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$\log x^2 = 3, x > 0$

If I enter a negative number like this
$\log ((-2)^2) = 3, x > 0$
It is valid right?

When it just says $\log x^2$ what is applied first? Log or the exponential?

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    $\log x^2$ means $\log(x^2)$. If you mean $(\log x)^2$, use parentheses.2011-10-05
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    Even so, $\log ((-4)^2)$ is still valid?2011-10-05
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    Yes: $\log((-4)^2) = \log 16$, which is fine.2011-10-05
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    $\log ((-4)^2) = \log (16) = 2\log (4)$ is valid as you are staying in the real numbers and taking a $\log$ of a positive number. $\log ((-4)^2) = 2\log (-4)$ is not valid because the [complex logarithm](http://en.wikipedia.org/wiki/Complex_logarithm) is multivalued.2011-10-05
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    @Algific: Yes, $\log ((-4)^2)$ is valid, in the sense that it represents a "construction scheme" for a real number that doesn't leave the realm of real numbers. The "construction scheme" is: $-4$ is inputted, then $(-4)^2 = 16$ is computed, then $\log(16)$ is computed. This can be compared with $(\sqrt{-4})^2$, which represents the real number $-4$, but the "construction scheme" I gave has you leaving the real numbers during the computation. (The phrase "construction scheme" is something I made up just now for the purpose of this note.)2011-10-05
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    http://www.wolframalpha.com/input/?i=lg+x%5E2+-3%3D0 why exclude it when negative numbers works just fine?2011-10-05
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    Perhaps the reason is that you want to be able to use properties of logarithms. Note that $\log x^2 = 2\log x$ is only valid if $x\gt 0$. For more general $x$, you would need to do $\log x^2 = 2\log|x|$. Also, notice that you don't actually have a function, what you have is an equation, and one is free to restrict the domain of consideration for an equation as one sees fit, for whatever reason one may wish.2011-10-05
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    @ArturoMagidin thanks! What about the other way? Example: $\log x^2 = 3$ and $10^{\log x^2} = 10^3$ gives $x^2=10^3$2011-10-05
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    @Algific: That's fine. Again: it depends what it is you want to do with the equation (which is not a function, it's an *equation*). Of course, notice that once you get to $x^2=10^3$, to "solve for $x$" you would need to do $|x|=\sqrt{10^3}$ (since $\sqrt{x^2}=|x|$). Perhaps the final solution only makes sense for positive $x$ so you can restrict from the beginning? Again: you are not being asked to consider *natural domain* of $\log(x^2)$, you are being asked to consider an *equation* with a particular restricted set of possible solutions.2011-10-05

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It's sometimes all about the restrictions. To solve: $\log (x^2) = 3$, $x > 0$, we interpret this as "Solve $\log(x^2) = 3$, but only give answers $x$ that are greater than $0$".

First take both sides as exponents on 10:

$$ \log (x^2) = 3 $$ $$10^{\log (x^2)} = 10^3 $$ $$x^2 = 1000, \textrm{by properties of $\log$}$$ $$x = \pm \sqrt{1000} \approx \pm 31.6227766$$

So without restrictions, we would say $x = \pm 31.6227766$, however, with the restriction that $x > 0$, we must only choose the positive result:

$$ x = 31.6227766. $$

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    I don't see why you interpret the equation that way. Clearly $-x$ is a solution also.2011-10-05
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    @FredrikMeyer: How would you solve "x^2 = 4, x > 2"? Again, the restriction must be satisfied as well as the equation. Think of it as a system of two expressions; one of the expressions happens to be an inequality.2011-10-14
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    Oh, sorry. I missed the additional restriction. My bad.2011-10-14