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

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

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

Find the roots of the equation $2^{x + 2} \cdot 27^{x/(x - 1)} = 9$.

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