Question

If logx + logy = log(x+y)
Then find the relation between x and y

Answer 1

hey!!!!
logx + logy = log(x+y)
log(xy) = log(x+y)
xy = x+y
xy - y = x
(x-1)y = x
y = x/(x-1)
hope you understand
if not understand inbox me

Answer 2

Answer:

$\frac{1}{y}+\frac{1}{x}$=1[/tex]

Step-by-step explanation:

Given logx+logy = log(x+y)

$\implies log(xy)= log(x+y)$

/* By Logarithmic Law :

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

$\implies xy = x+y$

Dived each term by xy , we get

$\implies \frac{xy}{xy}=\frac{x}{xy}+\frac{y}{xy}$

$\implies 1 = \frac{1}{y}+\frac{1}{x}$

Therefore,

Required relation :

$\frac{1}{y}+\frac{1}{x}$=1[/tex]

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