Evaluate the following:

log 2 + 16log $\dfrac{16}{15} + 12\text{log } \dfrac{25}{24} + 7\text{log }\dfrac{81}{80}$

Given,

$\Rightarrow \text{log 2 + 16log }\dfrac{16}{15} + 12\text{log } \dfrac{25}{24} + 7\text{log }\dfrac{81}{80} \\[1em] \Rightarrow \text{log 2 + 16(log 16 - log 15) + 12(log 25 - log 24) + 7(log 81 - log 80)} \\[1em] \Rightarrow \text{log 2 + 16log 16 - 16log 15 + 12log 25 - 12log 24 + 7log 81 - 7log 80} \\[1em] \Rightarrow \text{log 2 + 16log 2}^4 - \text{16log 3.5 + 12log 5}^2 - \text{12log 2}^3.3 + \text{7log 3}^4 - \text{7log 2}^4.5 \\[1em] \Rightarrow \text{log 2 + 16.4log 2 - 16(log 3 + log 5) + 12.2log 5 - 12(log 2}^3 + \text{log 3) + 7.4log 3} - \text{7(log 2}^4 + \text{log 5)} \\[1em] \Rightarrow \text{log 2 + 64log 2 - 16log 3 - 16log 5 + 24log 5 - 12.3log 2 - 12log 3 + 28log 3 - 28log 2 - 7log 5} \\[1em] \Rightarrow \text{65log 2 - 36log 2 - 28log 2 - 16log 3 - 12log 3 + 28log 3 - 16log 5 + 24log 5 - 7log 5} \\[1em] \Rightarrow \text{log 2 + log 5} \\[1em] \Rightarrow \text{log 2.5} \\[1em] \Rightarrow \text{log 10} = 1.$

Hence, $\text{log 2 + 16log }\dfrac{16}{15} + 12\text{log } \dfrac{25}{24} + 7\text{log }\dfrac{81}{80} = 1$.