Решите неравенства: 1)log2(x)>3 2)log1/3(2x)>-2 3)log5(3x-1)-1 6)log1/7(5x+3)>=-1/2 7)log2(x'2-2x)>=3 8)log3(13-4'x)>2 9)log1/5(26-3'x)>-2 10)log7(2-x)<=log7(3x+6)

Question

Решите неравенства: 1)log2(x)>3 2)log1/3(2x)>-2 3)log5(3x-1)-1 6)log1/7(5x+3)>=-1/2 7)log2(x'2-2x)>=3 8)log3(13-4'x)>2 9)log1/5(26-3'x)>-2 10)log7(2-x)<=log7(3x+6)

Осень892 21.02.2026 16:56

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Лебедев Дмитрий Сергеевич · Accepted Answer

Решения для представленных неравенств: 1) $x \in (8, + \infty) x is an element of open paren 8 comma positive infinity close paren$ ; 2) $x \in (0, 4.5) x is an element of open paren 0 comma 4.5 close paren$ ; 3) $x \in (1 / 3, 2) x is an element of open paren 1 / 3 comma 2 close paren$ ; 6) $x \in (-0.6, \frac{\sqrt{7} - 3}{5}] x is an element of open paren negative 0.6 comma the fraction with numerator the square root of 7 end-root minus 3 and denominator 5 end-fraction close bracket$ ; 7) $x \in (- \infty, -2] \cup [4, + \infty) x is an element of open paren negative infinity comma negative 2 close bracket union open bracket 4 comma positive infinity close paren$ ; 8) $x \in (- \infty, 1) x is an element of open paren negative infinity comma 1 close paren$ ; 9) $x \in (0, \log_{3} (26)) x is an element of open paren 0 comma log base 3 of 26 close paren$ ; 10) $x \in [-1, 2) x is an element of open bracket negative 1 comma 2 close paren$ . ️ Шаг 1: Решение простейших неравенств Для неравенств вида $\log_{a} (f (x)) > b log base a of f of x is greater than b$ учитываем область допустимых значений (ОДЗ) $f (x) > 0 f of x is greater than 0$ и изменение знака при $0 < a < 1 0 is less than a is less than 1$ .

$\log_{2} (x) > 3 x > 2^{3} x > 8 log base 2 of x is greater than 3 implies x is greater than 2 cubed implies x is greater than 8$ . ОДЗ $x > 0 x is greater than 0$ выполняется. $\log_{1 / 3} (2 x) > -2 log base 1 / 3 of 2 x is greater than negative 2$ . Так как основание $1 / 3 < 1 1 / 3 is less than 1$ , знак меняется: $2 x < (1 / 3)^{-2} 2 x < 9 x < 4.5 2 x is less than open paren 1 / 3 close paren to the negative 2 power implies 2 x is less than 9 implies x is less than 4.5$ . С учетом ОДЗ $x > 0 x is greater than 0$ : $x \in (0, 4.5) x is an element of open paren 0 comma 4.5 close paren$ . $\log_{5} (3 x - 1) < 1 3 x - 1 < 5 3 x < 6 x < 2 log base 5 of open paren 3 x minus 1 close paren is less than 1 implies 3 x minus 1 is less than 5 implies 3 x is less than 6 implies x is less than 2$ . ОДЗ: $3 x - 1 > 0 x > 1 / 3 3 x minus 1 is greater than 0 implies x is greater than 1 / 3$ . Итог: $x \in (1 / 3, 2) x is an element of open paren 1 / 3 comma 2 close paren$ .

