Question

Log(m/n)+log(n/m) =log(m+n)

Answer 1

Answer:

*/

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

\implies mn = (m+n)⟹mn=(m+n)

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

mn

mn

=

mn

m

+

mn

n

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

n

1

+

m

1

\implies 1-\frac{1}{n}=\frac{1}{m}⟹1−

n

1

=

m

1

\implies \frac{(n-1)}{n}=\frac{1}{m}⟹

n

(n−1)

=

m

1

\implies \frac{n}{(n-1)}=m⟹

(n−1)

n

=m

Therefore,

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

(n−1)

n