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 Problem: log(x)+log(1+x)=0 [ Calculator, Method and examples ]Solution:Your problem -> log(x)+log(1+x)=0log(x)+log(1+x)=0Simplify LHS =log(x)+log(1+x)=log(x)+log(1+x)=log(x xx (1+x))=log(x+x^2)Now, log(x+x^2)=0=>log(x+x^2)=0=>x+x^2=10^0=>x+x^2=1=>x+x^2-1=0Comparing the given equation with the standard quadratic equation ax^2+bx+c=0,we get, a=1,b=1,c=-1x_(1,2)=(-b+-sqrt(Delta))/(2a) and Delta=b^2-4acDelta=b^2-4ac:.Delta=1^2-4*1*-1:.Delta=1+4:.Delta=5:.sqrt(Delta)=sqrt(5):.sqrt(Delta)=2.2361x_1=(-b+sqrt(Delta))/(2a):.x_1=(-1+2.2361)/(2*1):.x_1=(1.2361)/2:.x_1=0.618x_2=(-b-sqrt(Delta))/(2a):.x_2=(-1-2.2361)/(2*1):.x_2=(-3.2361)/2:.x_2=-1.618

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