Question

if ((loga)/(b-c))=((logb)/(c-a))=((logc)/(a-b)), then prove that (a^a)(b^b)(c^c)=1 ​

Answer 1

ANSWER:

Let logab−c=logbc−a=logca−b=k

then considering 10 base logarithm we get

• a=10k(b−c)

• b=10k(c−a)

• c=10k(a−b)

So

• aa=10k(ab−ca)

• bb=10k(bc−ab)

• cc=10k(ca−bc)

Hence the numerical value of

aa⋅bb⋅cc

=10k(ab−ca)⋅10k(bc−ab)⋅10k(ca−bc)

=10k(ab−ca+bc−ab+ca−bc)

=100=1

EXPLANATION:

Set logab−c=logbc−a=logca−b=k.

∴loga=k(b−c),logb=k(c−a),and,logc=k(a−b).

Now, log(aa⋅bb⋅cc),

=aloga+nblogb+clogc,

=a{k(b−c)}+b{k(c−a)}+c{k(a−b)}.

=0,

i.e.,log(aa⋅bb⋅cc)=0.

⇒aa⋅bb⋅cc=1,

Answer 2

Answer:

Me Also.

This is correct answer