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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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    @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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    Oh, sorry. I missed the additional restriction. My bad.2011-10-14