Show that if $3^x+4^y=12^z \Rightarrow z=\frac{xy}{x+y}$

Question

Show that if $3^x+4^y=12^z \Rightarrow z=\frac{xy}{x+y}$

$\textbf{My Attempt}$ $$3^x+4^y=12^z \Rightarrow z=\frac{xy}{x+y}$$ $$\Rightarrow \log(3^x4^y)=\log(12^z)\Rightarrow \log \big(\frac{3^x4^y}{12^z}\big)=0\Rightarrow \frac{3^x4^y}{12^z}=1 \Rightarrow 3^x4^y=12^z$$ $$\Rightarrow 3^x4^y=3^x+4^y$$ How do you find what $z$ equals from this? Any help would be appreciated.

Answer 1

I believe the problem is $$3^x=4^y=12^z$$

Let each ratio $=k$

$$12=3\cdot4\implies k^{1/z}=k^{1/x}\cdot k^{1/y}=k^{1/x+1/y}$$

$$\implies 1/z=1/x+1/y$$

if $k\ne1\iff xyz\ne0$

Answer 2

It's wrong.

Try $x=z=1$ and $y=\log_23.$