Логарифмдік теңдеулер (Рустюмова 3.4.1)

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Логарифмнің анықтамасын қолданып, теңдеулерді шешіңіз.

№ 1 Теңдеуді шешіңіз: ${\log _x}{(\sqrt[4]{{500}})^3} = - \frac{3}{4}$

Шешуі:

$$\text{ММЖ:}\, x \gt 0;\quad x \ne 1$$ $${\log _x}{500^{\frac{3}{4}}} = - \frac{3}{4}$$ $${x^{ - \frac{3}{4}}} = {500^{\frac{3}{4}}}$$

$${\left( {{x^{ - 1}}} \right)^{\frac{3}{4}}} = {500^{\frac{3}{4}}}$$ $${x^{ - 1}} = 500$$ $$x = \frac{1}{{500}}$$

Жауабы: $ \frac{1}{{500}}$

№ 2 Теңдеуді шешіңіз: ${\log _{11}}{\log _3}{\log _2}\frac{2}{{1 - x}} = 0$

Шешуі:

$$\text{ММЖ:}\, \frac{2}{{1 - x}} \gt 0$$ $$1 - x \gt 0$$ $$x \lt 1$$

$${\log _3}{\log _2}\frac{2}{{1 - x}} = 1$$ $${\log _2}\frac{2}{{1 - x}} = 3$$ $$\frac{2}{{1 - x}} = {2^3}$$ $$\frac{2}{{1 - x}} = 8$$

$$8 - 8x = 2$$ $$8x = 6$$ $$x = \frac{6}{8}$$ $$x = \frac{3}{4}$$

Жауабы: $ \frac{3}{4}$

№ 3 Теңдеуді шешіңіз: ${\log _{\frac{{64}}{{7 + x}}}}8 - \frac{1}{2} = 0$

Шешуі:

$${\log _{\frac{{64}}{{7 + x}}}}8 = \frac{1}{2}$$ $$\left\{ {\begin{array}{*{20}{l}}{\frac{{64}}{{7 + x}} \gt 0}\\{\frac{{64}}{{7 + x}} \ne 1}\end{array}} \right. \Rightarrow $$ $$\left\{ {\begin{array}{*{20}{l}}{7 + x \gt 0}\\{7 + x \ne 64}\end{array}} \right. \Rightarrow $$

$$\left\{ {\begin{array}{*{20}{l}}{x \gt - 7}\\{x \ne 57}\end{array}} \right.$$ $${\left( {\frac{{64}}{{7 + x}}} \right)^{\frac{1}{2}}} = 8$$ $$\frac{{64}}{{7 + x}} = 64$$

$$\frac{1}{{7 + x}} = 1$$ $$7 + x = 1$$ $$x = - 6$$

Жауабы: $ - 6$

№ 4 Теңдеуді шешіңіз: ${\log _2}{\log _3}{\log _2}(10x + 12) = 1$

Шешуі:

$$\text{ММЖ:}\, 10x + 12 \gt 0$$ $$10x \gt - 12$$ $$x \gt - 1,2$$

$${\log _3}{\log _2}(10x + 12) = 2$$ $${\log _2}(10x + 12) = 9$$ $$10x + 12 = {2^9}$$

$$10x = 512 - 12$$ $$10x = 500$$ $$x = 50$$

Жауабы: $50$

№ 5 Теңдеуді шешіңіз: ${\log _3}{\log _2}\left( {{4^x} - 8} \right) = 1$

Шешуі:

$$\text{ММЖ:}\quad {4^x} - 8 \gt 0$$ $${4^x} \gt 8$$ $${2^{2x}} \gt {2^3}$$ $$2x \gt 3$$ $$x \gt 1,5$$

$${\log _2}\left( {{4^x} - 8} \right) = 3$$ $${4^x} - 8 = {2^3}$$ $${4^x} - 8 = 8$$

