Question

Evaluate the following:-

$\dfrac{1}{log_{a}bc + 1} + \dfrac{1}{log_{b}ac+1} + \dfrac{1}{log_{c}ab+1}$

Don't be greedy for points

Answer 1

Given Expression:-

$\\\bullet\quad\boxed{\sf \dfrac{1}{log_{a}bc+1} + \dfrac{1}{log_{b}ac+1}+\dfrac{1}{log_{c}ab+1} }\dag$

Solution:-

$\\\rightarrow\quad\sf \dfrac{1}{log_{a}bc+log_{a}a} + \dfrac{1}{log_{b}ac+log_{b}b}+\dfrac{1}{log_{c}ab+log_{c}c}$

$\\\rightarrow\quad\sf \dfrac{1}{log_{a}abc} + \dfrac{1}{log_{b}abc}+\dfrac{1}{log_{c}abc}$

$\\\rightarrow\quad\sf log_{abc}a+ log_{abc}b+log_{abc}c$

$\\\rightarrow\quad\sf log_{abc}abc$

$\\\rightarrow\quad\boxed{\sf 1} \bigstar$

___________________

Answer 2

Given :-

$\tt \dfrac{1}{log_{a}bc + 1} + \dfrac{1}{log_{b}ac+1} + \dfrac{1}{log_{c}ab+1}$

To Find :-

Value of the given Expression

Solution :-

At first see , the Denominator of all the Fractions clearly . You will find that their is a same thing in all i.e 1 . And Their are three Bases ( to the log ) a , b & c . In each fraction each of them is appeared ( in the base of log ) also their is a log in addition with 1 in all fractions . So , we are going to Change the 1 of every denominator in accordance with the base of the log !

We knows a Identity i.e ;

$\quad \qquad { \bigstar { \underline { \boxed { \red { \bf { log_{x}x = 1 }}}}}{\bigstar}}$

So , As I told Previous we will change the 1 in accordance with the base of the log . So , In 1st fraction we change 1 to log to base " a " in 2nd log to base " b " in 3rd log to base " c " .

Now we have ;

$\quad { : \longmapsto { \tt { \dfrac{1}{log_{a}bc + log_{a}a} + \dfrac{1}{log_{b}ac+log_{b}b} + \dfrac{1}{log_{c}ab+log_{c}c}}}}$

We also knows a logarithmic identity that ;

$\quad \qquad { \bigstar { \underline { \boxed { \red { \bf { log_{a}m + log_{a}n = log_{a} ( m . n ) }}}}}{\bigstar}}$

Using this the given Expression transforms to ;

$\quad { : \longmapsto { \tt { \dfrac{1}{log_{a}(bc.a)}+ \dfrac{1}{log_{a}(ac.b)} + \dfrac{1}{log_{c}(ab.c)} }}}$

$\quad { : \longmapsto { \tt { \dfrac{1}{log_{a}(abc)}+ \dfrac{1}{log_{a}(abc)} + \dfrac{1}{log_{c}(abc)} }}}$

Now , we knows another Logarithmic identity i.e ;

$\quad \qquad { \bigstar { \underline { \boxed { \red { \bf { \dfrac{1}{log_{a}b} = log_{b}a }}}}}{\bigstar}}$

Using this we have ;

$\quad { : \longmapsto { \tt { log_{abc}a + log_{abc}b + log_{abc}c }}}$

Now , Using the additive Identity used above we have ;

$\quad { : \longmapsto { \tt { log_{abc}(a.b.c)}}}$

$\quad { : \longmapsto { \tt { log_{abc}(abc)}}}$

Using the above same base identity we have ;

$\quad { : \longmapsto { \tt { 1 }}}$

${ \bigstar { \underline { \boxed { \red { \bf { \therefore { \dfrac{1}{log_{a}bc + 1} + \dfrac{1}{log_{b}ac+1} + \dfrac{1}{log_{c}ab+1} = 1 }}}}}{ \bigstar }}}$