Question

If `logx/(a^2+ab+b^2)=log y/(b^2+bc+c^2)=log z/(c^2+ca+a^2)` then find `x^(a-b)* y^(b-c)*z^(c-a)=`?

Answer 1

Let
`logx/(a^2+ab+b^2)=log y/(b^2+bc+c^2)=log z/(c^2+ca+a^2)=k`

so
`logx=k(a^2+ab+b^2)`

`log y=k(b^2+bc+c^2)`

`log z=k(c^2+ca+a^2)`

Now

`log[x^(a-b)* y^(b-c)*z^(c-a)]`

`=(a-b)logx+ (b-c)logy+(c-a)logz`

`=(a-b)xxk(a^2+ab+b^2)+ (b-c)lxxk(b^2+bc+c^2)+(c-a)lxxk(c^2+ca+a^2)`

`=k(a^3-b^3+b^3-c^3+c^3-a^3)=kxx0=0=log1`

Hence

`x^(a-b)* y^(b-c)*z^(c-a)=1`

Answer 2

`x^(a-b)*y^(b-c)*z^(c-a) =1`

Explanation:

We know that,

`(1)log_zA=P<=>A=z^P`

We take ,the base of log as `10`

Let ,

`logx/(a^2+ab+b^2)=logy/(b^2+bc+c^2)=logz/(c^2+ca+a^2)=k`

So ,

`logx=k(a^2+ab+b^2)=>x=10^(k(a^2+ab+b^2))`

`logy=k(b^2+bc+c^2)=>y=10^(k(b^2+bc+c^2)`

`logz=k(c^2+ca+a^2)=>z=10^(k(c^2+ca+a^2)`

Now ,

`x^(a-b)=10^(k(a-b)(a^2+ab+b^2))=10^(k(a^3-b^3)`

`y^(b-c)=10^(k(b-c)(b^2+bc+c^2))=10^(k(b^3-c^3)`

`z^(c-a)=10^(k(c-a)(c^2+ca+a^2))=10^(k(c^3-a^3)`

Hence ,

`x^(a-b)*y^(b-c)*z^(c-a) =10^(k(a^3-b^3) )*10^(k(b^3-c^3))*10^(k(c^3-a^3)`

`x^(a-b)*y^(b-c)*z^(c-a) =10^((k(a^3-b^3) )+(k(b^3-c^3))+(k(c^3-a^3)`

`x^(a-b)*y^(b-c)*z^(c-a) =10^(k(a^3-b^3+b^3-c^3+c^3-a^3)`

`x^(a-b)*y^(b-c)*z^(c-a) =10^(k(0)) =10^0=1`

`x^(a-b)*y^(b-c)*z^(c-a) =1`