$${4^x} = 8 + 8$$ $${4^x} = 16$$ $$x = 2$$

Жауабы: $2$

№ 6 Теңдеуді шешіңіз: ${\log _4}x + 3{\log _2}x = 7$

Шешуі:

$$\text{ММЖ:}\quad x \gt 0$$ $${\log _{{2^2}}}x + 3{\log _2}x = 7$$

$$\frac{1}{2}{\log _2}x + 3{\log _2}x = 7$$ $$3\frac{1}{2}{\log _2}x = 7$$

$$\frac{7}{2}{\log _2}x = 7$$ $${\log _2}x = 2$$ $$x = {2^2}=4$$

Жауабы: $4$

№ 7 Теңдеуді шешіңіз: ${\log _9}x + 2{\log _3}x = 5$

Шешуі:

$$\text{ММЖ:}\quad x \gt 0$$ $${\log _{{3^2}}}x + 2{\log _3}x = 5$$ $$\frac{1}{2}{\log _3}x + 2{\log _3}x = 5$$

$$2\frac{1}{2}{\log _3}x = 5$$ $$\frac{5}{2}{\log _3}x = 5$$

$${\log _3}x = 2$$ $$x = {3^2}$$ $$x = 9$$

Жауабы: $9$

№ 8 Теңдеуді шешіңіз: ${3^{{{\log }_2}1,5x}} = {\log _7}343$

Шешуі:

$$\text{ММЖ:} \quad x \gt 0$$ $${3^{{{\log }_2}1,5x}} = {\log _7}{7^3}$$ $${3^{{{\log }_2}1,5x}} = 3$$

$${\log _2}1,5x = 1$$ $$1,5x = 2$$ $$\frac{3}{2}x = 2$$

$$x = \frac{4}{3}$$ $$x = 1\frac{1}{3}$$

Жауабы: $1\frac{1}{3}$

№ 9 Теңдеуді шешіңіз: ${3^{{{\log }_4}( - 5x)}} = {\log _5}125$

Шешуі:

$$\text{ММЖ:} \quad - 5x \gt 0$$ $$x \lt 0$$

$${3^{{{\log }_4}( - 5x)}} = 3$$ $${\log _4}( - 5x) = 1$$

$$ - 5x = 4$$ $$x = - \frac{4}{5} = - 0,8$$

Жауабы: $- 0,8$

№ 10 Теңдеуді шешіңіз: $2{\log _{{{\log }_2}x}}2 = 1$

Шешуі:

$$\text{ММЖ:} \quad x \gt 0$$ $${\log _2}x \ne 1$$ $$x \ne 2$$

$${\log _{{{\log }_2}x}}{2^2} = 1$$ $${\log _2}x = 4$$

$$x = {2^4}$$ $$x = 16$$

Жауабы: $16$

№ 11 Теңдеуді шешіңіз: ${\log _{\sqrt {6 - x} }}3 - 2 = 0$

Шешуі:

$$\left\{ {\begin{array}{*{20}{l}}{6 - x \gt 0}\\{6 - x \ne 1}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x \lt 6}\\{x \ne 5}\end{array}} \right.$$

$${\log _{\sqrt {6 - x} }}3 = 2$$ $${\left( {\sqrt {6 - x} } \right)^2} = 3$$

$$6 - x = 3$$ $$x = 3$$

Жауабы: $ 3$

№ 12 Теңдеуді шешіңіз: ${\log _3}|2x - 1| = 2$

Шешуі: $$|2x - 1| = {3^2}$$ $$|2x - 1| = 9$$

I жағдай: $$\left\{ {\begin{array}{*{20}{l}}{2x - 1 \gt 0}\\{2x - 1 = 9}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}{2x \gt 1}\\{2x = 10}\end{array}} \right. $$ $$ \Rightarrow \left\{ {\begin{array}{*{20}{l}}{x \gt \frac{1}{2}}\\{x = 5}\end{array}} \right.$$ $$5 \in \left( {\frac{1}{2}; + \infty } \right)$$

