A) \[\frac{1}{2}\]
B) \[-\frac{19}{18}\]
C) \[\frac{3}{10}\]
D) \[\frac{12}{17}\]
Correct Answer: B
Solution :
We have, \[\frac{{{({{3}^{-2}})}^{2y-1}}{{({{3}^{4}}\cdot {{10}^{-4}})}^{1/3}}}{{{3}^{5/2}}}=\frac{{{3}^{-(2y-5)}}\cdot {{3}^{3\left( \frac{y-1}{3} \right)}}}{{{10}^{4/3}}}\] \[\Rightarrow \frac{{{3}^{-4y+2+\frac{4}{3}-\frac{5}{2}}}}{{{10}^{4/3}}}=\frac{{{3}^{-2y+5+y-1}}}{{{10}^{4/3}}}\] \[\Rightarrow {{3}^{-4y+\frac{5}{6}}}={{3}^{-y+4}}\] On comparing, we get \[-4y+\frac{5}{6}=-y+4\] \[\Rightarrow -3y=4-\frac{5}{6}\Rightarrow -3y=\frac{24-5}{6}\] \[\Rightarrow y=\frac{-19}{18}\]
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