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July 2022
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пожалуйста помогитее!!!!!решить уравнение а) 3^х+3^(x+1)=4, б) 5^(2x) - 6*5^(x)+5=0
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ShirokovP
Verified answer
3^x + 3^x*3 = 4
3^x (1 + 3) = 4
3^x * 4 = 4
3^x = 1
3^x = 3^0
x = 0
======================================
5^(2x) - 6*5^x + 5 = 0
5^x = t ==>
t^2 - 6t + 5 = 0
D = 36 - 20 = 16
t1 = (6 + 4)/2 = 10/2 = 5
t2 = (6 - 4)/2 = 2/2 = 1
5^x = 5
x = 1
5^x =1
x = 0
1 votes
Thanks 2
ShirokovP
3 - основание
uuuuuu2
а как записать правильно подскажите пожалуйста
ShirokovP
что?
uuuuuu2
решение как записать
ShirokovP
ну так и записать
ShirokovP
x = log(3, 7)
uuuuuu2
спасибо
ShirokovP
Прологарифмируем по основанию 3
ShirokovP
И вот x = log(3, 7)
uuuuuu2
хорошо спасибо
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Answers & Comments
Verified answer
3^x + 3^x*3 = 43^x (1 + 3) = 4
3^x * 4 = 4
3^x = 1
3^x = 3^0
x = 0
======================================
5^(2x) - 6*5^x + 5 = 0
5^x = t ==>
t^2 - 6t + 5 = 0
D = 36 - 20 = 16
t1 = (6 + 4)/2 = 10/2 = 5
t2 = (6 - 4)/2 = 2/2 = 1
5^x = 5
x = 1
5^x =1
x = 0