Question

Show that $\log \left(\frac{243}{343}\right)=5\log 3-3\log 7$

Answer 1

L.H.S$=\log \left(\frac{243}{343}\right)$

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

We know that $\log \left(\frac{a}{b}\right)=\log a-\log b$

$=\log \left(3^5\right)-\log \left(7^3\right)$

We know that $\log a^m=m\log a$

$=5\log 3-3\log 7$

$=$R.H.S

$\therefore \log \left(\frac{243}{343}\right)=5\log 3-3\log 7$

Hence, proved.