Question

(1)/(logab)(abc))+(1)/(logbc)(abc))+(1)/(log(ca)(abc))is equal to ​

Answer 1

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

Given expression is

$\rm :\longmapsto\:\dfrac{1}{ log_{ab}(abc) } + \dfrac{1}{ log_{bc}(abc) } + \dfrac{1}{ log_{ca}(abc) }$

We know,

$\boxed{ \bf{ \: log_{a}(b) = \frac{logb}{loga}}}$

So, using the above can be rewritten as

$\rm \:  =  \:  \: \dfrac{1}{\dfrac{logabc}{logab} } + \dfrac{1}{\dfrac{logabc}{logbc} } + \dfrac{1}{\dfrac{logabc}{logca} }$

$\rm \:  =  \:  \: \dfrac{logab}{logabc} + \dfrac{logbc}{logabc} + \dfrac{logca}{logabc}$

$\rm \:  =  \:  \: \dfrac{logab + logbc + logca}{logabc}$

We know,

$\boxed{ \bf{ \: logx + logy = logxy}}$

So, using this we get

$\rm \:  =  \:  \: \dfrac{log(ab \times bc \times ca)}{logabc}$

$\rm \:  =  \:  \: \dfrac{log {a}^{2} {b}^{2} {c}^{2} }{logabc}$

$\rm \:  =  \:  \: \dfrac{log {(abc)}^{2} }{logabc}$

We know,

$\boxed{ \bf{ \: log( {x}^{y} ) = y \: logx}}$

So, using this, we get

$\rm \:  =  \:  \: \dfrac{2 \: logabc}{logabc}$

$\rm \:  =  \:  \: 2$

Hence,

$\boxed{ \bf{ \: \rm :\longmapsto\:\dfrac{1}{ log_{ab}(abc) } + \dfrac{1}{ log_{bc}(abc) } + \dfrac{1}{ log_{ca}(abc) } = 2}}$

Additional Information :-

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

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

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

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

$\boxed{ \bf{ \: {e}^{logx} = x}}$

$\boxed{ \bf{ \: {e}^{ylogx} = {x}^{y} }}$

$\boxed{ \bf{ \: log1 = 0}}$