Rozwiąż równanie.- Zadanie 87: Matematyka poznać, zrozumieć 2. Zakres rozszerzony. Po gimnazjum

Rozwiązanie

a) Zał:

$x > 0 \land x + 8 > 0$

$x > 0 \land x > - 8$

$x \in (0, + \infty)$

$2 lo g_{5} x + lo g_{5} 2 = lo g_{5} 4 + lo g_{5} (x + 8)$

$lo g_{5} x^{2} + lo g_{5} 2 = lo g_{5} 4 + lo g_{5} (x + 8)$

$lo g_{5} (x^{2} \cdot 2) = lo g_{5} (4 \cdot (x + 8))$

$2 x^{2} = 4 x + 32 ∣ - 4 x - 32$

$2 x^{2} - 4 x - 32 = 0 ∣ : 2$

$x^{2} - 2 x - 16 = 0$

$Δ = (- 2)^{2} - 4 \cdot 1 \cdot (- 16) = 4 + 64 = 68$

$Δ = 68 = 4 \cdot 17 = 217$

$x_{1} = \frac{2 - 2 17}{2} = 1 - 17 < 0 sprzeczno \overset{s}{ˊ} \overset{c}{ˊ}$

$x_{2} = \frac{2 + 2 17}{2} = 1 + 17$

$36 \cdot 16^{\frac{1}{3} (x - \frac{2}{3})} = 3^{\frac{1}{2} x + 2} \cdot 2^{4 - x}$

$4 \cdot 9 \cdot (2^{4})^{\frac{1}{3} x - \frac{2}{9}} = 3^{\frac{1}{2} x + 2} \cdot 2^{4 - x}$

$2^{2} \cdot 3^{2} \cdot 2^{\frac{4}{3} x - \frac{8}{9}} = 3^{\frac{1}{2} x + 2} \cdot 2^{4 - x}$

$2^{2 + \frac{4}{3} x - \frac{8}{9}} \cdot 3^{2} = 3^{\frac{1}{2} x + 2} \cdot 2^{4 - x}$

$2^{\frac{4}{3} x + \frac{10}{9}} \cdot 3^{2} = 3^{\frac{1}{2} x + 2} \cdot 2^{4 - x} ∣ : 3^{2} 2^{4 - x}$

$\frac{2 ^{\frac{4}{3} x + \frac{10}{9}}}{2 ^{4 - x}} = \frac{3 ^{\frac{1}{2} x + 2}}{3 ^{2}}$

$2^{\frac{4}{3} x + \frac{10}{9} - 4 + x} = 3^{\frac{1}{2} x + 2 - 2}$

$2^{\frac{7}{3} x - \frac{26}{9}} = 3^{\frac{1}{2} x}$

Logarytmujemy stronami równanie wykładnicze.

$lo g_{2} 2^{\frac{7}{3} x - \frac{26}{9}} = lo g_{2} 3^{\frac{1}{2} x}$

$(\frac{7}{3} x - \frac{26}{9}) lo g_{2} 2 = \frac{1}{2} x lo g_{2} 3$

$\frac{7}{3} x - \frac{26}{9} = \frac{1}{2} x lo g_{2} 3 ∣ \cdot 54$

$126 x - 156 = 27 x lo g_{2} 3 ∣ : 3$

$42 x - 52 = 9 x lo g_{2} 3 ∣ - 9 x lo g_{2} 3 + 52$

$42 x - 9 x lo g_{2} 3 = 52$

$x (42 - 9 lo g_{2} 3) = 52 ∣ : (42 - 9 lo g_{2} 3)$

$x = \frac{52}{42 - 9 lo g _{2} 3}$

c) Zał:

$lo g_{4} (2 x - 5) > 0 \land 2 x - 5 > 0$

$lo g_{4} (2 x - 5) > lo g_{4} 1 \land 2 x > 5$

$2 x - 5 > 1 \land x > \frac{5}{2}$

$2 x > 6 \land x > \frac{5}{2}$

$x > 3 \land x > \frac{5}{2}$

$x \in (3, + \infty)$

$lo g_{4} (lo g_{4} (2 x - 5)^{16}) = 4$

$lo g_{4} (lo g_{4} (2 x - 5)^{16} = lo g_{4} 256$

$lo g_{4} (2 x - 5)^{16} = 256$

$16 lo g_{4} (2 x - 5) = 256 ∣ : 16$

$lo g_{4} (2 x - 5) = 16$

$4^{16} = 2 x - 5 ∣ + 5$

$4^{16} + 5 = 2 x ∣ : 2$

$\frac{4 ^{16} + 5}{2} = x$

$\frac{( 2 ^{2} ) ^{16} + 5}{2} = x$

$\frac{2 ^{32} + 5}{2} = x$

$\frac{2 ^{32}}{2} + \frac{5}{2} = x$

$2^{31} + \frac{5}{2} = x$

d) Zał:

