log 0,0625(5х-7/7) = -0,5
log 0,25²(5х-7/7) = -0,5
1/2 log 0,25(5х-7/7) = -0,5
log 0,25(5х-7/7) = -1
log 0,5²(5х-7/7) = -1
1/2 log0,5(5х-7/7) = -1
log 0,5(5х-7/7) = -2
log 1/2 (5х-7/7) = - 2
log2^-1 (5х-7/7) = - 2
- log 2 (5х-7/7) = - 2
log 2 (5х-7/7) = 2
log 2 (5х-7/7) = log 2 (4)
5х-7/7 = 4
5х-7 = 28
5х = 35
x = 7
Ответ: 7