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So I had this question in my Calc exam -(Simply solve for $x$ in $\log(1-2x)+\log(-x) =0$) and because the range of the equation on the right isn't the same as the equation of the left I made a mistake by assuming $x$ can be positive can someone explain to me why aren't these 2 equations equal? even though you can simplify the first one to get the second

Martin Argerami
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T.G
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  • The LHS is composed of negative logarithms, what is a multivalued function, and the RHS no. See here. Probably there is way to make them coincide through some branch cut. – Masacroso Nov 21 '16 at 05:04
  • They are equal. with the restriction x <0. And x < 1/2. So log (-x+2x^2)=0 means -x+2x^2=1, so x =-1/2. – fleablood Nov 21 '16 at 06:32
  • The domain of the RHS is larger (try $x=1$). –  Nov 23 '16 at 08:04

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If you have to find $x$ in $\log(1-2x)+\log(-x)=0$, then it is a given that $x<0$. Now you manipulate the logarithms to get $0=\log(-x+2x^2)$, which tells you that $$-x+2x^2=1.$$ This has solutions $1$ and $-1/2$, but only $-1/2$ is negative, so $x=-1/2$.

Martin Argerami
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