II жағдай: $$\left\{ {\begin{array}{*{20}{l}}{2x - 1 \lt 0}\\{2x - 1 = - 9}\end{array}} \right. \Rightarrow \left\{ {\begin{array}{*{20}{l}}{2x \lt 1}\\{2x = - 8}\end{array}} \right. $$ $$\Rightarrow \left\{ \begin{array}{l}x \lt \frac{1}{2}\\x = - 4\end{array} \right.$$ $$ - 4 \in \left( {\frac{1}{2}; + \infty } \right)$$

Жауабы: $ - 4;5$

№ 13 Теңдеуді шешіңіз: ${\log _{16}}x + {\log _4}x + {\log _2}x = 7$

Шешуі:

$$\text{ММЖ:} \quad x \gt 0$$ $${\log _{{2^4}}}x + {\log _{{2^2}}}x + {\log _2}x = 7$$ $$\frac{1}{4}{\log _2}x + \frac{1}{2}{\log _2}x + {\log _2}x = 7$$

$$\left( {\frac{1}{4} + \frac{1}{2} + 1} \right){\log _2}x = 7$$ $$1\frac{3}{4}{\log _2}x = 7$$ $$\frac{7}{4}{\log _2}x = 7$$

$${\log _2}x = 4$$ $$x = 2$$ $$x = 16$$

Жауабы: $16$

№ 14 Теңдеуді шешіңіз: ${\log _4}{\log _3}{\log _2}\left( {{x^2} - 1} \right) = 0$

Шешуі:

$$\text{ММЖ:} \quad {x^2} - 1 \gt 0$$ $${x^2} - 1 = 0$$ $${x^2} = 1$$ $$x = \pm 1$$

Логарифмдік теңдеулер (Рустюмова 3.4.1) $$x \in \left( { - \infty ; - 1} \right) \cup \left( { - 1; + \infty } \right)$$ $${\log _4}{\log _3}{\log _2}\left( {{x^2} - 1} \right)=0$$ $${\log _3}{\log _2}\left( {{x^2} - 1} \right) = 1$$

$${\log _2}\left( {{x^2} - 1} \right) = 3$$ $${x^2} - 1 = {2^3}$$ $${x^2} - 1 = 8$$ $${x^2} = 9$$ $$x = \pm 3$$

Жауабы: $-3;3$

№ 15 Теңдеуді шешіңіз: ${\log _{x + 2}}\left( {3{x^2} - 12} \right) = 2$

Шешуі:

$$\text{ММЖ:} \quad \left\{ \begin{array}{l}3{x^2} - 12 \gt 0\\x + 2 \gt 0\\x + 2 \ne 1\end{array} \right.$$ $$1)\quad 3{x^2} - 12 = 0$$ $$3{x^2} = 12$$ $${x^2} = 4$$ $$x = \pm 2$$ Логарифмдік теңдеулер (Рустюмова 3.4.1)$$x \in ( - \infty ; - 2) \cup (2; + \infty )$$

$$2)\quad x + 2 \gt 0$$ $$x \gt - 2$$ $$x \in ( - 2; + \infty )$$ $$3)\quad x + 2 \ne 1$$ $$x \ne - 1$$ Логарифмдік теңдеулер (Рустюмова 3.4.1) $$x \in (2; + \infty )$$ $${\log _{x + 2}}\left( {3{x^2} - 12} \right) = 2$$ $${(x + 2)^2} = 3{x^2} - 12$$

$${x^2} + 4x + 4 = 3{x^2} - 12$$ $$2{x^2} - 4x - 16 = 0$$ $${x^2} - 2x - 8 = 0$$ $$D = 4 + 32 = 36 \gt 0$$ $${x_1} = \frac{{2 + \sqrt {36} }}{2} = $$ $$\frac{{2 + 6}}{2} = 4$$ $${x_2} = \frac{{2 - \sqrt {36} }}{2} = $$ $$\frac{{2 - 6}}{2} = - 2$$ $$ - 2 \notin (2: + \infty )$$

