{x}} = \frac{1}{3}$ $3^y/27^x=1/3$

$\Rightarrow \frac{3^{y}}{{(3^{3})}^{x}} = 3^{- 1}$ $=> 3^y/(3^3)^x = 3^(-1)$

$\Rightarrow \frac{3^{y}}{3^{3 x}} = 3^{- 1}$ $=> 3^y/3^(3x) = 3^(-1)$

$\Rightarrow 3^{y} = 3^{3 x} \cdot 3^{- 1}$ $=> 3^y = 3^(3x)*3^(-1)$

$\Rightarrow 3^{y} = 3^{3 x - 1}$ $=> 3^y = 3^(3x-1)$

$\Rightarrow {log}_{3} (3^{y}) = {log}_{3} (3^{3 x - 1})$ $=> log_3(3^y) = log_3(3^(3x-1))$

$:. y = 3x-1$