Question

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

Answer 1

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`

Answer 2

`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!