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I'm doing problems in my book that say "Express as a single logarithm and, if possible, simplify". There's two I did that I'm not sure about, and they're even numbered unanswered problems.

The first is $\ln 2x + 3(\ln x - \ln y)$. I put it as $\ln( 2x^5y)$.

The second is $\log_a (a/\sqrt{x}) - \log_a \sqrt{ax}$ Not even sure how to put those together at all.

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  • $\begingroup$ By the second one, do you mean $\frac{\log(a)a}{\sqrt{x}} - \log(a)\sqrt{ax}$? $\endgroup$ – Mike Pierce Nov 30 '14 at 3:13
  • $\begingroup$ The first a in the subscript and then just the second a over the square root of x, and the third a also in the subscript is how it is in the book. $\endgroup$ – windy401 Nov 30 '14 at 3:15
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    $\begingroup$ So $\log_a(\frac{a}{\sqrt{x}}) - \log_a(\sqrt{ax})$? Also, you should check out how to typeset mathematics using LaTeX if you are going to post something difficult to understand. As a hint, using an underscore creates a subscript: \$\log_a(x)\$ --> $\log_a(x)$. $\endgroup$ – Mike Pierce Nov 30 '14 at 3:23
  • $\begingroup$ See math notation guide. $\endgroup$ – user147263 Nov 30 '14 at 4:02
  • $\begingroup$ Yea that's it.And yea, I will, thanks for the link. $\endgroup$ – windy401 Nov 30 '14 at 4:08
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the first one can be written as $$\ln(2)+\ln(x)+3\ln(x)-3\ln(y)=4\ln(x)-3\ln(y)+\ln(2)=$$ $$\ln\left(\frac{x^4}{y^3}\right)+\ln(2)=\ln\left(\frac{2x^4}{y^3}\right)$$ for the second one we obtain $$\log_a\frac{a}{\sqrt{x}\sqrt{ax}}=\log\frac{\sqrt{a}}{x}$$

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Going into detail for second one: $$\log_a\left(\frac{a}{\sqrt x}\right)-\log_a\sqrt{ax}=\log_a\left(\frac{a}{\sqrt{x}\cdot\sqrt{ax}}\right)$$ by the rule that $\log_ax-\log_ay=\log_a(x/y)$. Now we just have to simplify the expression, and we will get our answer. $$\begin{align} &= \log_a\left(\frac{a}{\sqrt{x}\cdot\sqrt{ax}}\right) = \log_a\left(\frac a{x\sqrt a}\right) \\ &= \log_a\left(\frac{\sqrt a}x\right) \end{align}$$

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For the second one,

$\log_a(a/\sqrt{x})-\log_a{\sqrt{ax}}\\=\log_a a-\log_a\sqrt{x}-\log_a\sqrt{a}-\log_a\sqrt{x}\\=\log_a a-\dfrac{1}{2}\log_a{x}-\dfrac{1}{2}\log_a{a}-\dfrac{1}{2}\log_a{x}\\=\dfrac{1}{2}\log_a a-\log_ax\\=\log_a\left({\dfrac{\sqrt{a}}{x}}\right)$

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