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

Question

69090 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-24T16:02:26

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

then considering 10 base logarithm we get

So

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#

Answer 2 · 2018-07-24T20:01:20

# 1#.

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!

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