№ 1 Теңдеуді шешіңіз: ${\left( {\frac{3}{7}} \right)^{3x - 7}} = {\left( {\frac{7}{3}} \right)^{7x - 3}}$
$${\left( {\frac{3}{7}} \right)^{3x - 7}} = {\left( {\frac{3}{7}} \right)^{3 - 7x}}$$ $$3x - 7 = 3 - 7x$$ $$3x + 7x = 3 + 7$$ $$10x = 10$$ $$x = 1$$
$${\left( {\frac{3}{7}} \right)^{3x - 7}} = {\left( {\frac{3}{7}} \right)^{3 - 7x}}$$ $$3x - 7 = 3 - 7x$$ $$3x + 7x = 3 + 7$$ $$10x = 10$$ $$x = 1$$
Жауабы: $1$
№ 2 Теңдеуді шешіңіз: $\sqrt[3]{{{2^{x - 1}}}} = \frac{2}{{\sqrt 2 }}$
$${2^{\frac{{x - 1}}{3}}} = {2^{1 - \frac{1}{2}}}$$ $${2^{\frac{{x - 1}}{3}}} = {2^{\frac{1}{2}}}$$ $$\frac{{x - 1}}{3} = \frac{1}{2}$$
$$x - 1 = \frac{3}{2}$$ $$x = \frac{3}{2} + 1 = \frac{5}{2}$$
Жауабы: $\frac{5}{2}$
№ 3 Теңдеуді шешіңіз: ${{2^x} \cdot {2^{x - 4}} = 16}$
$${{2^{x + x - 4}} = {2^4}}$$ $${2x - 4 = 4}$$
$${2x = 8}$$ $${x = 4}$$
Жауабы:$4$
№ 4 Теңдеуді шешіңіз: ${{2^{{x^2} - 6x - 2,5}} = 16\sqrt 2 }$
$${{2^{{x^2} - 6x - 2,5}} = {2^4} \cdot {2^{\frac{1}{2}}}}$$ $${{2^{{x^2} - 6x - 2,5}} = {2^{4\frac{1}{2}}}}$$
$${{x^2} - 6x - 2,5 = 4,5}$$ $${{x^2} - 6x - 7 = 0}$$ $${{x_1} = 7,\quad {x_2} = - 1}$$
Жауабы: ${ - 1;7}$
№ 5 Теңдеуді шешіңіз: ${2^{3x + 1}} = {32^{ - 1}}$
$${2^{3x + 1}} = {2^{ - 5}}$$ $$3x + 1 = - 5$$
$$3x = - 6$$ $$x = - 2$$
Жауабы: $ - 2$
№ 6 Теңдеуді шешіңіз: ${{{\left( {{{10}^{5 - x}}} \right)}^{6 - x}} = 100}$
$${{{10}^{(5 - x)(6 - x)}} = {{10}^2}}$$ $${(5 - x)(6 - x) = 2}$$ $${30 - 5x - 6x + {x^2} - 2 = 0}$$
$${{x^2} - 11x + 28 = 0}$$ $${{x_1} = 7,\quad {x_2} = 4}$$
Жауабы: ${4;7}$
№ 7 Теңдеуді шешіңіз: ${{{(2,5)}^{2x - 3}} = \frac{{125}}{8}}$
$${{{(2,5)}^{2x - 3}} = {{\left( {\frac{5}{2}} \right)}^3}}$$ $${{{\left( {\frac{5}{2}} \right)}^{2x - 3}} = {{\left( {\frac{5}{2}} \right)}^3}}$$
$${2x - 3 = 3}$$ $${2x = 6}$$ $${x = 3}$$
Жауабы: $3$
№ 8 Теңдеуді шешіңіз: ${{3^{{x^2} - 7,2x}} = \frac{1}{{3\sqrt[5]{9}}}}$
$${{3^{{x^2} - 7,2x}} = {3^{ - 1}} \cdot {3^{ - \frac{2}{5}}}}$$ $${{3^{{x^2} - 7,2x}} = {3^{ - 1 - \frac{2}{5}}}}$$ $${{x^2} - 7,2x + 1\frac{2}{5} = 0}$$
$${{x^2} - 7,2x + 1,4 = 0}$$ $${D = {{3,6}^2} - 1,4 = 11,56}$$ $${{x_{1,2}} = 3,6 \pm \sqrt {11,56} = 3,6 \pm 3,4}$$ $${x_1} = 7;\quad {x_2} = 0,2$$
Жауабы: ${\quad 7; 0,2}$
№ 9 Теңдеуді шешіңіз: ${{2^{\frac{x}{2}}} \cdot {3^{\frac{x}{2}}} = 36}$
$${{{(2 \cdot 3)}^{\frac{x}{2}}} = {6^2}}$$ $${{6^{\frac{x}{2}}} = {6^2}}$$
$${\frac{x}{2} = 2}$$ $${x = 4}$$
Жауабы: $4$
