Question

Prove that
log(m/n)=log(m)-log(n)

Answer 1

Answer:

Step-by-step explanation

loga(m/n) = x

then we have a^x=m/n (1)

loga(m)=b then we have a^b=m

loga(n)=p then we have a^p=n

m/n =a^b/a^p = a^b-p

m/n= a^(b-p); back to (1) have a^(b-p)=a^x

then x=b-p;

=> log(m/n)=log(m)-log(n)

Answer 2

Let,

$\rightarrow\ m=a^x\\\\\rightarrow\ n=a^y$

Taking their logs to the base 'a',

$\rightarrow\ \log_a(m)=x\\\\\rightarrow\ \log_a(n)=y$

So,

$\begin{aligned}&\text{LHS}\\\\\implies\ \ &\log_a\left(\dfrac{m}{n}\right)\\\\\implies\ \ &\log_a\left(\dfrac{a^x}{a^y}\right)\\\\\implies\ \ &\log_a(a^{x-y})\\\\\implies\ \ &x-y\\\\\implies\ \ &\log_a(m)-\log_a(n)\\\\\implies\ \ &\text{RHS}\end{aligned}$

Hence Proved!