!!!1) log3 8 - log3 2 + log3 9/4 =? 2) 2log2 6 + log2 24/9 - log2 24= ? 3) log6 2 - log6...

Решите задачу:

$\displaystyle log_38-log_32+log_3 \frac{9}{4}=log_3 \frac{8}{2}+log_3 \frac{9}{4}=log_3 \frac{8*9}{2*4}=log_39=2$

$\displaystyle 2log_26+log_2 \frac{24}{9}-log_224=log_2 \frac{6^2*24}{9*24}=log_24=2$

$\displaystyle log_62-log_62*log_26+log_618=log_62-log_62* \frac{1}{log_62}+log_618=$

$\displaystyle = log_62*18-1=log_236-1=2-1=1$