$x + 2 > 0$

$x > - 2$

$x \in (- 2, + \infty)$

$lo g^{2} (x + 2) - 4 = lo g 1$

$lo g^{2} (x + 2) - 4 = 0$

$(lo g (x + 2) - 2) (lo g (x + 2) + 2) = 0$

$lo g (x + 2) - 2 = 0 \lor lo g (x + 2) + 2 = 0$

$lo g (x + 2) = 2 \lor lo g (x + 2) = - 2$

$10^{2} = x + 2 \lor 10^{- 2} = x + 2$

$100 = x + 2 \lor 0, 01 = x + 2$

$98 = x \lor - 1, 99 = x$

e) Zał:

$x > 0$

$x \in (0, + \infty)$

$\frac{4 - lo g ^{2} x}{lo g x ^{3}} = 1 ∣ \cdot lo g x^{3}$

$4 - lo g^{2} x = lo g x^{3}$

$4 - lo g^{2} x = 3 lo g x ∣ - 3 lo g x$

$- lo g^{2} x - 3 lo g x + 4 = 0$

Podstawmy $t = lo g x$

$- t^{2} - 3 t + 4 = 0$

$Δ = (- 3)^{2} - 4 \cdot (- 1) \cdot 4 = 9 + 16 = 25$

$t_{1} = \frac{3 - 5}{- 2} = \frac{- 2}{- 2} = 1$

$t_{2} = \frac{3 + 5}{- 2} = \frac{8}{- 2} = - 4$

$lo g x = 1 \lor lo g x = - 4$

$10^{1} = x \lor 10^{- 4} = x$

$10 = x \lor 0, 0001 = x$

f) Zał:

$4 x^{2} - 24 > 0 ∣ + 24$

$4 x^{2} > 24 ∣ : 4$

$x^{2} > 6$

$x \in (- \infty, - 6) \cup (6, + \infty)$

$- 4 lo g^{2} (4 x^{2} - 24) + 20 lo g (4 x^{2} - 24) - 24 = 0$

Podstawmy $t = lo g (4 x^{2} - 24)$

$- 4 t^{2} + 20 t - 24 = 0 ∣ : (- 4)$

$t^{2} - 5 t + 6 = 0$

$Δ = (- 5)^{2} - 4 \cdot 1 \cdot 6 = 25 - 24 = 1$

$t_{1} = \frac{5 - 1}{2} = \frac{4}{2} = 2$

$t_{2} = \frac{5 + 1}{2} = \frac{6}{2} = 3$

Zatem otrzymujemy:

$lo g (4 x^{2} - 24) = 2 \lor lo g (4 x^{2} - 24) = 3$

$10^{2} = 4 x^{2} - 24 \lor 10^{3} = 4 x^{2} - 24$

$100 = 4 x^{2} - 24 \lor 1000 = 4 x^{2} - 24$

$124 = 4 x^{2} \lor 1024 = 4 x^{2}$

$31 = x^{2} \lor 256 = x^{2}$

$x = 31 \lor x = - 31 \lor x = 16 \lor x = - 16$

g) Zał:

$\frac{2 x - 3}{x + 1} > 0 ∣ \cdot (x + 1)^{2}$

$(2 x - 3) (x + 1) > 0$

$x \in (- \infty, - 1) \cup (\frac{3}{2}, + \infty)$

$- 2 lo g_{3}^{2} (\frac{2 x - 3}{x + 1}) - 2 lo g_{3} (\frac{2 x - 3}{x + 1}) + 12 = 0$

Podstawmy $t = lo g_{3} (\frac{2 x - 3}{x + 1})$

$- 2 t^{2} - 2 t + 12 = 0 ∣ : (- 2)$

$t^{2} + t - 6 = 0$

$Δ = 1^{2} - 4 \cdot 1 \cdot (- 6) = 1 + 24 = 25$

$t_{1} = \frac{- 1 - 5}{2} = \frac{- 6}{2} = - 3$

$t_{2} = \frac{- 1 + 5}{2} = \frac{4}{2} = 2$

Zatem otrzymujemy:

