How to find the value of $ 81^{1/\log_53} + 27^{\log_9{36}} + 3^{4/ \log_79}$?

Question

The question seems easy but is not. I tried making bases as 3 but log is in power and base of log too is different.

Answer 1

Bear in mind $\log_a b = \frac 1{\log_b a}$ and $\log_{a^k} b = \log_{a}b^{\frac 1k}$.

So $81^{1/\log_53} + 27^{\log_9{36}} + 3^{4/ \log_79}=$

$(3^4)^{\log_3 5} + (3^3)^{\log_{3^2}6^2} + 3^{4\log_{3^2} 7}=$

$(3^{\log_3 5})^4 + (3^{\log_3 6})^3 + (3^{\log_3 \sqrt 7})^4$

And from there you just finish it.

$5^4 + 6^3 + \sqrt{7}^4 =5^4 + 6^2 + 7^2$

$625 + 8*27 + 49$

$=.... $.

Well, heck, I don't do addition.