№ 10 Теңдеуді шешіңіз: ${{{\left( {\frac{4}{9}} \right)}^x} \cdot {{\left( {\frac{{27}}{8}} \right)}^{x - 1}} = \frac{2}{3}}$
$${{{\left( {\frac{4}{9}} \right)}^x} \cdot {{\left( {\frac{{27}}{8}} \right)}^x} \cdot {{\left( {\frac{{27}}{8}} \right)}^{ - 1}} = \frac{2}{3}}$$ $${{{\left( {\frac{4}{9} \cdot \frac{{27}}{8}} \right)}^x} = \frac{2}{3}:\frac{8}{{27}}}$$
$${{{\left( {\frac{3}{2}} \right)}^x} = \frac{2}{3} \cdot \frac{{27}}{8}}$$ $${{{\left( {\frac{3}{2}} \right)}^x} = {{\left( {\frac{3}{2}} \right)}^2}}$$ $${x = 2}$$
Жауабы: $2$
№ 11 Теңдеуді шешіңіз: ${{4^x} \cdot {5^{x + 1}} = 5 \cdot {{20}^{2 - x}}}$
$${{{20}^x} \cdot 5 = 5 \cdot {{20}^{2 - x}}}$$ $${{{20}^x} = {{20}^{2 - x}}}$$
$${x = 2 - x}$$ $${2x = 2}$$ $${x = 1}$$
Жауабы: $1$
№ 12 Теңдеуді шешіңіз: ${0,25^x} = \frac{{128}}{{{2^{x - 1}}}}$
$${\left( {\frac{1}{4}} \right)^x} = \frac{{{2^7}}}{{{2^{x - 1}}}}$$ $${2^{ - 2x}} = {2^{7 - x + 1}}$$
$$ - 2x = 8 - x$$ $$ - x = 8$$ $$x = - 8$$
Жауабы: $ - 8$
№ 13 Теңдеуді шешіңіз: ${{{(0,5)}^{{x^2} - 20x + 61,5}} = \frac{8}{{\sqrt 2 }}}$
$${{2^{ - {x^2} + 20x - 61,5}} = {2^{3 - \frac{1}{2}}}}$$ $${ - {x^2} + 20x - 61,5 = 2,5}$$
$${{x^2} - 20x + 64 = 0}$$ $${D = 100 - 64 = 36}$$ $${x_1} = 10 + 6 = 16$$ $${x_2} = 10 - 6 = 4$$
Жауабы: $4;16$
№ 14 Теңдеуді шешіңіз: ${{5^{\frac{{5 - 3x}}{{x - 2}}}} = \frac{1}{{25}}}$
$${{5^{\frac{{5 - 3x}}{{x - 2}}}} = {5^{ - 2}}}$$ $${\frac{{5 - 3x}}{{x - 2}} = - 2}$$
$${5 - 3x = - 2x + 4}, \quad x \ne 2$$ $${x = 1}$$
Жауабы: $1$
№ 15 Теңдеуді шешіңіз: ${\left( {\frac{5}{{12}}} \right)^{2x - 3}} = {(2,4)^{3x - 2}}$
$${\left( {\frac{5}{{12}}} \right)^{2x - 3}} = {\left( {\frac{{24}}{{10}}} \right)^{3x - 2}}$$ $${\left( {\frac{5}{{12}}} \right)^{2x - 3}} = {\left( {\frac{5}{{12}}} \right)^{2 - 3x}}$$
$$2x - 3 = 2 - 3x$$ $$5x = 5$$ $$x = 1$$
Жауабы: $1$
№ 16 Теңдеуді шешіңіз: ${\sqrt[x]{{256}} = {4^x}}$
$${{4^{\frac{4}{x}}} = {4^x},\quad x \in N}$$ $${\frac{4}{x} = x}$$
$${{x^2} = 4}$$ $${x = 2}$$
Жауабы: $2$
№ 17 Теңдеуді шешіңіз: $1000 \cdot \sqrt[x]{{0,1}} = {100^x}$
$$\boxed{\sqrt[q]{a^p} = a^{\frac{p}{q}},\quad a \ge 0,\,q \ne 1}$$ $${{{10}^3} \cdot {{10}^{ - \frac{1}{x}}} = {{10}^{2x}}}$$ $${{{10}^{3 - \frac{1}{x}}} = {{10}^2x}}$$ $${3 - \frac{1}{x} = 2x}$$
$${2x + \frac{1}{x} = 3}$$ $${2{x^2} - 3x + 1 = 0}$$ $${D = 9 - 8 = 1}$$ $${{x_{1,2}} = \frac{{3 \pm 1}}{4} = \left\langle \begin{array}{l}\frac{1}{2}\\1\end{array} \right.}$$ $${x \in N,\quad x \ne 1}$$
Жауабы: $\emptyset $
