Log16 (x^2-2x-3)^2-2 log16(x^2+x-2)=1/2

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Log16 (x^2-2x-3)^2-2 log16(x^2+x-2)=1/2

TheSystem 28.02.2026 20:51

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Лебедев Дмитрий Сергеевич · Accepted Answer

Для решения уравнения $\log_{16} (x^{2} - 2 x - 3)^{2} - 2 \log_{16} (x^{2} + x - 2) = \frac{1}{2} log base 16 of open paren x squared minus 2 x minus 3 close paren squared minus 2 log base 16 of open paren x squared plus x minus 2 close paren equals one-half$ воспользуемся свойствами логарифмов и определим область допустимых значений (ОДЗ). 1. Область допустимых значений (ОДЗ) Аргументы логарифмов должны быть строго больше нуля:

$(x^{2} - 2 x - 3)^{2} > 0 open paren x squared minus 2 x minus 3 close paren squared is greater than 0$ — это выполняется всегда, кроме случаев, когда основание равно нулю:
$x^{2} - 2 x - 3 \neq 0 ⟹ (x - 3) (x + 1) \neq 0 ⟹ x \neq 3, x \neq -1 x squared minus 2 x minus 3 is not equal to 0 ⟹ open paren x minus 3 close paren open paren x plus 1 close paren is not equal to 0 ⟹ bold x is not equal to 3 comma bold x is not equal to negative 1$ . $x^{2} + x - 2 > 0 x squared plus x minus 2 is greater than 0$
Находим корни уравнения $x^{2} + x - 2 = 0 x squared plus x minus 2 equals 0$ : $x_{1} = 1, x_{2} = -2 x sub 1 equals 1 comma x sub 2 equals negative 2$ .
Парабола ветвями вверх, следовательно: $x \in (- \infty; -2) \cup (1; + \infty) bold x is an element of open paren negative infinity ; negative 2 close paren union open paren 1 ; positive infinity close paren$ .

Итоговое ОДЗ: $x \in (- \infty; -2) \cup (1; 3) \cup (3; + \infty) x is an element of open paren negative infinity ; negative 2 close paren union open paren 1 ; 3 close paren union open paren 3 ; positive infinity close paren$ . 2. Преобразование уравнения Используем свойство $n \log_{a} b = \log_{a} b^{n} n log base a of b equals log base a of b to the n-th power$ для второго слагаемого: $\log_{16} (x^{2} - 2 x - 3)^{2} - \log_{16} (x^{2} + x - 2)^{2} = \frac{1}{2} log base 16 of open paren x squared minus 2 x minus 3 close paren squared minus log base 16 of open paren x squared plus x minus 2 close paren squared equals one-half$ Используем свойство разности логарифмов $\log_{a} b - \log_{a} c = \log_{a} \frac{b}{c} log base a of b minus log base a of c equals log base a of b over c end-fraction$ : $\log_{16} {(\frac{x^{2} - 2 x - 3}{x^{2} + x - 2})}^{2} = \frac{1}{2} log base 16 of open paren the fraction with numerator x squared minus 2 x minus 3 and denominator x squared plus x minus 2 end-fraction close paren squared equals one-half$ По определению логарифма: ${(\frac{x^{2} - 2 x - 3}{x^{2} + x - 2})}^{2} = 16^{1 / 2} open paren the fraction with numerator x squared minus 2 x minus 3 and denominator x squared plus x minus 2 end-fraction close paren squared equals 16 raised to the 1 / 2 power$ ${(\frac{x^{2} - 2 x - 3}{x^{2} + x - 2})}^{2} = 4 open paren the fraction with numerator x squared minus 2 x minus 3 and denominator x squared plus x minus 2 end-fraction close paren squared equals 4$ Разложим квадратные трехчлены на множители:

$x^{2} - 2 x - 3 = (x - 3) (x + 1) x squared minus 2 x minus 3 equals open paren x minus 3 close paren open paren x plus 1 close paren$ $x^{2} + x - 2 = (x + 2) (x - 1) x squared plus x minus 2 equals open paren x plus 2 close paren open paren x minus 1 close paren$

