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given $a^2+b^2=28ab$ what's $\log_{3} \left(\dfrac{(a+b)^2}{ab}\right)$?

$\log_{3} \left(\dfrac{(a+b)^2}{ab}\right)$

$\log_{3} \left(\dfrac{a^2+b^2+2ab}{ab}\right)$

$\log_{3} \left(\dfrac{a^2+b^2}{ab}+\dfrac{2ab}{ab}\right)$

$\log_{3} \left(\dfrac{28ab}{ab}+\dfrac{2ab}{ab}\right)$

$\log_{3} 30$

Here I tried using properties but couldn't manage to get trough.

---edit----

$\log_{3} 3 + \log_{3} 10 = 1 + \log_{3} 10$

$\log_{10} 3 = \dfrac{25}{12} = \dfrac{\log_{3} 3}{\log_{3} 10} = \dfrac{1}{\log_{3}10}$

$\dfrac{12}{25} = \dfrac{1}{\log_{3}10} \implies \log_{3}10 = \dfrac{25}{12}$

$\log_{3} 30 = 1 + \log_{3} 10 = 1 + \dfrac{25}{12}=\dfrac{12+25}{12}=\dfrac{37}{12}$

  • 2
    You're basically done. You could reduce it further to $\log_3 30 = \log_3 3 + \log_3 10 = 1 + \log_3 10$, but some may argue that is not even a simplication.2011-09-22
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    The answer is $37/12$ what labelled value I am entitled to know here?2011-09-22
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    The problem would work out simpler if $a^2+b^2=25ab$, but life isn't always easy.2011-09-22
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    No, $\log_{3}(30)$ is **not** equal to $37/12$. Perhaps they expect you to approximate it or calculate the answer numerically. Even then I have no idea why they are writing it as a fraction rather than as a decimal. (Note: $\log_3(30) = 3.096$ and $37/12 = 3.083$.)2011-09-22
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    $\log_3(30)$ is not rational, so the answer is not $37/12$. It is close:$$\begin{array}{rl}37/12&=3.08333333333333\\ \log_3(30)&=3.09590327428938\end{array}$$2011-09-22
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    @Kaeser, were you taught that $\log_{10} 3$ is approximately $\frac{12}{25}$?2011-09-22
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    No. I am trying using that now.2011-09-22
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    Well, if you use $\log_{10} 3 \approx \frac{12}{25}$, you should be able to get the text-book answer. But if you weren't told to use this approximation, then I do not see any point in doing so. (In fact, IMO the approximation isn't that great; $\log_{10} 3 = 0.477\ldots$ while $\frac{12}{25} = 0.48$.)2011-09-22
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    Got it, I will edit the question to answer it. Thanks.2011-09-22
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    You shouldn't write $\log_3 \frac{a^2+b^2}{ab} + \frac{2ab}{ab}$ when you mean $\log_3 \left(\frac{a^2+b^2}{ab} + \frac{2ab}{ab}\right)$.2011-09-22
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    Basically, $\log_3 30$ can't be simplified beyond $1 + \log_3 10$. So either plug that in a calculator or leave it in this form. ${37 \over 12}$ is just an approximation to the correct answer; it is slightly off.2011-09-22
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    @Srivatsan: Indeed, $\frac{10}{21}$ is better than $3\times$ as close as $\frac{12}{25}$. ($\frac{21}{44}$ and $\frac{73}{153}$ are the next two continued fraction approximants.)2011-09-22
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    @Srivatsan Narayanan:$\log_{10} 3 \approx \frac{12}{25}$ is this some thing that is supposed to be taught or useful to know? Like we sometime use $\pi \approx \frac{22}{7}$ for elementary mensuration problems,however I was not taught the first approximation either.2011-09-22
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    @Fool No, sorry if I misled. The only thing I was taught was to use log tables. And I remember the values of $\log_{10} 2 = .3010$ and $\log_{10} 3 = .4771$. I meant to clarify if the OP has been taught this and if so, then it would explain the 37/12 answer. Basically it was some reverse engineering the answer. I guess I didn't convey so in my comment. My apologies (to all, including the OP) if I was careless.2011-09-22
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    @Kaeser: It not *not good* to write $\log_{10} 3 = \frac{25}{12}$ as it is simply wrong. You could write $\log_{10} 3 \approx \frac{25}{12}$ but writing an equal-sign just hurts, especially on a math Q&A-site.2011-09-22

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