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

2 Answers

Let #loga/(b-c)=logb/(c-a)=logc/(a-b)=k#

then considering 10 base logarithm we get

  • #a=10^(k(b-c)#

  • #b=10^(k(c-a)#

  • #c=10^(k(a-b)#

So

  • #a^a=10^(k(ab-ca)#

  • #b^b=10^(k(bc-ab)#

  • #c^c=10^(k(ca-bc)#

Hence the numerical value of

#a^a*b^b*c^c#

#=10^(k(ab-ca))*10^(k(bc-ab))*10^(k(ca-bc))#

#=10^(k(ab-ca+bc-ab+ca-bc))#

#=10^0=1#

Jul 24, 2018

# 1#.

Explanation:

Set #loga/(b-c)=logb/(c-a)=logc/(a-b)=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!