Alg9.4.1

If a not equal to 1,m,n are positive real numbers, then show that $\log_a(\frac{m}{n})=\log_a(m)-\log_a(n)\,$

$\log_a(\frac{m}{n})=\log_a(m\cdot\frac{1}{n})\,$

Expanding the log,we get,

$\log_a(\frac{m}{n})=\log_a(m)+\log_a(\frac{1}{n})\,$

$\log_a(\frac{m}{n})=\log_a(m)+\log_a(n^{-1})\,$

Hence

$\log_a(\frac{m}{n})=\log_a(m)-\log_a(n)\,$