Va ro sa ma ajutati cu aceasta problema

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a fost răspuns

Răspuns :

Аноним Аноним · Answer 1 · 2021-07-15T09:54:18+03:00

Răspuns: [tex] \color{red} \boxed{x = 3}[/tex]

Explicație pas cu pas:

[tex] \bf log_{9}x + log_{27}x + log_{ \frac{1}{3} }x = - \frac{1}{6} [/tex]

[tex] \bf log_{ {3}^{2} }x + log_{ {3}^{3} }x + log_{ {3}^{ - 1} }x = - \frac{1}{6} [/tex]

[tex] \bf \frac{1}{2} log_{3}x + \frac{1}{3} log_{3}x - log_{3}x = - \frac{1}{6} [/tex]

[tex] \bf - \frac{1}{6} log_{3}x = - \frac{1}{6} [/tex]

[tex] \bf log_{3}x = 1[/tex]

[tex] \color{red} \boxed{x = 3}[/tex]

102533 102533 · Answer 2 · 2021-07-15T10:20:47+03:00

Răspuns:

Explicație pas cu pas:

log₉x +log₂₇x + log₁/₃x = -1/6

Conditie x > 0

log₉x = log₃x / log₃9 = log₃x / log₃3² = log₃x / 2log₃3 = log₃x /2

log₂₇x = log₃x / log₃27 = log₃x / log₃3³ = log₃x / 3log₃3 =log₃x/3

log₁/₃ x = log₃x / log₃1/3 = log₃x / log₃3⁻¹ = - log₃x

log₉x +log₂₇x + log₁/₃x = -1/6 <=>

³⁾log₃x /2 + ²⁾log₃x/3 - ⁶⁾log₃x = -1/6 <=>

3log₃x + 2log₃x - 6 log₃x = -1 <=>

log₃x³ + log₃x² - log₃x⁶ = -1 <=>

log₃(x³·x²:x⁶) = -1 <=>

log₃(1/x) = -1 <=>

- log₃x = -1 =>

log₃x = 1 <=>

log₃x = log₃3 =>

x = 3