Question

If $\log 2=0.3010,\log 3=0.4771,\log 7=0.8451$ and $\log 11=1.0414$ then find the value of the following
$\log \frac{121}{120}$

Answer 1

$\log \frac{121}{120}$

$=\log 121-\log 120$

$=2\log 11-\log (3\times 2^3\times 5)$

$=2\log 11-\log 3-3\log 2-\log 5$

$=2\log 11-\log 3-3\log 2-\log \left(\frac{10}{2}\right)$

$=2\log 11-\log 3-3\log 2-(\log 10-\log 2)$

$=2\log 11-\log 3-3\log 2-1+\log 2$

$=2\log 11-\log 3-2\log 2-1$

$=2\left(1.0414\right)-0.4771-2\times 0.3010-1$

$=3.604\times {10}^{-3}$