Question

(loga N).(logb N) + (logb N).(logc N) + (logc N).(loga N) = (loga N).(logb N).(logc N)/(logabc N)

⇒ Prove the identity
⇒ Show appropriate calculations​

Answer 1

★ Concept :-

Here the concept of Logarithmic Properties has been used. We see that we are given an identity where it is to be proved that L.H.S. = R.H.S. We see that the terms in L.H.S. are quite long and expanded. So we will work upon the R.H.S. term. Firstly we will try to make the bases of log of each term in fraction as same. Then we will apply the logarithmic property. After that we may try to seperate each term. Then we will factorise and prove it.

Let's do it !!

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★ Solution :-

Given to be proved,

$\;\bf{\mapsto\;\;\blue{\log_{a}N.\log_{b}N\:+\:\log_{b}N.\log_{c}N\:+\:\log_{c}N.\log_{a}N\;=\;\dfrac{\log_{a}N.\log_{b}N.\log_{c}N}{\log_{abc}N}}}$

From here we get,

$\;\bf{\odot\;\;L.H.S.\;=\;\green{\log_{a}N.\log_{b}N\:+\:\log_{b}N.\log_{c}N\:+\:\log_{c}N.\log_{a}N}}$

And,

$\;\bf{\odot\;\;R.H.S.\;=\;\orange{\dfrac{\log_{a}N.\log_{b}N.\log_{c}N}{\log_{abc}N}}}$

Here let's start working on R.H.S. first.

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\dfrac{\log_{a}N.\log_{b}N.\log_{c}N}{\log_{abc}N}}$

We see that N term is common in all log values. So we can change all the bases in form of N.

We know a property of logarithm that,

$\;\qquad\qquad\;\rm{\leadsto\;\;\log_{y}x\;=\;\dfrac{1}{\log_{x}y}}$

On applying this to each term at R.H.S., we get

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\dfrac{\bigg(\dfrac{1}{\log_{N}a}\bigg).\bigg(\dfrac{1}{\log_{N}b}\bigg).\bigg(\dfrac{1}{\log_{N}c}\bigg)}{\bigg(\dfrac{1}{\log_{N}abc}\bigg)}}$

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\dfrac{\bigg(\log_{N}abc\bigg)}{\bigg(\log_{N}a\bigg).\bigg(\log_{N}b\bigg).\bigg(\log_{N}c\bigg)}}$

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\dfrac{\log_{N}abc}{\log_{N}a.\log_{N}b.\log_{N}c}}$

Now we again know a property of logarithm that,

$\;\quad\;\rm{\leadsto\;\;\log_{y}(xza)\;=\;\log_{y}x\:+\:\log_{y}z\:+\:\log_{y}a}$

Clearly the base of a, b, and c is same, so we can expand it.

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\dfrac{\log_{N}a\:+\:\log_{N}b\:+\:\log_{N}c}{\log_{N}a.\log_{N}b.\log_{N}c}}$

Now let's seperate each of the terms.

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\bigg[\dfrac{\log_{N}a}{\log_{N}a.\log_{N}b.\log_{N}c}\bigg]\:+\:\bigg[\dfrac{\log_{N}b}{\log_{N}a.\log_{N}b.\log_{N}c}\bigg]\:+\:\bigg[\dfrac{\log_{N}c}{\log_{N}a\:.\log_{N}b.\log_{N}c}\bigg]}$

Cancellating the like terms from numerator and denominator, we get

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\bigg[\dfrac{\not{\log_{N}a}}{\not{\log_{N}a}.\log_{N}b.\log_{N}c}\bigg]\:+\:\bigg[\dfrac{\not{\log_{N}b}}{\log_{N}a\:.\not{\log_{N}b}.\log_{N}c}\bigg]\:+\:\bigg[\dfrac{\not{\log_{N}c}}{\log_{N}a.\log_{N}b.\not{\log_{N}c}}\bigg]}$

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\bigg[\dfrac{1}{\log_{N}b.\log_{N}c}\bigg]\:+\:\bigg[\dfrac{1}{\log_{N}a\:.\log_{N}c}\bigg]\:+\:\bigg[\dfrac{1}{\log_{N}a.\log_{N}b}\bigg]}$

Now again we can apply the reverse of the property that,

$\;\qquad\qquad\;\rm{\leadsto\;\;\log_{y}x\;=\;\dfrac{1}{\log_{x}y}}$

Then,

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\left[\dfrac{1}{\bigg(\dfrac{1}{\log_{b}N}\bigg).\bigg(\dfrac{1}{\log_{c}N}\bigg)}\right]\:+\:\left[\dfrac{1}{\bigg(\dfrac{1}{\log_{a}N}\bigg).\bigg(\dfrac{1}{\log_{c}N}\bigg)}\right]\:+\:\left[\dfrac{1}{\bigg(\dfrac{1}{\log_{a}N}\bigg).\bigg(\dfrac{1}{\log_{b}N}\bigg)}\right]}$

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\bigg[\bigg(\log_{b}N\bigg).\bigg(\log_{c}N\bigg)\bigg]\:+\:\bigg[\bigg(\log_{c}N\bigg).\bigg(\log_{a}N\bigg)\bigg]\:+\:\bigg[\bigg(\log_{a}N\bigg).\bigg(\log_{b}N\bigg)\bigg]}$

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\bigg[\log_{b}N.\log_{c}N\bigg]\:+\:\bigg[\log_{c}N.\log_{a}N\bigg]\:+\:\bigg[\log_{a}N.\log_{b}N\bigg]}$

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\bigg[\log_{b}N.\log_{c}N\bigg]\:+\:\bigg[\log_{c}N.\log_{a}N\bigg]\:+\:\bigg[\log_{a}N.\log_{b}N\bigg]}$

On rearranging the terms according to the qúestion, we get

$\;\tt{\rightarrowtail\;\;R.H.S.\;=\;\bigg[\log_{a}N.\log_{b}N\bigg]\:+\:\bigg[\log_{b}N.\log_{c}N\bigg]\:+\:\bigg[\log_{c}N.\log_{a}N\bigg]}$

Now on removing the brackets, we get

$\;\bf{\rightarrowtail\;\;\purple{R.H.S.\;=\;\log_{a}N.\log_{b}N\:+\:\log_{b}N.\log_{c}N\:+\:\log_{c}N.\log_{a}N}}$

We see that now R.H.S. = L.H.S. . Then,

$\;\bf{\rightarrowtail\;\;\red{L.H.S.\;=\;R.H.S.\;=\;\log_{a}N.\log_{b}N\:+\:\log_{b}N.\log_{c}N\:+\:\log_{c}N.\log_{a}N}}$

• Hence, proved

Answer 2

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