Question

Class 11 jee question

Answer 1

$\large\underline{\sf{Solution-}}$

$\sf \: \dfrac{loga}{b - c} = \dfrac{logb}{c - a} = \dfrac{logc}{a - b}$

$\sf \: \dfrac{(b + c)loga}{(b + c)(b - c)} = \dfrac{(c + a)logb}{(c - a)(c + a)} = \dfrac{(a + b)logc}{(a + b)(a - b)}$

$\sf \: \dfrac{log {a}^{b + c} }{ {b}^{2} - {c}^{2} } = \dfrac{log {b}^{c + a} }{ {c}^{2} - {a}^{2} } = \dfrac{log {c}^{a + b} }{ {a}^{2} - {b}^{2} }$

$\sf \: \dfrac{log {a}^{b + c} }{ {b}^{2} - {c}^{2} } = \dfrac{log {b}^{c + a} }{ {c}^{2} - {a}^{2} } = \dfrac{log {c}^{a + b} }{ {a}^{2} - {b}^{2} } = \dfrac{log {a}^{b + c} + log {b}^{c + a} + log {c}^{a + b} }{0}$

[ By using Addendo]

$\sf \: log {a}^{b + c} + log {b}^{c + a} + log {c}^{a + b} = 0$

$\sf \: log {a}^{b + c} + log {b}^{c + a} + log {c}^{a + b} = log1$

$\sf \: log ({a}^{b + c} {b}^{c + a}{c}^{a + b} )= log1$

So,

$\sf \: {a}^{b + c} {b}^{c + a}{c}^{a + b}= 1$

Hence, Proved.