If loga/(b-c)=logb/(c-a)=logc/(a-b)logabc=logbca=logcab then the numerical value of a^a*b^b*c^c=?aabbcc=?

2 Answers

Let loga/(b-c)=logb/(c-a)=logc/(a-b)=klogabc=logbca=logcab=k

then considering 10 base logarithm we get

  • a=10^(k(b-c)a=10k(bc)

  • b=10^(k(c-a)b=10k(ca)

  • c=10^(k(a-b)c=10k(ab)

So

  • a^a=10^(k(ab-ca)aa=10k(abca)

  • b^b=10^(k(bc-ab)bb=10k(bcab)

  • c^c=10^(k(ca-bc)cc=10k(cabc)

Hence the numerical value of

a^a*b^b*c^caabbcc

=10^(k(ab-ca))*10^(k(bc-ab))*10^(k(ca-bc))=10k(abca)10k(bcab)10k(cabc)

=10^(k(ab-ca+bc-ab+ca-bc))=10k(abca+bcab+cabc)

=10^0=1=100=1

Jul 24, 2018

11.

Explanation:

Set loga/(b-c)=logb/(c-a)=logc/(a-b)=klogabc=logbca=logcab=k.

:. loga=k(b-c), logb=k(c-a), and, logc=k(a-b).

Now, log(a^a*b^b*c^c),

=aloga+nblogb+clogc,

=a{k(b-c)}+b{k(c-a)}+c{k(a-b)}.

=0,

i.e., log(a^a*b^b*c^c)=0.

rArr a^a*b^b*c^c=1, as Respected P dilip_k has readily

derived!