<s>[INST] given $a^2+b^2=28ab$ what's $\log_{3}\left (\frac{(a+b)^2}{ab}\right)$? 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.
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$\log_{3} 3 + \log_{3} 10 = 1 + \log_{3} 10$
$\log_{10} 3 = \dfrac{25}{12} = \dfrac{\log_{3} 3}{\log_{3} 10} = \dfrac{1}{\log_{3}10}$
$\dfrac{12}{25} = \dfrac{1}{\log_{3}10} \implies \log_{3}10 = \dfrac{25}{12}$
$\log_{3} 30 = 1 + \log_{3} 10 = 1 + \dfrac{25}{12}=\dfrac{12+25}{12}=\dfrac{37}{12}$
 [/INST] 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. 
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