Rozwiąż układy równań ...- Zadanie 11: Matematyka 2. Zakres rozszerzony. Po gimnazjum

Rozwiązanie

${lo g x + lo g y = 12 lo g x^{2} - lo g y = 18$

Założenia:

$x > 0 \land y > 0$

Rozwiązujemy układ równań:

${lo g x + lo g y = 12 ∣ - lo g x lo g x^{2} - lo g y = 18$

${lo g y = 12 - lo g x 2 lo g x - lo g y = 18$

${lo g y = 12 - lo g x 2 lo g x - (12 - lo g x) = 18$

${lo g y = 12 - lo g x 2 lo g x - 12 + lo g x = 18$

${lo g y = 12 - lo g x 3 lo g x - 12 = 18 ∣ + 12$

${lo g y = 12 - lo g x 3 lo g x = 30 ∣ : 3$

${lo g y = 12 - lo g x lo g x = 10$

${lo g y = 12 - lo g x 10^{10} = x$

${lo g y = 12 - lo g 10^{10} x = 10^{10}$

${lo g y = 12 - 10 x = 10^{10}$

${lo g y = 2 x = 10^{10}$

${10^{2} = y x = 10^{10}$

${x = 10^{10} y = 10^{2}$

${8^{x} \cdot 4^{y} = 2 lo g_{2} x - lo g_{2} y = - 1$

Założenia:

$x > 0 \land y > 0$

Rozwiązujemy układ równań:

$⎩ ⎨ ⎧ (2^{3})^{x} \cdot (2^{2})^{y} = 2 lo g_{2} (\frac{x}{y}) = - 1$

${2^{3 x} \cdot 2^{2 y} = 2 2^{- 1} = \frac{x}{y}$

${2^{3 x + 2 y} = 2^{1} \frac{1}{2} = \frac{x}{y} ∣ \cdot 2$

${3 x + 2 y = 1 1 = 2 \frac{x}{y} ∣ \cdot y$

${3 x + 2 y = 1 y = 2 x$

${3 x + 2 \cdot 2 x = 1 y = 2 x$

${3 x + 4 x = 1 y = 2 x$

${7 x = 1 ∣ : 7 y = 2 x$

${x = \frac{1}{7} y = 2 x$

${x = \frac{1}{7} y = 2 \cdot \frac{1}{7}$

${x = \frac{1}{7} y = \frac{2}{7}$

${x^{y} = 8 lo g_{2} x = y - 2$

Założenia:

$x > 0$

Rozwiązujemy układ równań:

${x^{y} = 8 lo g_{2} x = y - 2$

${x^{y} = 8 2^{y - 2} = x$

${(2^{y - 2})^{y} = 8 x = 2^{y - 2}$

${2^{(y - 2) \cdot y} = 8 x = 2^{y - 2}$

${2^{y^{2} - 2 y} = 2^{3} x = 2^{y - 2}$

${y^{2} - 2 y = 3 ∣ - 3 x = 2^{y - 2}$

${y^{2} - 2 y - 3 = 0 x = 2^{y - 2}$

Rozwiązujemy równanie kwadratowe:

$Δ = (- 2)^{2} - 4 \cdot 1 \cdot (- 3) = 4 + 12 = 16$

$y_{1} = \frac{- ( - 2 ) - 16}{2 \cdot 1} = \frac{2 - 4}{2} = \frac{- 2}{2} = - 1$

$y_{2} = \frac{- ( - 2 ) + 16}{2 \cdot 1} = \frac{2 + 4}{2} = \frac{6}{2} = 3$

Mamy więc:

${y = - 1 x = 2^{y - 2} \lor {y = 3 x = 2^{y - 2}$

${y = - 1 x = 2^{- 1 - 2} \lor {y = 3 x = 2^{34 - 2}$

${y = - 1 x = 2^{- 3} \lor {y = 3 x = 2^{1}$

${y = - 1 x = (\frac{1}{2})^{3} \lor {y = 3 x = 2$

${y = - 1 x = \frac{1}{8} \lor {x = 2 y = 3$

${x = \frac{1}{8} y = - 1 \lor {x = 2 y = 3$

${lo g_{2} x + lo g_{4} y = - 1 lo g 2_{y} + lo g_{\frac{1}{2}} x = \frac{5}{2}$

Założenia:

$x > 0 \land y > 0$

Rozwiązujemy układ równań:

${lo g_{2} x + lo g_{4} y = - 1 lo g 2_{y} + lo g_{\frac{1}{2}} x = \frac{5}{2}$

$⎩ ⎨ ⎧ lo g_{2} x + \frac{l o g _{2} y}{l o g _{2} 4} = - 1 lo g 2_{y} + \frac{l o g _{2} x}{l o g _{2} ( \frac{1}{2} )} = \frac{5}{2}$

