Răspuns:
Explicație pas cu pas:
[tex]\log_3216 = \log_3 (6^3) = 3\cdot \log_36 = 3\cdot \log_3(3\cdot 2) = 3\cdot (\log_33 + \log_32) =\\ = 3\cdot (1 + a)\\\texttt{Am folosit identitatile: }\\\log_ab^n = n\cdot \log_ab\\log_a(b\cdot c) = \log_a b + \log_a c[/tex]
