Логарифмдік теңдеулер (Рустюмова 3.4.2)

Шешуі:

$$x – 8,5 \gt 0,\quad x \gt 8,5$$ $${2^{{{\log }_4}(x – 8,5)}} = {\log _3}81$$ $${2^{{{\log }_{{2^2}}}(x – 8,5)}} = {\log _3}{3^4}$$

$${2^{\frac{1}{2}{{\log }_2}(x – 8,5)}} = 4{\log _3}3$$ $${2^{{{\log }_2}{{(x – 8,5)}^{\frac{1}{2}}}}} = 4$$ $${(x – 8,5)^{\frac{1}{2}}} = 4$$

$$\sqrt {x – 8,5} = 4$$ $$x – 8,5 = 16$$ $$x = 16 + 8,5$$ $$x = 24,5$$

Жауабы: $24,5$

Шешуі: $x = {81^{\frac{1}{4}}} \cdot {81^{{{\log }_{81}}4}}$ $ = {\left( {{3^4}} \right)^{\frac{1}{4}}} \cdot 4 = 3 \cdot 4 = 12$

Жауабы: $12$

Шешуі:

$$3{x^2} + x = {\left( {{4^2}} \right)^{{{\log }_4}\sqrt {30} }}$$ $$3{x^2} + x = {4^{2{{\log }_4}\sqrt {30} }}$$ $$3{x^2} + x = {4^{{{\log }_4}{{\left( {\sqrt {30} } \right)}^2}}}$$

$$3{x^2} + x = 30$$ $$3{x^2} + x – 30 = 0$$ $$D = 1 + 360 = 361 = {19^2}$$

$${x_1} = \frac{{ – 1 + 19}}{6} = \frac{{18}}{6} = 3$$ $${x_2} = \frac{{ – 1 – 19}}{6} = – \frac{{20}}{6}$$ $$\text{ММЖ:} \quad x \gt 0$$

Жауабы: $3$

Шешуі:

$$36 – 4x \gt 0$$ $$4x \lt 36$$ $$x \lt 9$$ $$x \in ( – \infty ;9)$$

$${0,1^{\lg (36 – 4x)}} = \lg {10^{10}}$$ $${\left( {{{10}^{ – 1}}} \right)^{\lg (36 – 4x)}}10\lg 10$$ $${10^{ – \lg (36 – 4x)}} = 10$$ $$ – \lg (36 – 4x) = 1$$

$$\lg (36 – 4x) = – 1$$ $$36 – 4x = {10^{ – 1}}$$ $$36 – 4x = 0,1$$ $$4x = 36 – 0,1$$ $$4x = 35,9$$

$$4x = 35\frac{9}{{10}}$$ $$4x = \frac{{359}}{{10}}$$ $$x = \frac{{359}}{{40}}$$ $$x = 8\frac{{39}}{{40}}$$

Жауабы: $8\frac{{39}}{{40}}$

Шешуі:

$$\text{ММЖ:} \quad x \gt 0$$ $${9^{{{\log }_3}x}} – 12 \cdot {3^{{{\log }_3}x}} + {3^{{{\log }_3}27}} = 0$$ $${3^{2{{\log }_3}x}} – 12 \cdot x + 27 = 0$$

$${3^{{{\log }_3}{x^2}}} – 12x + 27 = 0$$ $${x^2} – 12x + 27 = 0$$

$$\left\{ {\begin{array}{*{20}{l}}{{x_1} + {x_2} = 12}\\{{x_1} \cdot {x_2} = 27}\end{array}} \right.$$ $${x_1} = 3;\quad {x_2} = 9$$

Жауабы: $3;\,9$

Шешуі:

$$\text{ММЖ:} \quad x \gt 0$$ $${\log _4}\left( {{2^{4x}}} \right) = {2^{{{\log }_2}4}}$$

$${\log _4}{\left( {{2^2}} \right)^{2x}} = 4$$ $${\log _4}\left( {{4^{2x}}} \right) = 4$$

$$2x = 4$$ $$x = 2$$

Жауабы: $2$

Шешуі:

$$\text{ММЖ:} \quad 1 – 2x \gt 0$$ $$2x \lt 1$$ $$x \lt \frac{1}{2}$$ $${9^{{{\log }_3}(1 – 2x)}} = 5{x^2} – 5$$ $${3^{2{{\log }_3}(1 – 2x)}} = 5{x^2} – 5$$

$${3^{{{\log }_3}{{(1 – 2x)}^2}}} = 5{x^2} – 5$$ $${(1 – 2x)^2} = 5{x^2} – 5$$ $$1 – 4x + 4{x^2} = 5{x^2} – 5$$ $${x^2} + 4x – 6 = 0$$ $$D = 16 + 24 = 40$$

$${x_1} = \frac{{ – 4 + \sqrt {40} }}{2} = $$ $$=\frac{{ – 4 + 2\sqrt {10} }}{2} = – 2 + \sqrt {10} $$ $${x_2} = \frac{{ – 4 – \sqrt {40} }}{2} = $$ $$\frac{{ – 4 – 2\sqrt {10} }}{2} =- 2 – \sqrt {10} $$ $$\{ – 2 + \sqrt {10} \} \notin \left( { – \infty ;\frac{1}{2}} \right)$$ $$\{ – 2 – \sqrt {10} \} \in \left( { – \infty ;\frac{1}{2}} \right)$$

