Question

Prove that $7\log \frac{16}{15}+5\log \frac{25}{24}+3\log \frac{81}{80}=\log 2$

Answer 1

L.H.S$=7\log \frac{16}{15}+5\log \frac{25}{24}+3\log \frac{81}{80}$

$=\log {\left(\frac{16}{15}\right)}^7+\log {\left(\frac{25}{24}\right)}^5+\log {\left(\frac{81}{80}\right)}^3$ using $m\log a=\log a^m$

$=\log {\left(\frac{16}{15}\right)}^7\times {\left(\frac{25}{24}\right)}^5\times {\left(\frac{81}{80}\right)}^3$

$=\log ({\left(\frac{2^4}{3\times 5}\right)}^7\times {\left(\frac{5^2}{2^3\times 3}\right)}^5\times {\left(\frac{3^4}{2^4\times 5}\right)}^3)$

$=\log \left(\frac{2^{28}\times 5^{10}\times 3^{12}}{3^7\times 5^7\times 2^{15}\times 3^5\times 2^{12}\times 5^3}\right)$

$=\log \left(2^{28-15-12}{5}^{10-7-3}{3}^{12-7-5}\right)$ using $\frac{a^m}{a^n}=a^{m-n}$

$=\log \left(2^{28-27}{5}^{10-10}{3}^{12-12}\right)$

$=\log \left(2^1{5}^0{3}^0\right)$

$=\log 2$

$=$R.H.S

Hence proved.