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?

I think the most straightforward approach is computing $a$ in terms of $b$.
$3^a=4^b$ so $a = \log_34^b=b\log_34$
$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 9^{ \log_34}+16^{\log_43}=3^{ 2\log_34}+4^{2\log_43}=16+9=25$