Жауабы: $4$

№ 16 Теңдеуді шешіңіз: ${\log _x}(64 \cdot \sqrt[3]{4}) = 1\frac{2}{3}$

Шешуі:

$$\text{ММЖ:} \quad x \gt 0;\quad x \ne 1$$ $${\log _x}(64 \cdot \sqrt[3]{4}) = 1\frac{2}{3}$$ $${\log _x}\left( {{2^6} \cdot {2^{\frac{2}{3}}}} \right) = 1\frac{2}{3}$$

$${\log _x}{2^{6\frac{2}{3}}} = 1\frac{2}{3}$$ $$6\frac{2}{3}{\log _x}2 = 1\frac{2}{3}$$ $$\log _x^2 = 1\frac{2}{3}:6\frac{2}{3}$$

$$\log _x^2 = \frac{5}{3} \cdot \frac{3}{{20}}$$ $$\log _x^2 = \frac{1}{4}$$ $${x^{\frac{1}{4}}} = 2$$

$${\left( {{x^{\frac{1}{4}}}} \right)^4} = {2^4}$$ $$x = 16$$

Жауабы: $ 16$

№ 17 Теңдеуді шешіңіз: ${\log _{x + 20}}(2x - \sqrt {x + 20} ) = \frac{1}{2}$

Шешуі:

$${(x + 20)^{\frac{1}{2}}} = 2x - \sqrt {x + 20} $$ $$\sqrt {x + 20} = 2x - \sqrt {x + 20} $$ $$2\sqrt {x + 20} = 2x$$

$$\sqrt {x + 20} = x$$ $$x + 20 = {x^2},\quad x \gt 0$$ $${x^2} - x - 20 = 0$$

$$D = 1 + 80 = 81$$ $${x_1} = \frac{{1 + \sqrt {81} }}{2} = $$ $$\frac{{1 + 9}}{2} = \frac{{10}}{2} = 5$$ $${x_2} = \frac{{1 - \sqrt {21} }}{2} = \frac{{1 - 9}}{2} $$ $$= \frac{{ - 8}}{2} = - 4 \quad{бөгде \, түбір}$$

Жауабы: $5$

№ 18 Теңдеуді шешіңіз: ${2^{\frac{1}{{{{\log }_8}x}}}} = \frac{1}{{64}}$

Шешуі:

$${2^{\frac{1}{{{{\log }_8}x}}}} = {2^{ - 6}}$$ $$\frac{1}{{{{\log }_8}x}} = - 6$$ $$ - 6 \cdot \log _8^x = 1$$

$$ - 6 \cdot {\log _{{2^3}}}x = 1$$ $$ - 6 \cdot \frac{1}{3}{\log _2}x = 1$$ $$ - 2{\log _2}x = 1$$

$${\log _2}x = - \frac{1}{2}$$ $$x = {2^{ - \frac{1}{2}}}$$ $$x = \frac{1}{{\sqrt 2 }}$$

Жауабы: $\frac{1}{{\sqrt 2 }}$

№ 19 Теңдеуді шешіңіз: $\ln (2x + 1) \cdot \ln (9 - 4x) = 0$

Шешуі:

$$\text{ММЖ:}$$ $$\left\{ {\begin{array}{*{20}{l}}{2x + 1 \gt 0}\\{9 - 4x \gt 0}\end{array}} \right. \Rightarrow $$ $$\left\{ {\begin{array}{*{20}{l}}{x \gt - \frac{1}{2}}\\{x \lt 2\frac{1}{4}}\end{array}} \right. \Rightarrow $$

$$\left\{ {\begin{array}{*{20}{l}}{2x \gt - 1}\\{4x \lt 9}\end{array}} \right. \Rightarrow $$ $$ - \frac{1}{2} \lt x \lt 2\frac{1}{4}$$

$${\ln (2x + 1) = 0}$$ $${2x + 1 = 1}$$ $${2x = 0}$$ $${x = 0}$$ $${0 \in \left( { - \frac{1}{2}:2\frac{1}{4}} \right)}$$

$${\ln (9 - 4x) = 0}$$ $${9 - 4x = 1}$$ $${4x = 8}$$ $${x = 2}$$ $${2 \in \left( { - \frac{1}{2};2\frac{1}{4}} \right)}$$

