# How to solve this equation $\sqrt{\log_{10}(-x)} = \log_{10}\sqrt{x^2}$

I am new to logarithms, I have the following equation

$\ \sqrt{\log_{10}(-x)} = \log_{10}\sqrt{x^2} \$

I tried by squaring both sides, which yielded me

$\ \log_{10}(-x) = ( \log_{10}\vert x \vert )^2 \$

So,

$\ \log_{10}(-x) = ( \log_{10}\vert x \vert )^2 \$

Or $\ \log_{10}(-x) = ( \log_{10}\vert x \vert ) ( \log_{10}\vert x \vert ) \$

Or $\ \log_{10}(-x) = ( \log_{10}2\vert x \vert ) \$

Which on further solving, doesn't gives me the solutions. What is the correct way?

• What should I do then? I'm not cheating at homework – Ravi Prakash Jul 4 '17 at 4:14
• Convince this community why we should help you? What makes your post useful to us? What do we gain by answering it? – The Great Duck Jul 4 '17 at 4:16
• @Typhon, that's a bit harsh, don't you think? I've seen many posts with far less work and/or thought slide. He has even shown some work and thoughts. Not every post on M.SE needs to be "useful to us" – Ravi Jul 4 '17 at 4:17
• @RaviPrakash do not worry about downvotes or rep; as long as your questions are answered, that is all that matters. More than that, do not worry about asking a lot of questions about a similar topic. Just do your best and show your efforts (as you have done so). – TomGrubb Jul 4 '17 at 4:28
• @RaviPrakash you're welcome, but I fail to see where I was being "kind" nor "un-kind". I am only voicing my complaint towards your behavior. Nothing more, nothing less. – The Great Duck Jul 4 '17 at 4:37

It seems like you mixed up your log rules. Here's one way to go about it: you know $log_{10}$ of a negative number is undefined, so $x$ must be negative for $\log_{10}(-x)$ to make sense. Further, you know $\sqrt{x^2}=|x|$; thus the question turns into solving (by setting $y=-x$) $$\sqrt{\log_{10}(y)}=\log_{10}(y)$$ Now when do we have that the square root of a number must be equal to that number? We get $\log_{10}(y)=0$ or $\log_{10}(y)=1$. Can you go from here?

• $\log_{10}(x)$ is very well defined for negative x's. Please explain where that assumption is ever justified? – The Great Duck Jul 4 '17 at 4:20
• Looks like a case of telepathy. Maybe I should have checked your answer while I was finishing mine off. – Lubin Jul 4 '17 at 4:21
• @Typhon The OP says they are "new to logarithms;" thus I am assuming that they have not heard of branch cuts and complex logs. – TomGrubb Jul 4 '17 at 4:22
• @Lubin hopefully this is a case of great minds thinking alike and not fools never differing :) – TomGrubb Jul 4 '17 at 4:23
• @Typhon, logarithms are definable for negative numbers, but they are not defined. That is, deciding what $\log(-1)$ shall be requires a choice, on which honest people of good will may easily disagree. – Lubin Jul 4 '17 at 4:23

First, you must remember that $\sqrt{x^2\,}=|x|$. Second, you must remember that logarithms are defined only for positive inputs, so on the left hand side, we need $x$ to be negative, which means that $|x|=-x$. As a result, your equation to solve becomes $\sqrt{\log_{10}(-x)}=\log_{10}(-x)$. Calling $z=\log_{10}(-x)$, the equation becomes $\sqrt z=z$, which happens only for $z=0,1$, as you can easily check. Thus, in case $z=0$, you have $-x=1$ and $x=-1$, while in case $z=1$, you have $-x=10$ and $x=-10$.

• logarithms are defined for negative inputs. – The Great Duck Jul 4 '17 at 4:21

First of all, $\log a \log b \neq \log ab$ in general.

$$\log (-x) = (\log |x|)^2$$

From LHS, $-x > 0 \implies x < 0$.

$$\log(-x) = (\log(-x))^2 \implies \log(-x) = 0 \text{ or } 1 \implies x = -1 \text{ or }-10$$

Edit: If the domain of $x$ is not restricted to real numbers then let $x = r e^{i\theta} \implies -x = re^{i(\theta+\pi)}$,

$$\log(re^{i(\theta+\pi)}) = (\log r)^2 \implies \log r + i(\theta+\pi) = (\log r)^2$$

$$\implies \theta = -\pi \wedge \log r = (\log r)^2 \implies \theta = -\pi \wedge r = 1 \text{ or } 10$$

• What prevents -x from being negative? – The Great Duck Jul 4 '17 at 4:17
• Domain of $\log x$ is $x > 0$. – Dhruv Kohli - expiTTp1z0 Jul 4 '17 at 4:18
• That is blatantly false. log(-1) is very well defined. – The Great Duck Jul 4 '17 at 4:21
• I think the OP needs a solution for $x \in \mathbb{R}$ – Dhruv Kohli - expiTTp1z0 Jul 4 '17 at 4:23
• I wouldn't make that assumption though without them saying so. I've seen stranger. Plus, are you sure allowing for negative x's forces a solution to be off of the real line? – The Great Duck Jul 4 '17 at 4:24