Question

Pls answer the Question in the Attachment...​

Answer 1

Answer:

Let,

log(a)/(b-c)=log(b)/(c-a)=log(c)/(a-b)=k

----------(1)

=> log(a) = k(b-c)

=> a•log(a) = a•k(b-c)

=> log(a^a) = k(ab-ac) ---------(2)

=> log(b) = k(c-a)

=> b•log(b) = b•k(c-a)

=> log(b^b) = k(bc-ba) ---------(3)

=> log(c) = k(a-b)

=> c•log(c) = c•k(a-b)

=> log(c^c) = k(ac-cb) ---------(4)

Now,

Adding eq-(2),(3) and (4) ,we get;

=> log(a^a) + log(b^b) + log(c^c)

= k(ab-ac) + k(bc-ba) + k(ac-cb)

=> log{(a^a)(b^b)(c^c)} = 0

=> {(a^a)(b^b)(c^c)} = e^0

=> (a^a)(b^b)(c^c) = 1

Hence, the required value of

(a^a)(b^b)(c^c) is 1.

Answer 2

Answer:

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refer to attachment