<s>[INST] what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac{2a}{a} } = 4^{ b } + 3^{ a }$$
Firstly is this right? Secondly how to complete?
 [/INST] Let $a = 1$. Then $4^b = 3 \Rightarrow b = \log_4 3 = \frac{\log 3}{\log 4}$.
Therefore:
$$9^{a/b} + 16^{b/a}$$
$$=9^{\log 4 / \log 3} + 16^{\log3 / \log 4}$$
$$=3^{2 \log 4 / \log 3} + 4^{2 \log 4 / \log 3}$$
$$=3^{2 \log_3 {4} } + 4^{2 \log_4 {3}}$$
$$= \left( 3^{\log_3 4} \right )^2 + \left( 4^{\log_4 3} \right)^2 $$
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