Question

Log9 (27c) если log3 c=0,5

Answer 1

$log_9 (27c)=\frac{log_3(27c)}{log_39}=\frac{log_327+log_3c}{log_33^2}=\frac{log_33^3+0,5}{2log_33}=\frac{3log_33+0,5}{2}=\frac{3+0,5}{2}=\\\\\frac{3,5}{2}=1,75$