A) \[lo{{g}_{2}}\left( \frac{2}{3} \right),-2\]
B) \[-3,3\]
C) \[-2,1-\frac{\log 3}{\log 2}\]
D) \[-2,1-{{\log }_{2}}3\]
Correct Answer: C
Solution :
[c] \[\because {{2}^{x+2}}{{.3}^{\frac{3x}{x-1}}}=9\] \[\Rightarrow {{2}^{x+2}}{{.3}^{\frac{3x}{x-1}}}={{3}^{2}}\] Taking log on both sides, we have \[\left( x+2 \right)log2+\frac{3x}{x-1}.log3=2log3\] \[\Rightarrow \left( x+2 \right)log2+\left( \frac{3x}{x-1}-2 \right).log3=0\] \[\Rightarrow \left( x+2 \right)log2+\left( \frac{x+2}{x-1} \right).log3=0\] \[\Rightarrow \left( x+2 \right)log2+\left[ \log 2+\left( \frac{1}{x-1} \right)\log 3 \right]=0\] If \[x+2=0\Rightarrow x=-2\] If \[log2+\left( \frac{1}{x-1} \right)log3=0\] \[x-1=-\frac{\log 3}{\log 2}\] \[x=1-\frac{\log 3}{\log 2}\] Hence, option [c] is correct.
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