Question #5119c

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Answer 1 · 2016-11-29T17:05:01

Let

$3^x=7^y=63^z=k$

So $3=k^(1/x)$

$7=k^(1/y)$

and

$63=k^(1/z)$

$=>3^2xx7=k^(1/z)$

$=>(k^(1/x))^2xxk^(1/y)=k^(1/z)$

$=>k^(2/x)xxk^(1/y)=k^(1/z)$

$=>k^(2/x+1/y)=k^(1/z)$

$=>2/x+1/y)=1/z$

$=>(2y+x)/(xy)=1/z$

$=>z(2y+x)=xy$