An algebra problem by Keerthi Reddy

$\Large81^{\frac{1}{\log_5 3}} + 27^{\log_9 36} + 3^{\frac{4}{\log_7 9}}\\ =\Large(3^4)^{\frac{1}{\left(\frac{\log_3 3}{\log_3 5}\right)}} + (3^3)^{\frac{\log_3 36}{\log_3 9}} + 3^{\frac{4}{\left(\frac{\log_3 9}{\log_3 7}\right)}}\\ =\Large3^{4 \times \frac{\log_3 5}{1}} + 3^{3 \times \frac{\log_3 36}{2}} + 3^{\frac{4 \log_3 7}{2}}\\ =\Large3^{4\log_3 5} + 3^{\frac{3}{2}\log_3 36} + 3^{2 \log_3 7}$

Notice that every term in the expression is in the form of $\large a^{b \log_a c}$

Now, let $\large y=a^{b \log_a c}$

$\large \log_a y = \log_a a^{b \log_a c}\\ \large \log_a y = b\log_a c \log_a a\\ \large \log_a y = \log_a c^b\\ \large y=c^b\\ \large a^{b \log_a c} = c^b$

We can substitute this identity into the expression:

$\Large3^{4\log_3 5} + 3^{\frac{3}{2}\log_3 36} + 3^{2 \log_3 7}\\ =\large 5^4 + 36^{\frac{3}{2}} + 7^2\\ =\large 625 + 216 + 49\\ =\large \boxed{890}$