Question

If log m+log n= log (m+n), then m=?

Answer 1

Answer:

$m = \frac{n}{(n-1)}$

Step-by-step explanation:

$Given\: logm+logn = log(m+n)$

/* we know that,

$\boxed {logm+logn = log(mn)}$ */

$\implies logmn = log(m+n)$

$\implies mn = (m+n)$

$\implies \frac{mn}{mn}=\frac{m}{mn}+\frac{n}{mn}$

$\implies 1 = \frac{1}{n}+\frac{1}{m}$

$\implies 1-\frac{1}{n}=\frac{1}{m}$

$\implies \frac{(n-1)}{n}=\frac{1}{m}$

$\implies \frac{n}{(n-1)}=m$

Therefore,

$m = \frac{n}{(n-1)}$

•••♪