The roots of the equation 2^{x + 2} cdot 27^{ | vedclass

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Question

(D) Given equation: $2^{x + 2} \cdot (3^3)^{x/(x - 1)} = 3^2$Taking $\log$ on both sides:$(x + 2)\log 2 + \frac{3x}{x - 1}\log 3 = 2\log 3$$(x + 2)\lo

Vedclass · Accepted Answer

$-2, 1 - \frac{\log 3}{\log 2}$

Vedclass · Answer

(D) Given equation: $2^{x + 2} \cdot (3^3)^{x/(x - 1)} = 3^2$Taking $\log$ on both sides:$(x + 2)\log 2 + \frac{3x}{x - 1}\log 3 = 2\log 3$$(x + 2)\log 2 = 2\log 3 - \frac{3x}{x - 1}\log 3$$(x + 2)\log 2 = \log 3 \left( 2 - \frac{3x}{x - 1} \right)$$(x + 2)\log 2 = \log 3 \left( \frac{2x - 2 - 3x}{x - 1} \right)$$(x + 2)\log 2 = \log 3 \left( \frac{-x - 2}{x - 1} \right)$$(x + 2)\log 2 = -\log 3 \left( \frac{x + 2}{x - 1} \right)$$(x + 2) \left( \log 2 + \frac{\log 3}{x - 1} \right) = 0$Case $1$: $x + 2 = 0 \implies x = -2$Case $2$: $\log 2 + \frac{\log 3}{x - 1} = 0 \implies \frac{\log 3}{x - 1} = -\log 2$$x - 1 = -\frac{\log 3}{\log 2} \implies x = 1 - \frac{\log 3}{\log 2}$

The roots of the equation $2^{x + 2} \cdot 27^{x/(x - 1)} = 9$ are given by

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