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
2026-04-28 12:38:37.1777379917
why isn't $\log(1-2x)+\log(-x) = \log(-x+2x^2)$?
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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 $x=-1/2$.