Решить логарифмическое уравнение. log(2; (9-2^x))=3^log(3; (3-x)

Решение:
^ - здесь степень
V - корень квадр.
Д4.12
log 5 (7-x) = log 5 (3-x) + 1
log 5 (7-x) = log 5 (3-x) + log 5 (5)
log 5 (7-x) = log 5 [5*(3-x)]
7-x = 5*(3-x)
7-x = 15 - 5x
5x = 8
x = 5/8

Д4.11
log (x-5) 49 = 2
(x-5)^2 = 49
x^2 - 10x + 25 = 49
x^2 - 10x - 24 = 0
x(1) = 12
x(2) = - 2

Д4.10
2^(3+x) = 0,4 * 5^(3+x)
2^3 * 2^x = 2/5 * 5^3 * 5^x
2^3 * 2^x = 2 * 5^2 * 5^x
2^x /5^x = 2/2^3 * 5^2
(2/5)^x = (5/2)^2
(2/5)^x = (2/5)^(-2)
x = -2

Д4.9
(1/3)^(3+x) = 9
[3^(-1)] ^(3+x) = 3^2
3^ (-3-x) = 3^2
-3-x = 2
x = -5

Д4.6
V(6+5x) = x
6+5x = x^2
x^2 - 5x - 6 = 0
x(1) = +6
x(2) = -1

Д4.5
V(1/(5-2x) = 1/3
1/(5-2x) = 1/9
5-2x = 9
2x = -4
x = -2

Д4.4
11x / (2x^2 + 5) = 1
11x = 2x^2 + 5
2x^2 - 11x + 5 = 0
x(1) = +5
x(2) = +1/2

Д4.3
x = (8x+25) / (x+8)
x^2 + 8x = 8x + 25
x^2 = 25
x(1) = +5
x(2) = -5

Д4.2
1/7 * x^2 = 9 1/7
1/7 * x^2 = 64/7
x^2 = 64
x(1) = +8
x(2) = -8

Д4.1
(2x+7)^2 = (2x-1)^2
4x^2 + 28x + 49 = 4x^2 - 4x + 1
24x = - 48
x = -2