Помогите решить

log1/2 (2x-1) - log1/2 (16) = 5

lg (5^x+2) = 1/2lg36+lg2

lg3 (4-2x) - log3=2

log3 (12-5x) = 2

log2 (7x*4) = 2+log2 (13)

1) log1/2 (2x-1) - log1/2 (16) = 5 ОДЗ: 2x-1> 0

log1/2 (2x-1) / 16 = 5 2x > 1

(2x-1) / 16 = (1/2) ^5 x > 0,5

(2x-1) / 16 = 1/32

2x-1=0,5

2x=1,5

x=7,5

2) lg (5x+2) = 1/2lg36+lg2

Одз 5x+2>0

lg (5x+2) = lg6+lg2

lg (5x+2) = lg12

5x+2='12

5x=10

x=2

3) log₃ (4-2x) - log₃2=2 ODZ 4-2x>0log₃ ((4-2x) / 2) = log₃3² - 2x>-4

(4-2x) / 2=9 x<2

4-2x=9*2

4-2x=18

2x=4-18

2x=-14

x=-14/2

x=-7

4) log3 (12-5x) = 2 ОДЗ: 12-5x>0; - 5x>-12; 5x<12; x<2,4

12-5x=3^2

12-5x=9

-5x=9-12

-5x=-3

x=3/5

5) log2 (7x - 4) = log2 4 + log2 13 log2 (7x-4) / 4 = log2 13

(7x - 4) / 4 = 13

7x - 4 = 52

x = 8