Prove the following identities:

$\dfrac{1}{\text{log}$a\text{ abc}} + \dfrac{1}{\text{log}b\text{ abc}} + \dfrac{1}{\text{log}_c\text{ abc}} = 1

Given,

$\dfrac{1}{\text{log}$a\text{ abc}} + \dfrac{1}{\text{log}b\text{ abc}} + \dfrac{1}{\text{log}_c\text{ abc}} = 1

Simplifying L.H.S. we get,

$\Rightarrow \dfrac{1}{\text{log}$a\text{ abc}} + \dfrac{1}{\text{log}b\text{ abc}} + \dfrac{1}{\text{log}c\text{ abc}} \\[1em] \Rightarrow \text{log}\text{abc}\text{ a} + \text{log}\text{abc}\text{ b} + \text{log}\text{abc}\text{ c} \\[1em] \Rightarrow \text{log}\text{abc}\space(\text{a }\times \text{b } \times \text{c}) \\[1em] \Rightarrow \text{log}\text{abc}\text{ abc} \\[1em] \Rightarrow 1.

Since, L.H.S. = R.H.S.,

Hence, proved that $\dfrac{1}{\text{log}$a\text{ abc}} + \dfrac{1}{\text{log}b\text{ abc}} + \dfrac{1}{\text{log}_c\text{ abc}} = 1.