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)=#?

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Answer 1 · 2018-08-11T13:43:02

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 · 2018-08-11T13:59:10

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

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#