решите уравнения
log0,5 x + 3 logx 0,5 = 4
5 log3 x - 3 logx 3 = 2
log0,3 x + 9 logx 0,3 = 10
Ответы на вопрос
Ответил zulkarnai
0
1) log₀.₅X + 3*
= 4
log₀.₅² X + 3 = 4log₀.₅X
log₀.₅X = t
t² - 4t + 3 =0
t₁ = 1
t₂ = 3
log₀.₅X = 1
x = 0.5
log₀.₅X = 3
x = 0.125
Ответ: 0,5; 0,125
2) 5*log₃X - 3*
= 2
5*log₃²X - 3 = 2log₃X
log₃X = t
5t² - 2t - 3=0
t₁ = 1
t₂ =
log₃X = 1
x = 3
log₃X=
x=![3^{- frac{3}{5} } = frac{1}{ sqrt[5]{27} } 3^{- frac{3}{5} } = frac{1}{ sqrt[5]{27} }](https://tex.z-dn.net/?f=+3%5E%7B-+frac%7B3%7D%7B5%7D+%7D+%3D++frac%7B1%7D%7B+sqrt%5B5%5D%7B27%7D+%7D+)
3) log₀.₃X + 9*
= 10
log₀.₃²X + 9 = 10*log₀.₃X
log₀.₃X = t
t² - 10t + 9=0
t₁ = 1
t₂ = 9
log₀.₃X = 1
x = 0.3
log₀.₃X = 9
x = (0.3)⁹
log₀.₅² X + 3 = 4log₀.₅X
log₀.₅X = t
t² - 4t + 3 =0
t₁ = 1
t₂ = 3
log₀.₅X = 1
x = 0.5
log₀.₅X = 3
x = 0.125
Ответ: 0,5; 0,125
2) 5*log₃X - 3*
5*log₃²X - 3 = 2log₃X
log₃X = t
5t² - 2t - 3=0
t₁ = 1
t₂ =
log₃X = 1
x = 3
log₃X=
x=
3) log₀.₃X + 9*
log₀.₃²X + 9 = 10*log₀.₃X
log₀.₃X = t
t² - 10t + 9=0
t₁ = 1
t₂ = 9
log₀.₃X = 1
x = 0.3
log₀.₃X = 9
x = (0.3)⁹
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