Show that:

$\dfrac{1}{\text{log}$2\space42} + \dfrac{1}{\text{log}3\space42} + \dfrac{1}{\text{log}_7\space42} = 1

Given,

$\dfrac{1}{\text{log}$2\space42} + \dfrac{1}{\text{log}3\space42} + \dfrac{1}{\text{log}_7\space42} = 1

Simplifying L.H.S. we get,

$\dfrac{1}{\text{log}$2\space42} + \dfrac{1}{\text{log}3\space42} + \dfrac{1}{\text{log}_7\space42} = log₄₂ 2 + log₄₂ 3 + log₄₂ 7

= log₄₂ (2 × 3 × 7)

= log₄₂ 42

= 1.

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

Hence, proved that $\dfrac{1}{\text{log}$2\space42} + \dfrac{1}{\text{log}3\space42} + \dfrac{1}{\text{log}_7\space42} = 1