Prove that :

$\text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} = \text{log 2}$

To prove:

$\text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} = \text{log 2}$

Solving L.H.S. of the equation, we get :

$\Rightarrow \text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} \\[1em] \Rightarrow \text{log } \Big(\dfrac{15}{18}\Big)^2 + \text{log } \dfrac{4}{9} - \text{log } \dfrac{25}{162} \\[1em] \Rightarrow \text{log } \dfrac{225}{324} + \text{log } \dfrac{4}{9} - \text{log } \dfrac{25}{162} \\[1em] \Rightarrow \text{log } \dfrac{\dfrac{225}{324} \times \dfrac{4}{9}}{\dfrac{25}{162}} \\[1em] \Rightarrow \text{log } \dfrac{\dfrac{900}{2916}}{\dfrac{25}{162}} \\[1em] \Rightarrow \text{log } \dfrac{900 \times 162}{25 \times 2916} \\[1em] \Rightarrow \text{log } \dfrac{145800}{72900} \\[1em] \Rightarrow \text{log } 2.$

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

Hence, proved that $\text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} = \text{log 2}$.