Question #5119c

Question

1195 views around the world

You can reuse this answer
Creative Commons License

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#