If $a >1$, $b >1$ and $p = \frac{\log_b{(\log_b{a})}}{\log_b{a}}$, then what's $a^p$?

Question

If $a >1$, $b >1$ and $p = \frac{\log_b{(\log_b{a})}}{\log_b{a}}$, then what's $a^p$?

My steps:

$$a^p=a^{\frac{\log_b{(\log_b{a})}}{\log_b{a}}}$$

But from then I had no idea what to do and would like assistance.

Answer 1

We have by change of base $$ \frac{\log_b(\log_b(a))}{\log_b(a)} = \log_a(\log_b(a)) $$ Therefore, $$ a^p = \log_b(a) $$

Answer 2

Hint: since the "outer" logarithms are $\log_b {(\text{stuff})}$ you might think that you need $b^{\text{something}}$ rather than $a^{\text{something}}$ - fortunately, using logarithms, you can write $a=b^{?}$

Answer 3

Hint: $p = \frac{\log_b{(\log_b{a})}}{\log_b{a}},$ so $p\log_b{a} = \log_b{(\log_b{a})},$ and finally $\log_b{a^p} = \log_b{(\log_b{a})}.$

Answer 4

$ p = \frac{\log_b{(\log_b{a})}}{\log_b{a}} => p = \log_a{(\log_b(a))} => a^p = \log_b{a} $ . Hope this helps .