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When I solved the following logarithmic equation:

log_2(2x)+log_2(x)=5

I got the answers x = -4 and x = 4, and evaluated each of these for extraneous solutions. When I plugged in both the negative and the positive solutions, I found that only the positive solution works without negative arguments. However, when I solved a different equation:

log(2x-3)-log(x-1)=log(5)

I got the answer 2/3, which when evaluated gives negative arguments log(-5/3) and log(-1/3). I thought at this point you know the solution is extraneous, but upon further evaluation it is still true. If you evaluated x = -4 for the previous problem further by multiplying the two arguments to make a positive argument, the equation would be true. How come x = -4 isn't a solution, but x = 2/3 is?

jared reed
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2 Answers2

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$\frac23$ is not, in fact, a solution to the second equation, precisely for the reason you mentioned: the left-hand side is not defined for $x=\frac23$. So at least that's consistent with the first example.

In the first example, it happens to be true that if both $\log_2(2x)$ and $\log_2(x)$ are defined, then the left-hand side equals $\log_2(2x^2)$. But that doesn't mean that $\log_2(2x)+\log_2(x)$ and $\log_2(2x^2)$ are the same function: the second one is defined at lots more $x$-values than the first one is. Similarly in the second example for $\log(2x-3)-\log(x-1)$ and $\log\frac{2x-3}{x-1}$.

Greg Martin
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  • Thank you for your response. So is it correct to assume that when evaluating solutions for logarithmic equations, that ANY negative arguments after only evaluating the argument will make the solution extraneous? Even when you multiply/divide the arguments of several logs and it becomes positive? – jared reed Dec 04 '22 at 02:42
  • A solution is extraneous if it is not a solution to the original equation. A value is a solution to the original equation if the original equation is true when it is put in for the variable. $\log A$ is undefined if $A\le0$, and a value that results in a term of the original equation that is undefined (such as $\log A$ where $A\le0$) is not a solution. Perhaps what you are missing is to understand that $\log A + \log B$ does not equal $\log (A\cdot B)$ for all real numbers $A$ and $B$. $\log A + \log B = \log (A\cdot B)$ only for positive real numbers $A$ and $B$. – Steve Kass Dec 04 '22 at 02:51
  • It might help to point out that extraneous solutions are an aspect of the method one uses to solve an equation, not something intrinsic to the equation itself. To take it to a ridiculous extreme: we can multiply the original equation by $0$ on both sides, resulting in $0=0$ which has every real number $x$ as a solution; but all of those solutions are extraneous (the method made them look like solutions but they don't satisfy the original equation). – Greg Martin Dec 04 '22 at 05:09
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There are no real solutions to the equation $\log(2x-3)-\log(x-1)=\log(5)$.

$\color{blue}{\log(2x-3)}-\log(x-1) \\\qquad < \color{blue}{\log(2x-2)}-\log(x-1) \\\qquad= \color{blue}{\log 2 + \log(x-1)}-\log(x-1)\\\qquad=\log 2\\ \qquad\qquad < \log 5$

therefore ${\log(2x-3)}-\log(x-1)<\log 5$.

Steve Kass
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