1)3х 2х>(1/36)2х-1 2)2х-1+2х-2+2х-3=896 3)32х+1+9=3х+3+3х 4)18,23х-2=0,272 5)34х -56х+29х=0 6)log1/3 (2-5x)> -2 7)log2 (2x+2)+ log(x-5)=7log75 8)log31/x +log3√x= -1 9)log16x +log4x+log2x=7 10)xlog3x-4=1/27 11)log√3(x-y)=2 2x-2 *5y-1=40 12)2log2(3x-4)=8 log9(x2-y2)-log9(x+y)=0,5

Question

1)3х 2х>(1/36)2х-1 2)2х-1+2х-2+2х-3=896 3)32х+1+9=3х+3+3х 4)18,23х-2=0,272 5)34х -56х+29х=0 6)log1/3 (2-5x)> -2 7)log2 (2x+2)+ log(x-5)=7log75 8)log31/x +log3√x= -1 9)log16x +log4x+log2x=7 10)xlog3x-4=1/27 11)log√3(x-y)=2 2x-2 *5y-1=40 12)2log2(3x-4)=8 log9(x2-y2)-log9(x+y)=0,5

CyborgJim 09.03.2026 10:51

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Лебедев Дмитрий Сергеевич · Accepted Answer

Ниже приведены подробные решения представленных уравнений и неравенств. 1) $3^{x} \cdot 2^{x} > {(\frac{1}{36})}^{2 x - 1} 3 to the x-th power center dot 2 to the x-th power is greater than open paren 1 over 36 end-fraction close paren raised to the 2 x minus 1 power$ Используем свойства степеней: $a^{x} \cdot b^{x} = (a b)^{x} a to the x-th power center dot b to the x-th power equals open paren a b close paren to the x-th power$ и $\frac{1}{36} = 6^{-2} 1 over 36 end-fraction equals 6 to the negative 2 power$ . $(3 \cdot 2)^{x} > (6^{-2})^{2 x - 1} open paren 3 center dot 2 close paren to the x-th power is greater than open paren 6 to the negative 2 power close paren raised to the 2 x minus 1 power$ $6^{x} > 6^{-4 x + 2} 6 to the x-th power is greater than 6 raised to the negative 4 x plus 2 power$ Так как основание $6 > 1 6 is greater than 1$ , знак неравенства сохраняется: $x > -4 x + 2 x is greater than negative 4 x plus 2$ $5 x > 2 ⟹ x > 0, 4 5 x is greater than 2 ⟹ bold x is greater than 0 comma 4$ 2) $2^{x - 1} + 2^{x - 2} + 2^{x - 3} = 896 2 raised to the x minus 1 power plus 2 raised to the x minus 2 power plus 2 raised to the x minus 3 power equals 896$ Вынесем за скобки степень с наименьшим показателем $2^{x - 3} 2 raised to the x minus 3 power$ : $2^{x - 3} (2^{2} + 2^{1} + 1) = 896 2 raised to the x minus 3 power open paren 2 squared plus 2 to the first power plus 1 close paren equals 896$ $2^{x - 3} (4 + 2 + 1) = 896 2 raised to the x minus 3 power open paren 4 plus 2 plus 1 close paren equals 896$ $2^{x - 3} \cdot 7 = 896 2 raised to the x minus 3 power center dot 7 equals 896$ $2^{x - 3} = 128 ⟹ 2^{x - 3} = 2^{7} 2 raised to the x minus 3 power equals 128 ⟹ 2 raised to the x minus 3 power equals 2 to the seventh power$ $x - 3 = 7 ⟹ x = 10 x minus 3 equals 7 ⟹ bold x equals 10$ 3) $3^{2 x + 1} + 9 = 3^{x + 3} + 3^{x} 3 raised to the 2 x plus 1 power plus 9 equals 3 raised to the x plus 3 power plus 3 to the x-th power$ Приведем к общему основанию $3^{x} 3 to the x-th power$ : $3 \cdot (3^{x})^{2} + 9 = 27 \cdot 3^{x} + 3^{x} 3 center dot open paren 3 to the x-th power close paren squared plus 9 equals 27 center dot 3 to the x-th power plus 3 to the x-th power$ Пусть $3^{x} = t 3 to the x-th power equals t$ ( $t > 0 t is greater than 0$ ): $3 t^{2} - 28 t + 9 = 0 3 t squared minus 28 t plus 9 equals 0$ Дискриминант $D = (-28)^{2} - 4 \cdot 3 \cdot 9 = 784 - 108 = 676 = 26^{2} cap D equals open paren negative 28 close paren squared minus 4 center dot 3 center dot 9 equals 784 minus 108 equals 676 equals 26 squared$ . $t_{1} = \frac{28 + 26}{6} = 9; t_{2} = \frac{28 - 26}{6} = \frac{1}{3} t sub 1 equals the fraction with numerator 28 plus 26 and denominator 6 end-fraction equals 9 ; space t sub 2 equals the fraction with numerator 28 minus 26 and denominator 6 end-fraction equals one-third$ Возвращаемся к замене:

$3^{x} = 9 ⟹ x_{1} = 2 3 to the x-th power equals 9 ⟹ bold x sub 1 equals 2$ $3^{x} = \frac{1}{3} ⟹ x_{2} = -1 3 to the x-th power equals one-third ⟹ bold x sub 2 equals negative 1$

4) $18, 2^{3 x - 2} = 0, 272 18 comma 2 raised to the 3 x minus 2 power equals 0 comma 272$ (Предположительно, в условии опечатка в числах, так как $18, 2 18 comma 2$ и $0, 272 0 comma 272$ не являются степенями одного числа. Прологарифмируем обе части): $(3 x - 2) \lg (18, 2) = \lg (0, 272) open paren 3 x minus 2 close paren l g open paren 18 comma 2 close paren equals l g open paren 0 comma 272 close paren$ $3 x - 2 = \frac{\lg (0, 272)}{\lg (18, 2)} 3 x minus 2 equals l g open paren 0 comma 272 close paren over l g open paren 18 comma 2 close paren end-fraction$ $x = \frac{1}{3} (\frac{\lg (0, 272)}{\lg (18, 2)} + 2) bold x equals one-third open paren l g open paren 0 comma 272 close paren over l g open paren 18 comma 2 close paren end-fraction plus 2 close paren$ 5) $3 \cdot 4^{x} - 5 \cdot 6^{x} + 2 \cdot 9^{x} = 0 3 center dot 4 to the x-th power minus 5 center dot 6 to the x-th power plus 2 center dot 9 to the x-th power equals 0$ Это однородное уравнение. Разделим на $9^{x} 9 to the x-th power$ : $3 \cdot {(\frac{4}{9})}^{x} - 5 \cdot {(\frac{6}{9})}^{x} + 2 = 0 3 center dot open paren four-nineths close paren to the x-th power minus 5 center dot open paren six-nineths close paren to the x-th power plus 2 equals 0$ $3 \cdot {(\frac{2}{3})}^{2 x} - 5 \cdot {(\frac{2}{3})}^{x} + 2 = 0 3 center dot open paren two-thirds close paren raised to the 2 x power minus 5 center dot open paren two-thirds close paren to the x-th power plus 2 equals 0$ Пусть $(\frac{2}{3})^{x} = t open paren two-thirds close paren to the x-th power equals t$ : $3 t^{2} - 5 t + 2 = 0 ⟹ t_{1} = 1, t_{2} = \frac{2}{3} 3 t squared minus 5 t plus 2 equals 0 ⟹ t sub 1 equals 1 comma t sub 2 equals two-thirds$

$(\frac{2}{3})^{x} = 1 ⟹ x_{1} = 0 open paren two-thirds close paren to the x-th power equals 1 ⟹ bold x sub 1 equals 0$ $(\frac{2}{3})^{x} = \frac{2}{3} ⟹ x_{2} = 1 open paren two-thirds close paren to the x-th power equals two-thirds ⟹ bold x sub 2 equals 1$

