Решение логарифмических уравнений

1. (2lg (2) + lg (x-3)) / (lg (7x+1) + lg (x-6) + lg (3)) = 1/2
ОДЗ:
(x-3) > 0 => x > 3
(7x+1) > 0 => x > - 1/7
(x-6) > 0 => x > 6
=> общее ОДЗ: x > 6
и
(lg (7x+1) + lg (x-6) + lg (3)) не = 0 или
lg ((7x+1)(x-6)*3)) не= 0 или
3*(7x+1)(x-6) не= 10^0 или не = 1
=>
___ (2lg (2) + lg (x-3)) = (lg (2^2) + lg (x-3)) = lg (4*(x-3) = lg (4x-12)
___ (lg (7x+1) + lg (x-6) + lg (3)) = lg ((7x+1)(x-6)*3)
=>
lg (4x-12) / lg ((7x+1)(x-6)*3) = 1/2
2*lg (4x-12) = lg ((7x+1)(x-6)*3)
lg (4x-12)^2 = lg ((7x+1)(x-6)*3)
(4x-12)^2 = 3*(7x+1)(x-6)
Решить уравнение

2. 5^[2(log5 2 + x)] - 2 = 5^(x + log5 2) (у тебя условие переписано неверно
5^(2log5 2 + 2x) - 2 = 5^(x + log5 2)
5^(2log5 2) * 5^(2x) - 2 = 5^(x) * 5^(log5 2)
5^(log5 2^2) * 5^(2x) - 2 = 5^(x) * 5^(log5 2)
5^(log5 4) * 5^(2x) - 2 = 5^(x) * 5^(log5 2)
4 * 5^(2x) - 2 = 5^(x) * 2 ---> 5^(x) = t ---> 5^(2t) = t^2
4t^2 - 2t - 2 = 0
2t^2 - t - 1 = 0 ---> t1 = - 1/2; t2 = 1
t1 = - 1/2 ---> 5^(x) = - 1/2 - быть не может < 0 => лишний корень
t2 = 1 ---> 5^(x) = 1 ---> 5^(x) = 5^(1) ---> x = 1

3. 7^lgx - 5^(lg x + 1) = 3*5^(lg x - 1) - 13*7^(lgx - 1)
7^(lg x) - 5^(lg x) *5 = 3*5^(lg x)/5 - 13*7^(lgx)/7 ---> (:) на 5^(lg x)
7^(lg x)/5^(lg x) - 5*5^(lg x)/5^(lg ) =
= 3/5*5^(lg x)/5^(lg x) - 13/7 * 7^(lgx)/5^(lg x)
=>
(7/5)^(lg x) - 5 = (3/5) - 13/7 * (7/5)^(lg x) ---> (7/5)^(lg x) = t
t - 5 = 3/5 - 13/7 * t
t * (7/7 + 13/7) = 25/5 + 3/5
t = 28/5 : 20/7 = 28/5 * 7/20 = (7/5)^2
=>
(7/5)^(lg x) = (7/5)^2
lg x = 2
x = 10^2 = 100