№ 18 Теңдеуді шешіңіз: ${{2^{\cos 2x}} = \sqrt 2 }$
$${{2^{\cos 2x}} = {2^{\frac{1}{2}}}}$$ $${\cos 2x = \frac{1}{2}}$$
$${2x = \pm \frac{\pi }{3} + 2\pi n}$$ $${x = \pm \frac{\pi }{6} + \pi n,\quad n \in Z}$$
Жауабы: ${ \pm \frac{\pi }{6} + \pi n,\quad n \in Z}$
№ 19 Теңдеуді шешіңіз: ${\sqrt[x]{{15}} = 225}$
$${{{15}^{\frac{1}{x}}} = {{15}^{ 2}}}$$ $${\frac{1}{x} = 2}$$
$$x = \frac{1}{2}$$ $$x \in N$$
Жауабы: $\emptyset $
№ 20 Теңдеуді шешіңіз: ${{2^{{x^2} - 3}} \cdot {5^{{x^2} - 3}} = 0,01 \cdot {{\left( {{{10}^{x - 1}}} \right)}^3}}$
$${{{10}^{{x^2} - 3}} = {{10}^{ - 2}} \cdot {{10}^{3x - 3}}}$$ $${{x^2} - 3 = 3x - 5}$$
$${{x^2} - 3x + 2 = 0}$$ $${{x_1} = 2,\quad {x_2} = 1}$$
Жауабы: ${1;2}$
№ 21 Теңдеуді шешіңіз: ${27 = {{\left( {0,(3)} \right)}^{6 - x}}}$
$${27 = {{\left( {\frac{1}{3}} \right)}^{6 - x}}}$$ $${{3^3} = {3^{x - 6}}}$$
$${3 = x - 6}$$ $${x = 9}$$
Жауабы: $9$
№ 22 Теңдеуді шешіңіз: ${25^{|1 - 2x|}} = {5^{4 - 6x}}$
$${5^2}^{|1 - 2x|} = {5^{4 - 6x}}$$ $$2|1 - 2x| = 4 - 6x$$ $$|1 - 2x| = 2 - 3x$$
$$\left\{ \begin{array}{l}\left[ {\begin{array}{*{20}{l}}{1 - 2x = 2 - 3x}\\{1 - 2x = 3x - 2}\end{array}} \right.\\2 - 3x \ge 0\end{array} \right. \Rightarrow $$ $$ \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 1\\x = \frac{3}{5}\end{array} \right.\\x \le \frac{2}{3}\end{array} \right.$$
Жауабы: $\frac{3}{5}$
№ 23 Теңдеуді шешіңіз: ${{2^{|x + 1|}} = {{(\sqrt 2 )}^{ - 2x + 3}}}$
$${{2^{|x + 1|}} = {2^{\frac{1}{2}( - 2x + 3)}}}$$ $${|x + 1| = - x + \frac{3}{2}}$$
$$\left\{ \begin{array}{l}\left[ \begin{array}{l}x + 1 = - x + \frac{3}{2}\\x + 1 = x - \frac{3}{2}\end{array} \right.\\ - x + \frac{3}{2} \ge 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = \frac{1}{4}\\\emptyset \end{array} \right.\\x \le \frac{3}{2}\end{array} \right.$$
Жауабы: $\frac{1}{4}$
№ 24 Теңдеуді шешіңіз: $\left( {{3^{{x^2} - 7,2x + 3,9}} - 9\sqrt 3 } \right) \cdot \sqrt {5 - x} = 0$
$${{x^2} - 7,2x + 3,9 = 2,5}$$ $$\left[ \begin{array}{l}{3^{{x^2} - 7,2x + 3,9}} - 9\sqrt 3 = 0,\\\sqrt {5 - x} = 0\\5 - x \ge 0 \Rightarrow x \le 5\end{array} \right. \Rightarrow $$ $$\left[ \begin{array}{l}{x^2} - 7,2x + 3,9 = 2,5\\x = 5\end{array} \right.$$
$${x^2} - 7,2x + 1,4 = 0$$ $$D = {3,6^2} - 1,4 = 11,56$$ $${{x_{1,2}} = 3,6 \pm 3,4 = \left\langle \begin{array}{l}7 -\text{бөгде түбір} \\0,2\end{array} \right.}$$
Жауабы: $0,2; 5$
№ 25 Теңдеуді шешіңіз: ${(\sqrt[4]{2})^{4x + 5}} = {\left( {\sin \frac{\pi }{4}} \right)^{\frac{{2x}}{3}}}$