6) $\log_{1 / 3} (2 - 5 x) > -2 log base 1 / 3 of open paren 2 minus 5 x close paren is greater than negative 2$ ОДЗ: $2 - 5 x > 0 ⟹ x < 0, 4 2 minus 5 x is greater than 0 ⟹ x is less than 0 comma 4$ . Так как основание $1 / 3 < 1 1 / 3 is less than 1$ , знак меняется: $2 - 5 x < (1 / 3)^{-2} 2 minus 5 x is less than open paren 1 / 3 close paren to the negative 2 power$ $2 - 5 x < 9 ⟹ -5 x < 7 ⟹ x > -1, 4 2 minus 5 x is less than 9 ⟹ negative 5 x is less than 7 ⟹ x is greater than negative 1 comma 4$ С учетом ОДЗ: $x \in (-1, 4; 0, 4) bold x is an element of open paren negative 1 comma 4 ; 0 comma 4 close paren$ 7) $\log_{2} (2 x + 2) + \log_{2} (x - 5) = 7 \log_{7} 5 log base 2 of open paren 2 x plus 2 close paren plus log base 2 of open paren x minus 5 close paren equals 7 log base 7 of 5$ (Вероятно, в правой части опечатка и должно быть основание 2. Если решать "как есть", $7 \log_{7} 5 = 5 7 log base 7 of 5 equals 5$ ): ОДЗ: $x > 5 x is greater than 5$ . $\log_{2} ((2 x + 2) (x - 5)) = 5 log base 2 of open paren open paren 2 x plus 2 close paren open paren x minus 5 close paren close paren equals 5$ $(2 x + 2) (x - 5) = 2^{5} = 32 open paren 2 x plus 2 close paren open paren x minus 5 close paren equals 2 to the fifth power equals 32$ $2 x^{2} - 10 x + 2 x - 10 - 32 = 0 ⟹ 2 x^{2} - 8 x - 42 = 0 2 x squared minus 10 x plus 2 x minus 10 minus 32 equals 0 ⟹ 2 x squared minus 8 x minus 42 equals 0$ $x^{2} - 4 x - 21 = 0 ⟹ (x - 7) (x + 3) = 0 x squared minus 4 x minus 21 equals 0 ⟹ open paren x minus 7 close paren open paren x plus 3 close paren equals 0$ С учетом ОДЗ ( $x > 5 x is greater than 5$ ): $x = 7 bold x equals 7$ 8) $\log_{3} \frac{1}{x} + \log_{3} \sqrt{x} = -1 log base 3 of 1 over x end-fraction plus log base 3 of the square root of x end-root equals negative 1$ ОДЗ: $x > 0 x is greater than 0$ . $- \log_{3} x + \frac{1}{2} \log_{3} x = -1 negative log base 3 of x plus one-half log base 3 of x equals negative 1$ $- \frac{1}{2} \log_{3} x = -1 ⟹ \log_{3} x = 2 negative one-half log base 3 of x equals negative 1 ⟹ log base 3 of x equals 2$ $x = 3^{2} ⟹ x = 9 x equals 3 squared ⟹ bold x equals 9$ 9) $\log_{16} x + \log_{4} x + \log_{2} x = 7 log base 16 of x plus log base 4 of x plus log base 2 of x equals 7$ Приведем к основанию 2: $\frac{1}{4} \log_{2} x + \frac{1}{2} \log_{2} x + \log_{2} x = 7 one-fourth log base 2 of x plus one-half log base 2 of x plus log base 2 of x equals 7$ $(\frac{1}{4} + \frac{2}{4} + \frac{4}{4}) \log_{2} x = 7 open paren one-fourth plus two-fourths plus four-fourths close paren log base 2 of x equals 7$ $\frac{7}{4} \log_{2} x = 7 ⟹ \log_{2} x = 4 seven-fourths log base 2 of x equals 7 ⟹ log base 2 of x equals 4$ $x = 2^{4} ⟹ x = 16 x equals 2 to the fourth power ⟹ bold x equals 16$ 10) $x^{\log_{3} x - 4} = \frac{1}{27} x raised to the log base 3 of x minus 4 power equals 1 over 27 end-fraction$ Прологарифмируем обе части по основанию 3: $\log_{3} (x^{\log_{3} x - 4}) = \log_{3} 3^{-3} log base 3 of open paren x raised to the log base 3 of x minus 4 power close paren equals log base 3 of 3 to the negative 3 power$ $(\log_{3} x - 4) \log_{3} x = -3 open paren log base 3 of x minus 4 close paren log base 3 of x equals negative 3$ Пусть $\log_{3} x = t log base 3 of x equals t$ : $t^{2} - 4 t + 3 = 0 ⟹ t_{1} = 3, t_{2} = 1 t squared minus 4 t plus 3 equals 0 ⟹ t sub 1 equals 3 comma t sub 2 equals 1$

