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