Problem: If a>1,b>1a>1, b>1a>1,b>1 and p=b(ba)bap=\dfrac{\log _{b}\left(\log _{b} a\right)}{\log _{b} a}p=logbalogb(logba), then apa^{p}ap equals
Answer Choices:
A. 111
B. bbb
C. ab\log _{a} blogab
D. ba\log _{b} alogba
E. abaa^{\log _{b a}}alogba
Solution:
We have,
p(ba)=b(ba)b(aP)=b(ba),aP=ba.\begin{aligned} p\left(\log _{b} a\right)&=\log _{b}\left(\log _{b} a\right) \\ \log _{b}\left(a^{P}\right) & =\log _{b}\left(\log _{b} a\right), \\ a^{P} & =\log _{b} a . \end{aligned} p(logba)logb(aP)aP=logb(logba)=logb(logba),=logba.