$lo g_{3} (\frac{2 x - 3}{x + 1}) = - 3 \lor lo g_{3} (\frac{2 x - 3}{x + 1}) = 2$

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

$\frac{1}{27} = \frac{2 x - 3}{x + 1} \lor 9 = \frac{2 x - 3}{x + 1}$

$x + 1 = 27 (2 x - 3) \lor 9 (x + 1) = 2 x - 3$

$x + 1 = 54 x - 81 \lor 9 x + 9 = 2 x - 3$

$82 = 53 x \lor 7 x = - 12$

$\frac{82}{53} = x \lor x = - \frac{12}{7}$

h) Zał:

$2 x + 2 > 0 \land 3 x - 3 > 0 \land x - 1 > 0 \land x + 1 > 0$

$2 x ≻ 2 \land 3 x > 3 \land x > 1 \land x > - 1$

$x > - 1 \land x > 1$

$x \in (1, + \infty)$

$\frac{lo g _{6} ( 2 x + 2 ) + lo g _{6} ( 3 x - 3 )}{lo g _{6} ( x - 1 )} = - lo g_{6} (x + 1)$

$\frac{lo g _{6} ( ( 2 x + 2 ) ( 3 x - 3 ) )}{lo g _{6} ( x - 1 )} = - lo g_{6} (x + 1)$

$\frac{lo g _{6} ( 6 x ^{2} - 6 x + 6 x - 6 )}{lo g _{6} ( x - 1 )} = - lo g_{6} (x + 1)$

$\frac{lo g _{6} ( 6 ( x ^{2} - 1 ) )}{lo g _{6} ( x - 1 )} = - lo g_{6} (x + 1)$

$\frac{lo g _{6} ( 6 ( x - 1 ) ( x + 1 ) )}{lo g _{6} ( x - 1 )} = - lo g_{6} (x + 1)$

$\frac{lo g _{6} 6 + lo g _{6} ( x - 1 ) + lo g _{6} ( x + 1 )}{lo g _{6} ( x - 1 )} = - lo g_{6} (x + 1)$

Podstawmy $a = lo g_{6} (x - 1), b = lo g_{6} (x + 1)$

$\frac{1 + a + b}{a} = - b ∣ \cdot a$

$1 + a + b = - a b ∣ + a b$

$1 + a + b + a b = 0$

$(a + 1) (b + 1) = 0$

$a + 1 = 0 \lor b + 1 = 0$

$a = - 1 \lor b = - 1$

$lo g_{6} (x - 1) = - 1 \lor lo g_{6} (x + 1) = - 1$

$6^{- 1} = x - 1 \lor 6^{- 1} = x + 1$

$\frac{1}{6} = x - 1 \lor \frac{1}{6} = x + 1$

$\frac{7}{6} = x \lor - \frac{5}{6} = x sprzeczno \overset{s}{ˊ} \overset{c}{ˊ}$

$x = \frac{7}{6}$

i) Zał:

$3 x - 1 > 0 ∣ + 1$

$3 x > 1 ∣ : 3$

$x > \frac{1}{3}$

$(3 x - 1)^{l o g_{15} (3 x - 1) + 2} = 3375$

Podstawmy $t = 3 x - 1$

$t^{l o g_{15} t + 2} = 3375$

$lo g_{15} t^{l o g_{15} t + 2} = lo g_{15} 3375$

$(lo g_{15} t + 2) lo g_{15} t = 3$

Podstawmy $s = lo g_{15} t$

$(s + 2) s = 3$

$s^{2} + 2 s - 3 = 0$

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

$s_{1} = \frac{- 2 - 4}{2} = \frac{- 6}{2} = - 3$

$s_{2} = \frac{- 2 + 4}{2} = \frac{2}{2} = 1$

$lo g_{15} t = 1 \lor lo g_{15} t = - 3$

$15 = t \lor 15^{- 3} = t$

$15 = t \lor \frac{1}{3375} = t$

$15 = 3 x - 1 \lor \frac{1}{3375} = 3 x - 1$

$16 = 3 x \lor \frac{3376}{3375} = 3 x$

$\frac{16}{3} = x \lor \frac{3376}{10125} = x$

Magdalena Matusik

Nauczycielka matematyki

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