wykaz ze:  [latex]log_ab+log_ac=log_abc[/latex]    dokladne obliczenia 

[latex]a^{log_a bc} = bc \ \ \ bc = a ^ {log_a b} cdot a ^ {log_a c} = a ^ {log_a b + log_a c} \ \ \ a^{log_a bc} = a ^ {log_a b + log_a c} \ \ \ log_a b + log_a c = log_a bc\[/latex]