Explicație pas cu pas:
[tex]log_{3}(9) + log_{3}(2) - log_{3}(6) = log_{3}( {3}^{2} ) + log_{3}( \frac{2}{6} ) \\ = 2log_{3}(3) + log_{3}( \frac{1}{3} ) = 2 + log_{3}( {3}^{ - 1} ) = 2 - log_{3}(3) \\ = 2 - 1 = 1[/tex]
[tex]\it log_39+log_32-log_36=log_3\dfrac{9\cdot2}{6}=log_33=1[/tex]