Жауабы: $0;\,2$

№ 20 Теңдеуді шешіңіз: $\log _x^{\sqrt[3]{4}} = 0,1(6)$

Шешуі:

$$\text{ММЖ:}$$ $$x \gt 0;\quad x \ne 1$$ $${\log _x}\sqrt[3]{4} = 0,1(6)$$ $${\log _x}\sqrt[3]{{{2^2}}} = \frac{{15}}{{90}}$$

$${\log _x}\sqrt[3]{{{2^2}}} = \frac{{15}}{{90}}$$ $${\log _x}{2^{\frac{2}{3}}} = \frac{{15}}{{90}}$$

$$\log _x^2 = \frac{{15}}{{90}} \cdot \frac{3}{2}$$ $$\log _x^2 = \frac{1}{4}$$

$${x^{\frac{1}{4}}} = 2$$ $$x = {2^4}$$ $$x = 16$$

Жауабы: $16$

№ 21 Теңдеуді шешіңіз: ${\log _x}(36 \cdot \sqrt[3]{{36}}) = 2,(6)$

Шешуі:

$$\text{ММЖ:}$$ $$x \gt 0;\quad x \ne 1$$ $${\log _x}(36 \cdot \sqrt[3]{{36}}) = 2,(6)$$ $${\log _x}\left( {{6^2} \cdot {6^{\frac{2}{3}}}} \right) = 2\frac{6}{9}$$

$${\log _x}{6^{2\frac{2}{3}}} = 2\frac{6}{9}$$ $$2\frac{2}{3}{\log _x}6 = 2\frac{6}{9}$$ $${\log _x}6 = 2\frac{6}{9}:2\frac{2}{3}$$

$${\log _x}6 = \frac{{24}}{9}:\frac{8}{3}$$ $${\log _x}6 = \frac{{24}}{9} \cdot \frac{3}{8}$$ $${\log _x}6 = 1$$ $$x = 6$$

Жауабы: $6$

№ 22 Теңдеуді шешіңіз: ${\log _x}(8 \cdot \sqrt[5]{{0,25}}) = \frac{{13}}{5}$

Шешуі:

$${\log _x}\left( {{2^3} \cdot \sqrt[5]{{\frac{1}{4}}}} \right) = \frac{{13}}{5}$$ $${\log _x}\left( {{2^3} \cdot {2^{ - \frac{2}{5}}}} \right) = \frac{{13}}{5}$$

$${\log _x}{2^{3 - \frac{2}{5}}} = \frac{{13}}{5}$$ $${\log _x}{2^{\frac{{13}}{5}}} = \frac{{13}}{5}$$

$$\frac{{13}}{5}{\log _x}2 = \frac{{13}}{5}$$ $${\log _x}2 = 1$$ $$x = 2$$

Жауабы: $2$

№ 23 Теңдеуді шешіңіз: $\left( {3{x^2} - 5x - 2} \right){\log _3}(5 - 4x) = 0$

Шешуі:

$$\left\{ \begin{array}{l}3{x^2} - 5x - 2 = 0\\{\log _3}(5 - 4x) = 0\\5 - 4x \gt 0\end{array} \right.$$

$$3{x^2} - 5x - 2 = 0$$ $$D = 25 + 24 = 49 \gt 0$$ $${x_{1/2}} = \frac{{5 \pm \sqrt {49} }}{6} = \frac{{5 \pm 7}}{6}$$ $${x_1} = 2;\quad {x_2} = - \frac{1}{3}$$

