A) \[\frac{3}{2}\]
B) \[{{\left( \frac{4}{3} \right)}^{7}}\]
C) \[\frac{1}{3}\]
D) \[\frac{1}{2}\]
Correct Answer: B
Solution :
(b) Let no be \[{{\left( \frac{4}{3} \right)}^{a}}\] Then, \[{{\left( \frac{4}{3} \right)}^{-5}}\times {{\left( \frac{4}{3} \right)}^{a}}=\frac{16}{9}or{{\left( \frac{4}{3} \right)}^{2}}\] \[\Rightarrow {{\left( \frac{4}{3} \right)}^{(-5+a)}}={{\left( \frac{4}{3} \right)}^{2}}\] \[\Rightarrow a-5=2\Rightarrow a=7\] \[={{\left( \frac{4}{3} \right)}^{7}}\]. Mind of Mathematician: See the ingenuity, as we have not started by saying that let the no. be ?a?. Rather, we have started by saying, ?Let the no be \[{{\left( \frac{4}{3} \right)}^{n}}\].
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