if f(x)=log 1+x/1-x then show that f(x)+f(y)=f(x+y/1+xy)
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here is your answer .
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Solution :-
f(x) = ln (1-x/1+x)
Given that f(x) +f (y) = f(x+y /1 +xy)
ln (1-x/1+x) + ln (1-y/1+y) = ln (1 -(x+y)/1 +x+y)
ln((1-x)× (1-y)/(1+x)× (1+y)) = ln (1-(x+y)/1+x+y)
Removing log
And. Then on solving
We get
2xy(x+y) = 0
Hence xy = 0 or x+y = 0
For xy = 0, solution is but obvious
For x+y = 0, x = -y
Putting in the given equation
ln((1+x)× (1-x)/(1-x)(1+x)) = ln1
LHS and RHS contradicts
Hence
x = 0 = y
Also
1-x >0
x 1
hence
x belongs to (0, 1)
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@GauravSaxena01
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