$${\log _3}(5 - 4x) = 0$$ $$5 - 4x = 1$$ $$4x = 4$$ $$x = 1$$

$$5 - 4x \gt 0$$ $$4x \lt 5$$ $$x \lt 1,25$$

Жауабы: $ - \frac{1}{3};\,1$

№ 24 Теңдеуді шешіңіз: $\left( {4{x^2} + 5x - 6} \right){\log _2}(2x - 6) = 0$

Шешуі:

$$\left\{ {\begin{array}{*{20}{l}}{2x - 6 \gt 0}\\{4{x^2} + 5x - 6 = 0}\\{{{\log }_2}(2x - 6) = 0}\end{array}} \right.$$ $$2x - 6 \gt 0$$ $$2x \gt 6, \quad x \gt 3$$

$$4{x^2} + 5x - 6 = 0$$ $$D = 25 + 96 = 121 \gt 0$$ $${x_1} = \frac{{ - 5 + \sqrt {121} }}{8} =$$ $$= \frac{{ - 5 + 11}}{8} = \frac{3}{4}$$ $${x_2} = \frac{{ - 5 - \sqrt {721} }}{8} =$$ $$= \frac{{ - 5 - 11}}{8} = - 2$$

$${\log _2}(2x - 6) = 0$$ $${\mkern 1mu} 2x - 6 = 1$$ $$2x = 7$$ $$x = 3,5$$

Жауабы: $3,5$

№ 25 Теңдеуді шешіңіз: ${{{\log }_3}\left( {{3^x} - 2} \right) = 1 - x}$

Шешуі:

$$\text{ММЖ:}$$ $${{3^x} - 2 \gt 0}$$ $${{3^x} \gt 2}$$ $${x \gt {{\log }_3}2}$$

$${{{\log }_3}\left( {{3^x} - 2} \right) = 1 - x}$$ $${{3^x} - 2 = {3^{1 - x}}}$$ $${{3^x} - 2 = \frac{3}{{{3^x}}}}$$ $${{3^{2x}} - 2 \cdot {3^x} - 3 = 0}$$ $${{3^x} = t}$$

$${t^2} - 2t - 3 = 0$$ $${D = 4 + 12 = 16}$$ $${{t_1} = \frac{{2 + 4}}{2} = 3}$$ $${{t_2} = \frac{{2 - 4}}{2} = - 1}$$

$${3^x} = 3$$ $$x = 1$$ $${3^x} \ne - 1$$

Жауабы: $1$

№ 26 Теңдеуді шешіңіз: ${\log _2}\left( {9 - {2^x}} \right) = {10^{\lg (3 - x)}}$

Шешуі:

$$\text{ММЖ:} \quad \left\{ \begin{array}{l}9 - {2^x} \gt 0\\3 - x \gt 0\end{array} \right. \Rightarrow $$ $$\left\{ \begin{array}{l}{2^x} \lt 9\\x \lt 3\end{array} \right. \Rightarrow \left\{ \begin{array}{l}x \lt {\log _2}9\\x \lt 3\end{array} \right.$$Логарифмдік теңдеулер (Рустюмова 3.4.1)$$x \in ( - \infty ;3)$$

$${\log _2}\left( {9 - {2^x}} \right) = {10^{\lg (3 - x)}}$$ $${\log _2}\left( {9 - {2^x}} \right) = 3 - x$$ $$9 - {2^x} = {2^{3 - x}}$$ $$9 - {2^x} = \frac{8}{{{2^x}}}$$ $$9 \cdot {2^x} - {2^{2x}} - 8 = 0$$ $${2^{2x}} - 9 \cdot {2^x} + 8 = 0$$ $${2^x} = t$$

$${t^2} - 9t + 8 = 0$$ $$D = 21 - 32 = 49$$ $${t_1} = \frac{{9 + 7}}{2} = 8$$ $${t_2} = \frac{{9 - 7}}{2} = 1$$ $${2^x} = 8;\quad {2^x} = 1$$ $$x = 3;\quad x = 0$$ $$x = 3 - \text{бөгде түбір}$$

Жауабы: $0$

№ 27 Теңдеуді шешіңіз: ${\log _5}\left( {2 + 3 \cdot {5^{ - x}}} \right) = x + 1$