Жауабы: $ – 2 – \sqrt {10} $

Шешуі:

$$\text{ММЖ:} \quad x \gt 0$$ $$2 \cdot {4^{{{\log }_2}x}} = 7x + 4$$ $$2 \cdot {2^{{{\log }_2}x}} = 7x + 4$$ $$2 \cdot {2^{{{\log }_2}{x^2}}} = 7x + 4$$

$$2{x^2} = 7x + 4$$ $$2{x^2} – 7x – 4 = 0$$ $$D = 49 + 32 = 81$$ $${x_1} = \frac{{7 + 9}}{4} = \frac{{16}}{4} = 4$$ $${x_2} = \frac{{7 – 9}}{4} = – \frac{2}{4} = – \frac{1}{2}$$

$$ – \frac{1}{2} \notin (0; + \infty )$$ $$4 \in (0; + \infty )$$

Жауабы: $4$

Шешуі:

$$\text{ММЖ:}$$ $$\left\{ \begin{array}{l}{x^2} – 9 \gt 0\\21 – {x^2} \gt 0\end{array} \right.$$ $$1)\quad {x^2} – 9 \gt 0$$ $${x^2} – 9 = 0$$ $${x^2} = 9$$ $$x = \pm 3$$ $$x \in ( – \infty ; – 3) \cup (3; + \infty )$$

$$2)\quad 21 – {x^2} \gt 0$$ $$21 – {x^2} = 0$$ $${x^2} = 21$$ $$x = \pm \sqrt {21} $$ $$x \in \left( { – \sqrt {21} ;\sqrt {21} } \right)$$ $$x \in \left( { – \sqrt {21} ; – 3} \right) \cup \left( {3;\sqrt {21} } \right)$$

$${7^{{{\log }_7}\sqrt {{x^2} – 9} }} = {2^{{{\log }_2}\left( {21 – {x^2}} \right)}}$$ $$\sqrt {{x^2} – 9} = 21 – {x^2}$$ $$\sqrt {{x^2} – 9} = 12 – {x^2} + 9$$ $$\sqrt {{x^2} – 9} = 12 – \left( {{x^2} – 9} \right)$$ $$\sqrt {{x^2} – 9} = t,\quad t \gt 0$$ $$t = 12 – {t^2}$$ $${t^2} + t – 12 = 0$$

$$\left\{ {\begin{array}{*{20}{l}}{{t_1} + {t_2} = – 1}\\{{t_1} \cdot {t_2} = – 12}\end{array}} \right.$$ $${t_1} = 3;\\ {t_2} = – 4$$ $$\sqrt {{x^2} – 9} = 3$$ $${x^2} – 9 = 9$$ $${x^2} = 18$$ $$x = \pm \sqrt {18} $$ $$x = \pm 3\sqrt 2 $$

Жауабы: $\pm 3\sqrt 2 $

Шешуі:

$$1 – 2{x^2} \gt 0$$ $$1 – 2{x^2} = 0$$ $$2{x^2} = 1$$ $${x^2} = \frac{1}{2}$$ $$x = \pm \frac{1}{{\sqrt 2 }}$$ $$x \in \left( { – \frac{1}{{\sqrt 2 }};\frac{1}{{\sqrt 2 }}} \right)$$

$${2^{ – 2{{\log }_{{2^{ – 2}}}}\left( {1 – 2{x^2}} \right)}} = \frac{1}{2}{\log _{{3^{ – 1}}}}{3^{2x}}$$ $${2^{ – 2 \cdot \left( { – \frac{1}{2}} \right){{\log }_2}\left( {1 – 2{x^2}} \right)}} = – \frac{1}{2}{\log _3}{3^{2x}}$$ $${2^{{{\log }_2}\left( {1 – 2{x^2}} \right)}} = – \frac{1}{2} \cdot 2x$$ $$1 – 2{x^2} = – x$$ $$1 – 2{x^2} + x = 0$$ $$2{x^2} – x – 1 = 0$$

$$D = 1 + 8 = 9$$ $${x_1} = \frac{{1 + \sqrt 9 }}{4} = \frac{{1 + 3}}{4} = 1$$ $${x_2} = \frac{{1 – \sqrt 9 }}{4} = \frac{{1 – 3}}{4} = $$ $$ – \frac{2}{4} = – \frac{1}{2}$$ $$1 \notin \left( { – \frac{1}{{\sqrt 2 }};\frac{1}{{\sqrt 2 }}} \right)$$ $$ – \frac{1}{2} \in \left( { – \frac{1}{{\sqrt 2 }};\frac{1}{{\sqrt 2 }}} \right)$$

Жауабы: $ – \frac{1}{2}$

Шешуі:

$$\text{ММЖ:}$$ $$x \gt 0;\quad x \ne 1$$ $${x^{{{\log }_{\sqrt x }}2x}} = 4$$ $${x^{{{\log }_{{x^{\frac{1}{2}}}}}2x}} = 4$$

$${x^{2{{\log }_x}2x}} = 4$$ $${x^{{{\log }_x}{{(2x)}^2}}} = 4$$ $${(2x)^2} = 4$$

$$4{x^2} = 4$$ $${x^2} = 1$$ $$x = \pm 1$$ $$ \pm 1 \notin (0;1) \cup (1: + \infty )$$

Жауабы: $\emptyset $