Question

log2 x- log4 9 + 1/log3 2 + 2 =6
Пожалуйста! !!!!!!!

Answer 1

формулы :
loga b -loga c=loga b/c
loga b+loga c=loga b*c
1/logb a=loga b
loga^n (b^n) =loga b
Решение :
log2 x-log2 3 +log2 3 +2=6
log2 ((x/3)*3)=6-2
log2 x =4
x=16

Answer 2

6=log2 64 2=log2 4 log2 x- log2 3 +log2 3 +log2 4= log2 64 log2 x*4 =log2 64 x*4=64 x=64/4=16