Question

Prove that :
$4\log \frac{24}{25}-16\log \frac{9}{10}+7\log \frac{81}{80}=\log 5$

Answer 1

$4\log \frac{24}{25}-16\log \frac{9}{10}+7\log \frac{81}{80}=\log 5\Rightarrow \log (\frac{24}{25}{)}^4-\log (\frac{9}{10}{)}^{16}+\log (\frac{81}{80}{)}^7=\log \frac{(\frac{24}{25}{)}^4\ast (\frac{81}{80}{)}^7}{(\frac{9}{10}{)}^{16}}=\log \frac{{24}^4\ast {81}^7\ast {10}^{16}}{{25}^4\ast {80}^7\ast 9^{16}}=\log \frac{3^4{8}^4\ast 3^{28}\ast 2^{16}\ast 5^{16}}{5^8\ast 8^7\ast 2^7\ast 5^7\ast 3^{32}}=\log 5$