Question

what is the value of log a^2/bc + log b^2/ca + log c^2/ab​

Answer 1

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

Given expression is

$\rm :\longmapsto\: \: log\bigg[\dfrac{ {a}^{2} }{bc} \bigg] + log\bigg[\dfrac{ {b}^{2} }{ca} \bigg] + log\bigg[\dfrac{ {c}^{2} }{ab} \bigg]$

We know,

$\red{\rm :\longmapsto\:\boxed{\tt{ logx + logy = log(xy) \: }}}$

$\rm \:  =  \: log\bigg[\dfrac{ {a}^{2} }{bc} \times \dfrac{ {b}^{2} }{ca} \times \dfrac{ {c}^{2} }{ab} \bigg]$

$\rm \:  =  \: log\bigg[\dfrac{ {a}^{2} {b}^{2} {c}^{2} }{ {a}^{2} {b}^{2} {c}^{2} } \bigg]$

$\rm \:  =  \: log1$

$\rm \:  =  \: 0$

Hence,

$\rm \implies\:\boxed{\tt{ log\bigg[\dfrac{ {a}^{2} }{bc} \bigg] + log\bigg[\dfrac{ {b}^{2} }{ca} \bigg] + log\bigg[\dfrac{ {c}^{2} }{ab} \bigg] = 0}}$

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Explore More :-

$\boxed{\tt{ logx - logy = log \frac{x}{y} \: }}$

$\boxed{\tt{ log {x}^{y} = y \: logx \: }}$

$\boxed{\tt{ log_{x}(y) = \frac{logx}{logy} \: }}$

$\boxed{\tt{ log_{x}(x) = 1\: }}$

$\boxed{\tt{ log_{ {x}^{a} }( {x}^{b} ) = \frac{b}{a} \: }}$

$\boxed{\tt{ log_{ {x}^{a} }( {y}^{b} ) = \frac{b}{a} log_{x}(y) \: }}$

$\boxed{\tt{ {a}^{ log_{a}(x) } = x \: }}$

$\boxed{\tt{ {a}^{y log_{a}(x) } = {x}^{y} \: }}$