Шешуі:

$$2 + 3 \cdot {5^{ - x}} = {5^{x + 1}}$$ $$2 + \frac{3}{{{5^x}}} = 5 \cdot {5^x}$$ $$5 \cdot {5^{2x}} = 2 \cdot {5^x} + 3$$ $$5 \cdot {5^{2x}} - 2 \cdot {5^x} - 3 = 0$$

$${5^x} = t$$ $$5{t^2} - 2t - 3 = 0$$ $$D = 4 + 60 = 64$$ $${t_1} = \frac{{2 + 8}}{{10}} = 1$$ $${t_2} = \frac{{2 - 8}}{{10}} = - \frac{3}{5}$$

$${5^x} = 1$$ $$x = 0$$ $${5^x} \ne - \frac{3}{5}$$

Жауабы: $0$

№ 28 Теңдеуді шешіңіз: ${\log _2}\left( {5 \cdot {2^x} + 3} \right) = 2x + 1$

Шешуі:

$$5 \cdot {2^x} + 3 = {2^{2x + 1}}$$ $$5 \cdot {2^x} + 3 - {2^{2x + 1}} = 0$$ $$5 \cdot {2^x} + 3 - 2 \cdot {2^{2x}} = 0$$ $$2 \cdot {2^{2x}} - 5 \cdot {2^x} - 3 = 0$$

$${2^x} = t$$ $$2{t^2} - 5t - 3 = 0$$ $${t_1} = \frac{{5 + 7}}{4} = 3$$ $${t_2} = \frac{{5 - 7}}{4} = - \frac{1}{2}$$

$${2^x} = 3$$ $$x = {\log _2}3$$ $${2^x} \ne - \frac{1}{2}$$

Жауабы: ${\log _2}3$

№ 29 Теңдеуді шешіңіз: ${\log _7}\left( {6 + {7^{ - x}}} \right) = x + 1$

Шешуі:

$$6 + {7^{ - x}} = {7^{x + 1}}$$ $$6 + \frac{1}{{{7^x}}} = 7 \cdot {7^x}$$ $$7 \cdot {7^{2x}} - 6 \cdot {7^x} - 1 = 0$$ $${7^x} = t$$

$$7{t^2} - 6t - 1 = 0$$ $$D = 36 + 28 = 64$$ $${t_1} = \frac{{6 + 8}}{{14}} = 1$$ $${t_2} = \frac{{6 - 8}}{{14}} = - \frac{1}{7}$$

$${7^x} = 1$$ $$x = 0$$ $${7^x} \ne - \frac{1}{7}$$

Жауабы: $0$

№ 30 Теңдеуді шешіңіз: $\lg \left( {81 \cdot \sqrt[3]{{{3^{{x^2} - 8x}}}}} \right) = 0$

Шешуі:

$$81 \cdot \sqrt[3]{{{3^{{x^2} - 8x}}}} = 1$$ $${3^4} \cdot {3^{\frac{{{x^2} - 8x}}{3}}} = 1$$ $${3^{4 + \frac{{{x^2} - 8x}}{3}}} = 1$$ $${3^{4 + \frac{{{x^2} - 8x}}{3}}} = {3^0}$$

$$4 + \frac{{{x^2} - 8x}}{3} = 0$$ $$\frac{{12 + {x^2} - 8x}}{3} = 0$$ $$12 + {x^2} - 8x = 0$$

$${x^2} - 8x + 12 = 0$$ $$\left\{ \begin{array}{l}{x_1} + {x_2} = 8\\{x_1} \cdot {x_2} = 12\end{array} \right.$$ $${x_1} = 2;\quad {x_2} = 6$$

Жауабы: $2;\,6$

№ 31 Теңдеуді шешіңіз: ${\log _{2x + 7}}(2\sqrt {2x + 7} - 2x + 5) = 0,5$

Шешуі:

