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?

$$(9)^{ \displaystyle\frac{a}{b} } + (16)^{ \displaystyle\frac{b}{a} } = \left(3^{\displaystyle \color{magenta}{2}}\right)^{ \displaystyle\frac{\color{red}{a}}{b} } + \left(4^{\displaystyle \color{brown}{2}}\right)^{ \displaystyle\frac{\color{blue}{b}}{a} } =\left(3^{\displaystyle \color{red}a}\right)^{ \displaystyle\frac{\color{magenta}2}{b} } + \left(4^{\displaystyle \color{blue}b}\right)^{ \displaystyle\frac{\color{brown}2}{a} }  $$
Now if $3^{\displaystyle a} = 4^{\displaystyle b}$, then 
$$\left(3^{\displaystyle a}\right)^{ \displaystyle \frac{2}{b} } + \left(4^{\displaystyle b}\right)^{ \displaystyle \frac{2}{a} } = \left(4^{\displaystyle b}\right)^{ \displaystyle \frac{2}{b} } + \left(3^{\displaystyle a}\right)^{ \displaystyle \frac{2}{a} }  =  25 $$
The rules are: 


*

*If you have three numbers $x,y$ and $z$, then $(x^y)^z = x^{y~\cdot ~z}= x^{z~\cdot~ y}=(x^z)^y$.

*If you have three numbers $x,y$ and $z\neq 0$, then $\frac{x~\cdot ~y}{z} = x~\cdot~ \frac{y}{z}=  y~\cdot~ \frac{x}{z} =\frac{y~\cdot~ x}{z}$.

