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

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

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

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(D) Given equation: ${2^{x + 2}} \cdot {3^{3x/(x - 1)}} = {3^2}$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 $1$: $x + 2 = 0 \implies x = -2$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}}$Thus,the roots are $-2$ and $1 - \frac{{\log 3}}{{\log 2}}$.

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

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