A) \[\frac{1}{4}\]
B) 4
C) \[\frac{1}{2}\]
D) 2
Correct Answer: B
Solution :
[b] If \[{{27}^{k}}=\frac{9}{{{3}^{k}}}\] \[\Rightarrow {{3}^{3k}}=\frac{9}{{{3}^{k}}}\Rightarrow {{3}^{4k}}=9\] \[[{{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}]\] \[\Rightarrow {{9}^{2k}}=9\Rightarrow k=\frac{1}{2}\] \[\left[ {{a}^{m}}={{a}^{n}}\,\,then\,\,m=n \right]\] \[\Rightarrow \frac{1}{{{k}^{2}}}=4\]You need to login to perform this action.
You will be redirected in
3 sec