If loga/(b-c)=logb/(c-a)=logc/(a-b)logab−c=logbc−a=logca−b then the numerical value of a^a*b^b*c^c=?aa⋅bb⋅cc=?
2 Answers
Jul 24, 2018
Let
then considering 10 base logarithm we get
-
a=10^(k(b-c)a=10k(b−c) -
b=10^(k(c-a)b=10k(c−a) -
c=10^(k(a-b)c=10k(a−b)
So
-
a^a=10^(k(ab-ca)aa=10k(ab−ca) -
b^b=10^(k(bc-ab)bb=10k(bc−ab) -
c^c=10^(k(ca-bc)cc=10k(ca−bc)
Hence the numerical value of
Jul 24, 2018
Explanation:
Set
Now,
derived!