$${{2^{\frac{{4x + 5}}{4}}} = {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^{\frac{{2x}}{3}}}}$$ $${{2^{\frac{{4x + 5}}{4}}} = {2^{ - \frac{1}{2} \cdot \frac{{2x}}{3}}}}$$ $${x + \frac{5}{4} = - \frac{x}{3}}$$
$${\frac{4}{3}x = - \frac{5}{4}}$$ $${x = - \frac{5}{4} \cdot \frac{3}{4} = - \frac{{15}}{{16}}}$$
Жауабы: ${ - \frac{{15}}{{16}}}$
№ 26 Теңдеуді шешіңіз: ${49^{\frac{1}{6}}} \cdot {7^{2,5}} = {7^{\frac{1}{2}}} \cdot {7^{ - \frac{2}{3}}} \cdot 49 \cdot {x^{0,5}}$
$${7^{\frac{1}{3}}} \cdot {7^{2,5}} = {7^{\frac{1}{2} - \frac{2}{3}}} \cdot {7^2} \cdot {x^{0,5}}$$ $${\frac{{{7^{2,5 + \frac{1}{3}}}}}{{{7^{\frac{1}{2} - \frac{2}{3} + 2}}}} = {x^{\frac{1}{2}}}}$$
$${{7^{\frac{5}{2} + \frac{1}{3} - \frac{1}{2} + \frac{2}{3} - 2}} = {x^{\frac{1}{2}}}}$$ $${{7^{2 + 1 - 2}} = {x^{\frac{1}{2}}}}$$ $${x = 49}$$
Жауабы: ${49}$
№ 27 Теңдеуді шешіңіз: $16 \cdot {2^{\frac{1}{8}}} \cdot {8^{\frac{1}{{40}}}} \cdot x = {4^3} \cdot {2^{\frac{6}{5}}} \cdot {\left( {\frac{1}{4}} \right)^2}$
$${{2^{4 + \frac{1}{8} + \frac{3}{{40}}}} \cdot x = {2^{6 + \frac{6}{5} - 4}}}$$ $${x = {2^{2 + \frac{6}{5} - 4 - \frac{1}{8} - \frac{3}{{40}}}}}$$
$${x = {2^{ - 2 + 1}} = {2^{ - 1}} = \frac{1}{2}}$$
Жауабы: ${\frac{1}{2}}$
№ 28 Теңдеуді шешіңіз: ${3^{\left| {{x^2} - 3x + 2} \right|}} = {9^{x + 1}}$
$$\left| {{x^2} - 3x + 2} \right| = 2(x + 1)$$ $$\left\{ \begin{array}{l}\left[ \begin{array}{l}{x^2} - 3x + 2 = 2x + 2\\{x^2} - 3x + 2 = - 2x - 2\end{array} \right.\\2x + 2 \ge 0\end{array} \right.$$
$$ \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}{x_1} = 0,\quad {x_2} = 5\\{x^2} - x + 4 = 0,\quad x \in \emptyset \end{array} \right.\\x \ge - 1\end{array} \right.$$
Жауабы: $0;\,5$
№ 29 Теңдеуді шешіңіз: ${{{3,24}^{2\sqrt x - 5}} = {{\left( {\frac{5}{9}} \right)}^{5\sqrt x + 1}}}$
$${{{1,8}^{4\sqrt x - 10}} = {{\left( {\frac{5}{9}} \right)}^{5\sqrt x + 1}}}$$ $${{{\left( {\frac{5}{9}} \right)}^{10 - 4\sqrt x }} = {{\left( {\frac{5}{9}} \right)}^{5\sqrt x + 1}}}$$
$${10 - 4\sqrt x = 5\sqrt x + 1}$$ $${9 = 9\sqrt x }$$ $${x = 1}$$
Жауабы: $1$
№ 30 Теңдеуді шешіңіз: ${\frac{1}{{27}}\sqrt[4]{{{9^{3x - 1}}}} = {{27}^{ - \frac{2}{3}}}}$
$${{3^{ - 3}} \cdot {3^{\frac{{6x - 2}}{4}}} = {3^{ - 2}}}$$ $${{3^{ - 3 + \frac{{3x - 1}}{2}}} = {3^{ - 2}}}$$ $${\frac{{ - 6 + 3x - 1}}{2} = - 2}$$
$${ - 6 + 3x - 1 = - 4}$$ $${3x = 3}$$ $${x = 1}$$
Жауабы: $1$
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