$\log_{3} x = 3 ⟹ x_{1} = 27 log base 3 of x equals 3 ⟹ bold x sub 1 equals 27$ $\log_{3} x = 1 ⟹ x_{2} = 3 log base 3 of x equals 1 ⟹ bold x sub 2 equals 3$

11) Система: $\log_{\sqrt{3}} (x - y) = 2 log base the square root of 3 end-root of open paren x minus y close paren equals 2$ и $2^{x - 2} \cdot 5^{y - 1} = 40 2 raised to the x minus 2 power center dot 5 raised to the y minus 1 power equals 40$

Из первого: $x - y = (\sqrt{3})^{2} = 3 ⟹ x = y + 3 x minus y equals open paren the square root of 3 end-root close paren squared equals 3 ⟹ x equals y plus 3$ . Подставим во второе: $2^{y + 3 - 2} \cdot 5^{y - 1} = 40 2 raised to the y plus 3 minus 2 power center dot 5 raised to the y minus 1 power equals 40$
$2^{y + 1} \cdot 5^{y - 1} = 40 ⟹ 2^{y - 1} \cdot 2^{2} \cdot 5^{y - 1} = 40 2 raised to the y plus 1 power center dot 5 raised to the y minus 1 power equals 40 ⟹ 2 raised to the y minus 1 power center dot 2 squared center dot 5 raised to the y minus 1 power equals 40$ $4 \cdot (2 \cdot 5)^{y - 1} = 40 ⟹ 10^{y - 1} = 10^{1} 4 center dot open paren 2 center dot 5 close paren raised to the y minus 1 power equals 40 ⟹ 10 raised to the y minus 1 power equals 10 to the first power$ $y - 1 = 1 ⟹ y = 2; x = 5 y minus 1 equals 1 ⟹ bold y equals 2 ; space bold x equals 5$

12) Система: $2 \log_{2} (3 x - 4) = 8 2 log base 2 of open paren 3 x minus 4 close paren equals 8$ и $\log_{9} (x^{2} - y^{2}) - \log_{9} (x + y) = 0, 5 log base 9 of open paren x squared minus y squared close paren minus log base 9 of open paren x plus y close paren equals 0 comma 5$

Из первого: $\log_{2} (3 x - 4) = 4 ⟹ 3 x - 4 = 16 ⟹ 3 x = 20 ⟹ x = 6 \frac{2}{3} log base 2 of open paren 3 x minus 4 close paren equals 4 ⟹ 3 x minus 4 equals 16 ⟹ 3 x equals 20 ⟹ bold x equals 6 and two-thirds$ . Из второго: $\log_{9} \frac{(x - y) (x + y)}{x + y} = 0, 5 log base 9 of the fraction with numerator open paren x minus y close paren open paren x plus y close paren and denominator x plus y end-fraction equals 0 comma 5$
$\log_{9} (x - y) = 0, 5 ⟹ x - y = 9^{0, 5} = 3 log base 9 of open paren x minus y close paren equals 0 comma 5 ⟹ x minus y equals 9 raised to the 0 comma 5 power equals 3$ $6 \frac{2}{3} - y = 3 ⟹ y = 3 \frac{2}{3} 6 and two-thirds minus y equals 3 ⟹ bold y equals 3 and two-thirds$

Могу ли я помочь с решением других математических задач?

1)3х *2х>(1/36)2х-1 2)2х-1+2х-2+2х-3=896 3)32х+1+9=3х+3+3х 4)18,23х-2=0,272 5)3*4х -5*6х+2*9х=0 6)log1/3 (2-5x)> -2 7)log2 (2x+2)+ log(x-5)=7log75 8)log31/x +log3√x= -1 9)log16x +log4x+log2x=7 10)xlog3x-4=1/27 11)log√3(x-y)=2 2x-2 *5y-1=40 12)2log2(3x-4)=8 log9(x2-y2)-log9(x+y)=0,5

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1)3х 2х>(1/36)2х-1 2)2х-1+2х-2+2х-3=896 3)32х+1+9=3х+3+3х 4)18,23х-2=0,272 5)34х -56х+29х=0 6)log1/3 (2-5x)> -2 7)log2 (2x+2)+ log(x-5)=7log75 8)log31/x +log3√x= -1 9)log16x +log4x+log2x=7 10)xlog3x-4=1/27 11)log√3(x-y)=2 2x-2 *5y-1=40 12)2log2(3x-4)=8 log9(x2-y2)-log9(x+y)=0,5