$2\sqrt {2x + 7} - 2x + 5 = $ ${(2x + 7)^{0,5}}$ $$2\sqrt {2x + 7} - 2x + 5 = \sqrt {2x + 7} $$ $$\sqrt {2x + 7} - 2x + 5 = 0$$ $$\sqrt {2x + 7} - 2x - 7 + 12 = 0$$ $$\sqrt {2x + 7} - (2x + 7) + 12 = 0$$

$$\sqrt {2x + 7} = t,\quad t \gt 0$$ $$t - {t^2} + 12 = 0$$ $$\left\{ \begin{array}{l}{t_1} + {t_2} = 1\\{t_1} \cdot {t_2} = - 12\end{array} \right.$$ $${t_1} = -3,\quad {t_2} = 4$$

$$\sqrt {2x + 7} = 4$$ $$2x + 7 = 16$$ $$2x = 9$$ $$x = 4,5$$

Жауабы: $4,5$

№ 32 Теңдеуді шешіңіз: ${\log _6}\left( {5 + {6^{ - x}}} \right) = x + 1$

Шешуі:

$$5 + {6^{ - x}} = {6^{x + 1}}$$ $$5 + \frac{1}{{{6^x}}} = 6 \cdot {6^x}$$ $$\frac{{5 \cdot {6^x} + 1}}{{{6^x}}} = 6 \cdot {6^x}$$ $$6 \cdot {6^{2x}} = 5 \cdot {6^x} + 1$$ $$6 \cdot {6^{2x}} - 5 \cdot {6^x} - 1 = 0$$

$${6^x} = t$$ $$6{t^2} - 5t - 1 = 0$$ $$D = 25 + 24 = 49 \gt 0$$ $${t_1} = \frac{{5 + 7}}{{12}} = 1$$ $${t_2} = \frac{{5 - 7}}{{12}} = - \frac{1}{6}$$

$${6^x} = 1;\quad {6^x} \ne - \frac{1}{6}$$ $$x = 0$$

Жауабы: $0$

№ 33 Теңдеуді шешіңіз: ${\log _{256}}{x^2} + {\log _{16}}{x^2} + {\log _4}{x^2} =$ $ 7{\log _7}7$

Шешуі:

$${\log _{{4^4}}}{x^2} + {\log _{{4^2}}}{x^2} + {\log _4}{x^2} = 7$$ $\frac{1}{4}{\log _4}{x^2} + \frac{1}{2}{\log _4}{x^2} + {\log _4}{x^2} $ $= 7$ $$\left( {\frac{1}{4} + \frac{1}{2} + 1} \right) \cdot {\log _4}{x^2} = 7$$

$$\frac{7}{4}{\log _4}{x^2} = 7$$ $${\log _4}{x^2} = 7 \cdot \frac{4}{7}$$ $${\log _4}{x^2} = 4$$

$${x^2} = {4^4}$$ $${x^2} = 256$$ $$x = \pm 16$$

Жауабы: $\pm 16$

№34 Теңдеуді шешіңіз: ${\log _8}\left( {4 - {{\log }_6}(5 - x)} \right) = \frac{1}{3}$

Шешуі:

$$\text{ММЖ:}$$ $$5 - x \gt 0,\quad x \lt 5$$ $$4 - {\log _6}(5 - x) \gt 0$$ $${\log _6}(5 - x) \lt 4$$ $$5 - x \lt 1296,\quad x \gt - 1291$$ $$x \in ( - 1291;5)$$

$${\log _8}\left( {4 - {{\log }_6}(5 - x)} \right) = \frac{1}{3}$$ $$4 - {\log _6}(5 - x) = {8^{\frac{1}{3}}}$$ $$4 - {\log _6}(5 - x) = {\left( {{2^3}} \right)^{\frac{1}{3}}}$$ $$4 - {\log _6}(5 - x) = 2$$

$${\log _6}(5 - x) = 2$$ $$5 - x = {6^2}$$ $$5 - x = 36$$ $$x = - 31$$

Жауабы: $-31$

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