Получаем: ${(\frac{(x - 3) (x + 1)}{(x + 2) (x - 1)})}^{2} = 4 open paren the fraction with numerator open paren x minus 3 close paren open paren x plus 1 close paren and denominator open paren x plus 2 close paren open paren x minus 1 close paren end-fraction close paren squared equals 4$ 3. Решение через совокупность уравнений Извлекая квадратный корень из обеих частей, получаем два случая: Случай А: $\frac{x^{2} - 2 x - 3}{x^{2} + x - 2} = 2 the fraction with numerator x squared minus 2 x minus 3 and denominator x squared plus x minus 2 end-fraction equals 2$ $x^{2} - 2 x - 3 = 2 (x^{2} + x - 2) x squared minus 2 x minus 3 equals 2 open paren x squared plus x minus 2 close paren$ $x^{2} - 2 x - 3 = 2 x^{2} + 2 x - 4 x squared minus 2 x minus 3 equals 2 x squared plus 2 x minus 4$ $x^{2} + 4 x - 1 = 0 x squared plus 4 x minus 1 equals 0$ $D = 16 - 4 (1) (-1) = 20 cap D equals 16 minus 4 open paren 1 close paren open paren negative 1 close paren equals 20$ $x = \frac{-4 \pm \sqrt{20}}{2} = \frac{-4 \pm 2 \sqrt{5}}{2} = -2 \pm \sqrt{5} x equals the fraction with numerator negative 4 plus or minus the square root of 20 end-root and denominator 2 end-fraction equals the fraction with numerator negative 4 plus or minus 2 the square root of 5 end-root and denominator 2 end-fraction equals negative 2 plus or minus the square root of 5 end-root$ Случай Б: $\frac{x^{2} - 2 x - 3}{x^{2} + x - 2} = -2 the fraction with numerator x squared minus 2 x minus 3 and denominator x squared plus x minus 2 end-fraction equals negative 2$ $x^{2} - 2 x - 3 = -2 (x^{2} + x - 2) x squared minus 2 x minus 3 equals negative 2 open paren x squared plus x minus 2 close paren$ $x^{2} - 2 x - 3 = -2 x^{2} - 2 x + 4 x squared minus 2 x minus 3 equals negative 2 x squared minus 2 x plus 4$ $3 x^{2} = 7 3 x squared equals 7$ $x^{2} = \frac{7}{3} ⟹ x = \pm \sqrt{\frac{7}{3}} x squared equals seven-thirds ⟹ bold x equals plus or minus the square root of seven-thirds end-root$ 4. Проверка корней по ОДЗ Напомним ОДЗ: $x < -2 x is less than negative 2$ или $x > 1 x is greater than 1$ (при этом $x \neq 3 x is not equal to 3$ ).

$x = -2 + \sqrt{5} \approx -2 + 2.23 = 0.23 x equals negative 2 plus the square root of 5 end-root is approximately equal to negative 2 plus 2.23 equals 0.23$ (Не подходит: $0.23 < 1 0.23 is less than 1$ ) $x = -2 - \sqrt{5} \approx -2 - 2.23 = -4.23 x equals negative 2 minus the square root of 5 end-root is approximately equal to negative 2 minus 2.23 equals negative 4.23$ (Подходит: $-4.23 < -2 negative 4.23 is less than negative 2$ ) $x = \sqrt{\frac{7}{3}} \approx \sqrt{2.33} \approx 1.52 x equals the square root of seven-thirds end-root is approximately equal to the square root of 2.33 end-root is approximately equal to 1.52$ (Подходит: $1.52 > 1 1.52 is greater than 1$ ) $x = - \sqrt{\frac{7}{3}} \approx -1.52 x equals negative the square root of seven-thirds end-root is approximately equal to negative 1.52$ (Не подходит: попадает в интервал $[-2; 1] open bracket negative 2 ; 1 close bracket$ )

Ответ: $x_{1} = -2 - \sqrt{5}, x_{2} = \sqrt{\frac{7}{3}} x sub 1 equals negative 2 minus the square root of 5 end-root comma space x sub 2 equals the square root of seven-thirds end-root$ Хотите, чтобы я проверил решение аналогичного уравнения с другими основаниями логарифмов?

Log16 (x^2-2x-3)^2-2 log16(x^2+x-2)=1/2

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