️ Шаг 2: Квадратичные и иррациональные выражения

$\log_{1 / 7} (5 x + 3) \geq -1 / 2 log base 1 / 7 of open paren 5 x plus 3 close paren is greater than or equal to negative 1 / 2$ . Знак меняется: $5 x + 3 \leq (1 / 7)^{-1 / 2} = \sqrt{7} 5 x plus 3 is less than or equal to open paren 1 / 7 close paren raised to the negative 1 / 2 power equals the square root of 7 end-root$ . Тогда $5 x \leq \sqrt{7} - 3 x \leq \frac{\sqrt{7} - 3}{5} 5 x is less than or equal to the square root of 7 end-root minus 3 implies x is less than or equal to the fraction with numerator the square root of 7 end-root minus 3 and denominator 5 end-fraction$ . ОДЗ: $x > -0.6 x is greater than negative 0.6$ . $\log_{2} (x^{2} - 2 x) \geq 3 x^{2} - 2 x \geq 8 x^{2} - 2 x - 8 \geq 0 log base 2 of open paren x squared minus 2 x close paren is greater than or equal to 3 implies x squared minus 2 x is greater than or equal to 8 implies x squared minus 2 x minus 8 is greater than or equal to 0$ . Корни уравнения $x_{1} = 4, x_{2} = -2 x sub 1 equals 4 comma x sub 2 equals negative 2$ . Решение: $x \in (- \infty, -2] \cup [4, + \infty) x is an element of open paren negative infinity comma negative 2 close bracket union open bracket 4 comma positive infinity close paren$ .

️ Шаг 3: Показательные функции и сравнение

$\log_{3} (13 - 4^{x}) > 2 13 - 4^{x} > 9 4^{x} < 4 x < 1 log base 3 of open paren 13 minus 4 to the x-th power close paren is greater than 2 implies 13 minus 4 to the x-th power is greater than 9 implies 4 to the x-th power is less than 4 implies x is less than 1$ . $\log_{1 / 5} (26 - 3^{x}) > -2 26 - 3^{x} < 25 3^{x} > 1 x > 0 log base 1 / 5 of open paren 26 minus 3 to the x-th power close paren is greater than negative 2 implies 26 minus 3 to the x-th power is less than 25 implies 3 to the x-th power is greater than 1 implies x is greater than 0$ . ОДЗ: $26 - 3^{x} > 0 3^{x} < 26 x < \log_{3} (26) 26 minus 3 to the x-th power is greater than 0 implies 3 to the x-th power is less than 26 implies x is less than log base 3 of 26$ . $\log_{7} (2 - x) \leq \log_{7} (3 x + 6) log base 7 of open paren 2 minus x close paren is less than or equal to log base 7 of open paren 3 x plus 6 close paren$ . Переходим к системе: $2 - x \leq 3 x + 6 2 minus x is less than or equal to 3 x plus 6$ , $2 - x > 0 2 minus x is greater than 0$ (ОДЗ) и $3 x + 6 > 0 3 x plus 6 is greater than 0$ (ОДЗ).
Из неравенства: $4 x \geq -4 x \geq -1 4 x is greater than or equal to negative 4 implies x is greater than or equal to negative 1$ . Из ОДЗ: $x < 2 x is less than 2$ и $x > -2 x is greater than negative 2$ .

Ответ:

$x \in (8, + \infty) x is an element of open paren 8 comma positive infinity close paren$ $x \in (0, 4.5) x is an element of open paren 0 comma 4.5 close paren$ $x \in (1 / 3, 2) x is an element of open paren 1 / 3 comma 2 close paren$ $x \in (-0.6, \frac{\sqrt{7} - 3}{5}] x is an element of open paren negative 0.6 comma the fraction with numerator the square root of 7 end-root minus 3 and denominator 5 end-fraction close bracket$ $x \in (- \infty, -2] \cup [4, + \infty) x is an element of open paren negative infinity comma negative 2 close bracket union open bracket 4 comma positive infinity close paren$ $x \in (- \infty, 1) x is an element of open paren negative infinity comma 1 close paren$ $x \in (0, \log_{3} (26)) x is an element of open paren 0 comma log base 3 of 26 close paren$ $x \in [-1, 2) x is an element of open bracket negative 1 comma 2 close paren$

Нужно ли графическое представление решений для любого из этих логарифмических неравенств?

Решите неравенства: 1)log2(x)>3 2)log1/3(2x)>-2 3)log5(3x-1)-1 6)log1/7(5x+3)>=-1/2 7)log2(x'2-2x)>=3 8)log3(13-4'x)>2 9)log1/5(26-3'x)>-2 10)log7(2-x)<=log7(3x+6)

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