Find $x, y .$ If $27^{x + 1} = 9^{x + 3} = 3^{y}$

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Solution

$3^{3 x + 3} = 3^{2 x + 6} = 3^{y}$

$3^{3 x + 3} = 3^{y}$ and $3^{3 x + 3} = 3^{2 x + 6}$

Equating the powers with the same base, we get

$\Rightarrow 3 x + 3 = y$ and $3 x + 3 = 2 x + 6$

$\Rightarrow 3 x + 3 = y$ and $3 x - 2 x = 6 - 3 = 3$

$\Rightarrow 3 x + 3 = y$ and $x = 3$

Put $x = 3$ in $y = 3 x + 3$ we get $y = 3 \times 3 + 3 = 9 + 3 = 12$

$\Rightarrow y = 12$

$∴ y = 12$

Hence $x = 3, y = 12$

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