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Question:-
$\pink{\displaystyle \sf \frac{1}{1 + log_{a}bc} + \frac{1}{ 1 + \log_{b}ac} + \frac{1}{1 + log_{c}ab}1+loga​bc1​+1+logb​ac1​+1+logc​ab1​}\ \textless \ br /\ \textgreater$


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Answer 1

Appropriate Question

Evaluate the following :

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

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

Given Logarithmic 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,

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

So, using this identity, we get

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

We know that

$\green{\rm :\longmapsto\:\boxed{\tt{ log_{a}(x) + log_{a}(y) = log_{a}(xy) \: \: }}}$

So, using this identity we get

$\rm \:  =  \: \dfrac{1}{ log_{a}(abc) } + \dfrac{1}{ log_{b}(abc) } + \dfrac{1}{ log_{c}(abc) }$

can be rewritten as

$\rm \:  =  \: log_{abc}(a) + log_{abc}(b) + log_{abc}(c)$

$\rm \:  =  \: log_{abc}(abc)$

$\rm \:  =  \: 1$

Hence,

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

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ADDITIONAL INFORMATION

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

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

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

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

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

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

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