Prove the following:

$2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} = \text{log} \space 2.$

Given,

$2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} = \text{log} \space 2.$

Simplifying L.H.S. of the above equation we get,

$\Rightarrow 2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 130 - \text{log} \space 77) - (\text{log} \space 55 - \text{log} \space 91) \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 13.10 - \text{log} \space 11.7) - (\text{log} \space 11.5 - \text{log} \space 13.7) \\[1em] \Rightarrow 2(\text{log} \space 11 - \text{log} \space 13) + (\text{log} \space 13 + \text{log} \space 10 - (\text{log} \space 11 + \text{log} \space 7) - (\text{log} \space 11 + \text{log} \space 5 - (\text{log} \space 13 + \text{log} \space 7))) \\[1em] \Rightarrow 2\text{log} \space 11 - 2\text{log} \space 13 + \text{log} \space 13 + \text{log} \space 10 - \text{log} \space 11 - \cancel{\text{log} \space 7} - \text{log} \space 11 - \text{log} \space 5 + \text{log} \space 13 + \cancel{\text{log} \space 7} \\[1em] \Rightarrow \cancel{2\text{log} \space 11} - \cancel{2\text{log} \space 11} + \cancel{2\text{log} \space 13} - \cancel{2\text{log} \space 13} + \text{log} \space 10 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 10 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 2.5 - \text{log} \space 5 \\[1em] \Rightarrow \text{log} \space 2 + \cancel{\text{log} \space 5} - \cancel{\text{log} \space 5} \\[1em] \Rightarrow \text{log} \space 2.$

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

Hence, proved that $2\text{log} \space \dfrac{11}{13} + \text{log} \space \dfrac{130}{77} - \text{log} \space \dfrac{55}{91} = \text{log} \space 2.$