$⎩ ⎨ ⎧ lo g_{2} x + \frac{l o g _{2} y}{l o g _{2} 2 ^{2}} = - 1 lo g 2_{y} + \frac{l o g _{2} x}{l o g _{2} 2 ^{- 1}} = \frac{5}{2}$

${lo g_{2} x + \frac{l o g _{2} y}{2} = - 1 ∣ \cdot 2 lo g 2_{y} + \frac{l o g _{2} x}{- 1} = \frac{5}{2}$

${2 lo g_{2} x + lo g_{2} y = - 2 lo g 2_{y} - lo g_{2} x = \frac{5}{2}$

${lo g_{2} x^{2} + lo g_{2} y = - 2 lo g_{2} (\frac{y}{x}) = \frac{5}{2}$

${lo g_{2} (x^{2} \cdot y) = - 2 lo g_{2} (\frac{y}{x}) = \frac{5}{2}$

${2^{- 2} = x^{2} \cdot y 2^{\frac{5}{2}} = \frac{y}{x}$

${(\frac{1}{2})^{2} = x^{2} \cdot y 2^{\frac{5}{2}} = \frac{y}{x}$

${\frac{1}{4} = x^{2} \cdot y ∣ : x^{2} 2^{\frac{5}{2}} = \frac{y}{x}$

${\frac{1}{4 x ^{2}} = y 2^{\frac{5}{2}} = \frac{y}{x}$

${y = \frac{1}{4 x ^{2}} 2^{\frac{5}{2}} = \frac{\frac{1}{4 x ^{2}}}{x}$

${y = \frac{1}{4 x ^{2}} 2^{\frac{5}{2}} = (\frac{1}{4 x ^{2}}) \cdot \frac{1}{x}$

${y = \frac{1}{4 x ^{2}} 2^{\frac{5}{2}} = \frac{1}{4 x ^{3}}$

${y = \frac{1}{4 x ^{2}} 2^{\frac{5}{2}} = \frac{1}{4 x ^{3}} ∣ \cdot 4$

${y = \frac{1}{4 x ^{2}} 4 \cdot 2^{\frac{5}{2}} = \frac{1}{x ^{3}}$

${y = \frac{1}{4 x ^{2}} 2^{2} \cdot 2^{\frac{5}{2}} = \frac{1}{x ^{3}}$

${y = \frac{1}{4 x ^{2}} 2^{2 + \frac{5}{2}} = \frac{1}{x ^{3}}$

${y = \frac{1}{4 x ^{2}} 2^{\frac{9}{2}} = \frac{1}{x ^{3}} ∣ \cdot x^{3}$

${y = \frac{1}{4 x ^{2}} x^{3} \cdot 2^{\frac{9}{2}} = 1 ∣ : 2^{\frac{9}{2}}$

${y = \frac{1}{4 x ^{2}} x^{3} = \frac{1}{2 ^{\frac{9}{2}}}$

${y = \frac{1}{4 x ^{2}} x^{3} = \frac{1}{2 ^{3 + \frac{3}{2}}}$

$⎩ ⎨ ⎧ y = \frac{1}{4 x ^{2}} x^{3} = \frac{1}{2 ^{3} \cdot 2 ^{\frac{3}{2}}} ∣ (\dots)^{\frac{1}{3}}$

$⎩ ⎨ ⎧ y = \frac{1}{4 x ^{2}} x = \frac{1}{( 2 ^{3} \cdot 2 ^{\frac{3}{2}} ) ^{\frac{1}{3}}}$

$⎩ ⎨ ⎧ y = \frac{1}{4 x ^{2}} x = \frac{1}{( 2 ^{3} ) ^{\frac{1}{3}} \cdot ( 2 ^{\frac{3}{2}} ) ^{\frac{1}{3}}}$

${y = \frac{1}{4 x ^{2}} x = \frac{1}{2 ^{3 \cdot \frac{1}{3}} \cdot 2 ^{\frac{3}{2} \cdot \frac{1}{3}}}$

${y = \frac{1}{4 x ^{2}} x = \frac{1}{2 \cdot 2 ^{\frac{1}{2}}}$

${y = \frac{1}{4 x ^{2}} x = \frac{1}{2 2}$

${y = \frac{1}{4 \cdot ( \frac{1}{2 2} ) ^{2}} x = \frac{1}{2 2}$

${y = \frac{1}{4 \cdot \frac{1}{2 ^{2} \cdot 2 ^{2}}} x = \frac{1}{2 2}$

${y = \frac{1}{4 \cdot \frac{1}{4 \cdot 2}} x = \frac{1}{2 2}$

${y = \frac{1}{4 \cdot \frac{1}{8}} x = \frac{1}{2 2}$

${y = \frac{1}{\frac{1}{2}} x = \frac{1}{2 2}$

${y = 1 \cdot 2 x = \frac{1}{2 2}$

${y = 2 x = \frac{1}{2 2}$

${x = \frac{1}{2 2} y = 2$

Agnieszka Sermak

Nauczycielka matematyki

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