1 equation 3 variables

I think mine looks stupid compared to other solutions but it's worth a try!

$2^{x} = 3^{y} = 12^{z}$

By multiplying, therefore,

$2^{x} \times 3^{y} = 12^{2z}$

and since 12 can be broken up into $2^{2} , 3$

then

$2^{x} \times 3^{y} = 2^{4z} \times 3^{2z}$

things getting clearer now,

therefore,

$2^{x} = 2^{4z} , 3^{y} = 3^{2z}$

then $x = 4z , y = 2z$

by substituting in the required problem,

$\frac{z (x + 2y)}{x y} = \frac{z (4z + 2(2z))}{4z \times 2z} = \frac{8z^2}{8z^2} = \boxed{1}$

The other way is to use $\log{}$ .

$2^x=3^y =12^z\quad \Rightarrow x\log{2} = y\log{3} = z\log{12} \\ \Rightarrow y = \dfrac {x\log{2}}{\log{3}} \quad \Rightarrow z = \dfrac {x\log{2}}{\log{12} }$

$\begin{aligned} \frac {z(x+2y)}{xy} & = \frac {\frac {x\log{2}}{\log{12}}\left(x + 2\frac {x\log{2}}{\log{3}} \right)}{x\frac {x\log{2}}{\log{3}}} \\ & = \frac {\log{2}}{\log{12}}\left(\frac {\log{3}+2\log{2}}{\log{3}} \right) \times \frac{\log{3}}{\log{2}} \\ & = \frac{\log{2}\times \log{12}\times \log{3}}{\log{12}\times \log{3}\times \log{2}} \\& = \boxed{1} \end{aligned}$

AN easy solution for this could be GIVEN that $\boxed{2^x}$ = $\boxed{3^y}$ = $\boxed{12^z}$ $\Rightarrow$ LET $\boxed{2^x}$ = $\boxed{3^y}$ = $\boxed{12^z}$ = k then $k^\frac{1}{x}$ = 2 , $k^\frac{1}{y}$ = 3 and $k^\frac{1}{z}$ = 12 "if you are careful observant you would write it as- $k^\frac{1}{x}$ $k^\frac{1}{x}$ $k^\frac{1}{y}$ = $k^\frac{1}{z}$ Further since the bases are same power could be compared and the answer would be $\boxed{1}$

Let $2^x=3^y=12^z$ . This can be solved using some general term $a$ where

$x=\frac{\ln a}{\ln 2}$

$y= \frac{\ln a}{\ln 3}$

$z= \frac{\ln a}{\ln12}$

Such that $2^x=3^y=12^z$ yields

$2^{\frac{\ln a}{\ln 2}}=3^{\frac{\ln a}{\ln 3}}=12^{ \frac{\ln a}{\ln 12}}=a$

Then substitute $x, y, z$ in terms of $a$ so $\frac{z(x+2y)}{xy}$ becomes $\frac{\frac{\ln a}{\ln12}(\frac{\ln a}{\ln 2}+2\frac{\ln a}{\ln3})}{\frac{\ln a}{\ln 2 } \cdot \frac{\ln a}{\ln3}}$

Then just simplify

$\frac{\frac{1}{\ln12}(\frac{1}{\ln 2}+2\frac{1}{\ln3})}{\frac{1}{\ln 2} \cdot \frac{1}{\ln3}}$

$\frac{1}{\ln12}$ $\cdot$ $\frac {\frac{2\ln2+\ln3 }{\ln2 \cdot \ln3} }{ \frac{1}{\ln2} \cdot \frac{1}{\ln3} }$

$\frac{2\ln2 + \ln 3}{\ln 12}$ = $\frac{\ln4 +\ln3}{\ln12}$ = $\frac{\ln12}{\ln12}=1$

$2^x = 3^y = 12^z = (2^2\times3)^z$

$2^2 = (2^x)^{\frac{2}{x}} = (12^z)^{\frac{2}{x}} = 12^{\frac{2z}{x}}$

$3 = (3^y)^{\frac{1}{y}} = (12^z)^{\frac{1}{y}} = 12^{\frac{z}{y}}$

$2^2\times3 = 12^{\frac{2z}{x}}\times12^{\frac{z}{y}} = 12^{{\frac{2z}{x}} + {\frac{z}{y}}} = 12^{\frac{z(x+2y)}{xy}} = (2^2\times3)^{\frac{z(x+2y)}{xy}}$

$(2^2\times3)^1 = (2^2\times3)^{\frac{z(x+2y)}{xy}} \Longrightarrow\frac{z(x+2y)}{xy} = \boxed{1}$

Let's work a bit the equation: $2^x=12^z \rightarrow \textit{Lower the power } x \textit{ times}$ $2^\frac{x}{x}=12^\frac{z}{x}$ $2=12^\frac{z}{x} \rightarrow \textit{Raise to power 2}$ $4=12^\frac{2z}{x} \text{ }(1)$ Now let's work the second equation as well: $3^y=12^z \rightarrow \textit{Lower the power } y \textit{ times}$ $3^\frac{y}{y}=12^\frac{z}{y}$ $3=12^\frac{z}{y} \textit{ }(2)$ Upon mutiplying the pairs of left and right sides of equations $(1)$ and $(2)$ we get: $4×3=12^\frac{2z}{x}×12^\frac{z}{y}$ $12=12^{\frac{2z}{x}+\frac{z}{y}}$ $12^1=12^\frac{2yz+xz}{xy}$ Since we have equal bases we can solve only for the exponents, which, as well, have to be equal: $1=\frac{2yz+xz}{xy}$ $1= \frac{z(x+2y)}{xy}$ That's exactly what we are asked for and so the answer is $\frac{z(x+2y)}{xy}=\fbox{1}$ .

Great problem Swapnil Das

I found from data x and y in terms of z and then substituted to get 1.But this is not a very elegent solution

(2^x)×(3^y)=(12^z)×(12^z) =[2^(4z)]×[3^(2z)] Comparing both side powers.... x=4z,y=2z. Put the value of x and y in given equation, z(4z+2×2z)÷(4z×2z)=1.