Question

Evaluate
${\log }_{10}\left(\frac{351}{539}\right)+2{\log }_{10}\left(\frac{91}{110}\right)=3{\log }_{10}\left(\frac{39}{110}\right)$

Answer 1

Consider,

${\log }_{10}\left(\frac{351}{539}\right)+2{\log }_{10}\left(\frac{91}{110}\right)-3{\log }_{10}\left(\frac{39}{110}\right)$

$={\log }_{10}\left(\frac{3^3\times 13}{7^2\times 11}\right)+2{\log }_{10}\left(\frac{7\times 13}{2\times 5\times 11}\right)-3{\log }_{10}\left(\frac{3\times 13}{2\times 5\times 11}\right)$

Applying inverse log property to the third term, ie., $\log \left(x^{-1}\right)=\log \frac{1}{x}=-\log x$, we get

$={\log }_{10}\left(\frac{3^3\times 13}{7^2\times 11}\right)+2{\log }_{10}\left(\frac{7\times 13}{2\times 5\times 11}\right)+3{\log }_{10}\left(\frac{2\times 5\times 11}{3\times 13}\right)$

Using the exponents property on log, ie., $n\log b=\log b^n$, we can write;

$={\log }_{10}\left(\frac{3^3\times 13}{7^2\times 11}\right)+{\log }_{10}{\left(\frac{7\times 13}{2\times 5\times 11}\right)}^2+{\log }_{10}{\left(\frac{2\times 5\times 11}{3\times 13}\right)}^3$

$={\log }_{10}\left(\frac{3^3\times 13}{7^2\times 11}\right)+{\log }_{10}\left(\frac{7^2\times {13}^2}{2^2\times 5^2\times {11}^2}\right)+{\log }_{10}\left(\frac{2^3\times 5^3\times {11}^3}{3^3\times {13}^3}\right)$

Using the products property on log, ie., $\log a+\log b=\log ab$, we can write;

$={\log }_{10}\left(\frac{3^3\times 13\times 7^2\times {13}^2\times 2^3\times 5^3\times {11}^3}{7^2\times 11\times 2^2\times 5^2\times {11}^2\times 3^3\times {13}^3}\right)$

$={\log }_{10}2\times 5={\log }_{10}10$

We know that, ${\log }_{a}a=1$

$=1$