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artyomya1
@artyomya1
March 2022
1
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помогите пожалуйста посчитать логарифмические и показательные уравнения
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amax777
1)
(1/3)^(x^2-5x+8) = 1/9
(1/3)^(x^2-5x+8) = (1/3)^2
x^2 - 5x + 8 = 2
x^2 - 5x + 6 = 0
D = 5^2 - 4 6 = 25 - 24 = 1
x1 = ( 5 - 1)/2 = 2
x2 = ( 5 +1)/2 = 3
2)
4^(x-1) - 1 / 2^(3-x) = 62
[2^(x-1)]^2 - (1/4) 2^(x-1) = 62
Обозначим 2^(x-1) = t > 0
t^2 - t/4 - 62 = 0
D = (1/4)^2 + 4 62 = 3969/16 = (63/4)^2
t1 = (1/4 - 63/4)/2 < 0 - не удв. условию (t>0)
t2 = (1/4 +63/4)/2 = 8
2^(x-1) = 8
2^(x-1)=2^3
x-1=3
x=4
3)
log (5) (3x^2 + 12x + 5) = 1
log (5) (3x^2 + 12x + 5) = log (5) (5)
3x^2 + 12x + 5 = 5
3x(x + 4) = 0
x1 = 0
x2 = -4
3 x^2 + 12x + 5>0, т.к. находится в логарифме. Проверим:
3 x1^2 + 12 x1 + 5 = 5 > 0
3x2^2 + 12 x2 + 5 = 5 > 0
Ответ:
x = 0, x = -4
4)
Найдем допустимые значения:
x>0 т.к. стоит в логарифме
недопустимо: 3 - lg(x) = 0
3 = lg(x)
3 lg(10) = lg(x)
lg(10^3) = lg(x)
x = 10^3
недопустимо: lg(x) - 1 = 0
lg(x) = 1
lg(x) = lg(10)
x= 10
Теперь займемся уравнением:
1 / (3 - lg(x) ) + 2 / (lg(x) - 1) = 3
обозначим: lg(x) - 1 = t
1 / (2 -t) + 2 / t = 3
t + 2(2-t) = 3 t (2-t)
t + 4 - 2t = 6t - 3t^2
3t^2 - 7t + 4 = 0
D = 49 - 48 = 1
t1 = (7 - 1)/6 = 1
t2 = (7 + 1)/6 = 4/3
lg(x) - 1 = t1
lg(x) - 1 = 1
lg(x) = 2
lg(x) = 2 lg(10)
lg(x) = lg(10^2)
x = 10^2
lg(x) - 1 = t2
lg(x) - 1 = 4/3
lg(x) = 7/3
lg(x) = (7/3) lg(10)
lg(x) = lg(10^(7/3))
x = 10^(7/3)
Ответ: x = 10^2, x = 10^(7/3)
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Answers & Comments
(1/3)^(x^2-5x+8) = 1/9
(1/3)^(x^2-5x+8) = (1/3)^2
x^2 - 5x + 8 = 2
x^2 - 5x + 6 = 0
D = 5^2 - 4 6 = 25 - 24 = 1
x1 = ( 5 - 1)/2 = 2
x2 = ( 5 +1)/2 = 3
2)
4^(x-1) - 1 / 2^(3-x) = 62
[2^(x-1)]^2 - (1/4) 2^(x-1) = 62
Обозначим 2^(x-1) = t > 0
t^2 - t/4 - 62 = 0
D = (1/4)^2 + 4 62 = 3969/16 = (63/4)^2
t1 = (1/4 - 63/4)/2 < 0 - не удв. условию (t>0)
t2 = (1/4 +63/4)/2 = 8
2^(x-1) = 8
2^(x-1)=2^3
x-1=3
x=4
3)
log (5) (3x^2 + 12x + 5) = 1
log (5) (3x^2 + 12x + 5) = log (5) (5)
3x^2 + 12x + 5 = 5
3x(x + 4) = 0
x1 = 0
x2 = -4
3 x^2 + 12x + 5>0, т.к. находится в логарифме. Проверим:
3 x1^2 + 12 x1 + 5 = 5 > 0
3x2^2 + 12 x2 + 5 = 5 > 0
Ответ:
x = 0, x = -4
4)
Найдем допустимые значения:
x>0 т.к. стоит в логарифме
недопустимо: 3 - lg(x) = 0
3 = lg(x)
3 lg(10) = lg(x)
lg(10^3) = lg(x)
x = 10^3
недопустимо: lg(x) - 1 = 0
lg(x) = 1
lg(x) = lg(10)
x= 10
Теперь займемся уравнением:
1 / (3 - lg(x) ) + 2 / (lg(x) - 1) = 3
обозначим: lg(x) - 1 = t
1 / (2 -t) + 2 / t = 3
t + 2(2-t) = 3 t (2-t)
t + 4 - 2t = 6t - 3t^2
3t^2 - 7t + 4 = 0
D = 49 - 48 = 1
t1 = (7 - 1)/6 = 1
t2 = (7 + 1)/6 = 4/3
lg(x) - 1 = t1
lg(x) - 1 = 1
lg(x) = 2
lg(x) = 2 lg(10)
lg(x) = lg(10^2)
x = 10^2
lg(x) - 1 = t2
lg(x) - 1 = 4/3
lg(x) = 7/3
lg(x) = (7/3) lg(10)
lg(x) = lg(10^(7/3))
x = 10^(7/3)
Ответ: x = 10^2, x = 10^(7/3)