[tex]a=\log_{72}48=\dfrac{\log_{2} 48}{\log_2 72}=\dfrac{\log_2 16\cdot3}{\log_2 8\cdot9}=\dfrac{\log_2 16 +\log_2 3}{\log_2 8 +\log_2 3^2} =\dfrac{4+\log_2 3}{3+2\log_2 3}[/tex]
[tex]Notam\ \ \log_2 3 = k\ \ si\ \ a = \dfrac{k+4}{2k+3}\ \ \ (1)[/tex]
Analog, trecand la baza 2 si folosind notatia precedenta, obtinem:
[tex]b = \dfrac{k+3}{k+1} \ \ \ (2)[/tex]
Trebuie sa verificam :
a(b + 3) - 3b + 1 = 0 ⇔ a = (3b-1)/(b+3)
Tinand seama de (2), evaluam (3b-1)/(b+3) si vom obtine, la final, ca expresia este egala cu a.
