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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----

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

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

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

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

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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

1 Answers 1

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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

$\sqrt{a^2+b^2}=28ab$.

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.