If logxa2+ab+b2=logyb2+bc+c2=logzc2+ca+a2 then find xabybczca=?

2 Answers
Aug 11, 2018

Let
logxa2+ab+b2=logyb2+bc+c2=logzc2+ca+a2=k

so
logx=k(a2+ab+b2)

logy=k(b2+bc+c2)

logz=k(c2+ca+a2)

Now

log[xabybczca]

=(ab)logx+(bc)logy+(ca)logz

=(ab)×k(a2+ab+b2)+(bc)l×k(b2+bc+c2)+(ca)l×k(c2+ca+a2)

=k(a3b3+b3c3+c3a3)=k×0=0=log1

Hence

xabybczca=1

Aug 11, 2018

xabybczca=1

Explanation:

We know that,

(1)logzA=PA=zP

We take ,the base of log as 10

Let ,

logxa2+ab+b2=logyb2+bc+c2=logzc2+ca+a2=k

So ,

logx=k(a2+ab+b2)x=10k(a2+ab+b2)

logy=k(b2+bc+c2)y=10k(b2+bc+c2)

logz=k(c2+ca+a2)z=10k(c2+ca+a2)

Now ,

xab=10k(ab)(a2+ab+b2)=10k(a3b3)

ybc=10k(bc)(b2+bc+c2)=10k(b3c3)

zca=10k(ca)(c2+ca+a2)=10k(c3a3)

Hence ,

xabybczca=10k(a3b3)10k(b3c3)10k(c3a3)

xabybczca=10(k(a3b3))+(k(b3c3))+(k(c3a3)

xabybczca=10k(a3b3+b3c3+c3a3)

xabybczca=10k(0)=100=1

xabybczca=1