Question

$\displaystyle \sf \frac{1}{1 + log_{a}bc} + \frac{1}{ 1 + \log_{b}ac} + \frac{1}{1 + log_{c}ab}$​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\displaystyle \sf \frac{1}{1 + log_{a}bc} + \frac{1}{ 1 + \log_{b}ac} + \frac{1}{1 + log_{c}ab}$

We know,

$\rm :\longmapsto\:\boxed{\tt{ log_{a}(b) = \frac{logb}{loga}}}$

So, using this identity, we get

$\rm \:  =  \: \displaystyle \sf \frac{1}{1 + \dfrac{logbc}{loga}} + \frac{1}{ 1 + \dfrac{logac}{logb} } + \frac{1}{1 + \dfrac{logab}{logc} }$

$\rm \:  =  \: \displaystyle \sf \frac{1}{ \dfrac{loga + logbc}{loga}} + \frac{1}{\dfrac{logb + logac}{logb} } + \frac{1}{\dfrac{logc + logab}{logc} }$

We know,

$\rm :\longmapsto\:\boxed{\tt{ \: \: logx + logy = logxy \: \: }}$

So, using this identity, we get

$\rm \:  =  \: \displaystyle \sf \frac{1}{ \dfrac{logabc}{loga}} + \frac{1}{\dfrac{logabc}{logb} } + \frac{1}{\dfrac{logabc}{logc} }$

$\rm \:  =  \: \dfrac{loga}{logabc} + \dfrac{logb}{logabc} + \dfrac{logc}{logabc}$

$\rm \:  =  \: \dfrac{loga + logb + logc}{logabc}$

$\rm \:  =  \: \dfrac{logabc}{logabc}$

$\rm \:  =  \: 1$

Hence,

$\boxed{\tt{ \displaystyle \sf \frac{1}{1 + log_{a}bc} + \frac{1}{ 1 + \log_{b}ac} + \frac{1}{1 + log_{c}ab} = 1}}$

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ADDITIONAL INFORMATION

$\boxed{\tt{ log \frac{a}{b} = loga - logb}}$

$\boxed{\tt{ log {a}^{b} = bloga \: }}$

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

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

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

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

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

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

Answer 2

★ To Find :

The value of expression

$\displaystyle \sf \frac{1}{1 + \log_{a}bc} + \frac{1}{ 1 + \log_{b}ac} + \frac{1}{1 + \log_{c}ab} = \: ?$

$\:$

★ Solution :

We use the following result,

$\displaystyle \sf \log_{c} x \: = \: \frac{ \log_{} \: x }{ \log_ \: c} \: and \log \: xy \: = \log x + \log y$

$\:$

Consider,

$\displaystyle \sf \frac{1}{1 + log_{a}bc} + \frac{1}{ 1 + \log_{b}ac} + \frac{1}{1 + log_{c}ab}$

$\displaystyle \sf \: : ⟹ \: \frac{1} { 1 + \frac{ \log \: bc}{a} } + \frac{1}{ 1 + \frac{ \log \: ca}{b} } + \frac{1}{ 1 + \frac{ \log \: ab}{c} }$

$\displaystyle \sf\: : ⟹ \: \frac{ \log \: a}{ \log \: a \: + \log \: bc} + \frac{ \log \: b}{ \log \: b\: + \log \: ca} \: + \frac{ \log \: c}{ \log \: c\: + \log \: ab}$

$\displaystyle \sf\: : ⟹ \: \frac{ \log \: a}{ \log \: a \: + \log \: b + \log \: c} + \frac{ \log \: b}{ \log \: b\: + \log \: c + \log \: a} \: + \frac{ \log \: c}{ \log \: c\: + \log \: b + \log \: a}$

$\displaystyle \sf\: : ⟹ \: \frac{ \log \: a \: + \log \: b + \log \: c}{ \log \: b \: + \log \: c \: + \log \: a}$

$\displaystyle \sf\: : ⟹ \: 1$

$\:$

Hence,

$\boxed{\displaystyle \sf \frac{1}{1 + log_{a}bc} + \frac{1}{ 1 + \log_{b}ac} + \frac{1}{1 + log_{c}ab} = 1 }$