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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.


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

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

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

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

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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. – TMM Sep 22 '11 at 16:01
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 $37/12 = 3.083$.) – Srivatsan Sep 22 '11 at 16:08
@Kaeser, were you taught that $\log_{10} 3$ is approximately $\frac{12}{25}$? – Srivatsan Sep 22 '11 at 16:13
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 $\frac{12}{25} = 0.48$.) – Srivatsan Sep 22 '11 at 16:25
@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.) – robjohn Sep 22 '11 at 19:18

The unwritten core of the problem is that it is possible to determine $C = (a+b)^2/ab$ (and hence also its logarithm to the base 3) given the first equation on $a$ and $b$. This would not be true if the condition had been modified to

$a^2 + b^2 = 28ab + 5, \quad$ or

$a^3 + b^3 = 28ab, \quad$ or


Here the condition can be expressed as $C=30$. But it is not true in general that given some condition on $a$ and $b$, every other function of $a$ and $b$ can be calculated.

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