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)=?

2 Answers
Aug 11, 2018

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

Aug 11, 2018

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