<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 $3^a=4^b=k^{ab}\implies 3=k^b,4=k^a$
$$9^{a/b}+16^{a/b}$$
$$=(3)^{2a/b}+4^{2b/a}$$
$$=(k^b)^{2a/b}+(k^a)^{2b/a}$$
$$=k^{2b}+k^{2a}$$
$$=(k^b)^2+(k^a)^2=?$$
</s>