3.1.1. Екі жағын бірдей дәрежеге шығару

$${\left( {\frac{3}{7}} \right)^{3x – 7}} = {\left( {\frac{3}{7}} \right)^{3 – 7x}}$$ $$3x – 7 = 3 – 7x$$ $$3x + 7x = 3 + 7$$ $$10x = 10$$ $$x = 1$$

$${\left( {\frac{3}{7}} \right)^{3x – 7}} = {\left( {\frac{3}{7}} \right)^{3 – 7x}}$$ $$3x – 7 = 3 – 7x$$ $$3x + 7x = 3 + 7$$ $$10x = 10$$ $$x = 1$$

Жауабы: $1$

$${2^{\frac{{x – 1}}{3}}} = {2^{1 – \frac{1}{2}}}$$ $${2^{\frac{{x – 1}}{3}}} = {2^{\frac{1}{2}}}$$ $$\frac{{x – 1}}{3} = \frac{1}{2}$$

$$x – 1 = \frac{3}{2}$$ $$x = \frac{3}{2} + 1 = \frac{5}{2}$$

Жауабы: $\frac{5}{2}$

$${{2^{x + x – 4}} = {2^4}}$$ $${2x – 4 = 4}$$

$${2x = 8}$$ $${x = 4}$$

Жауабы:$4$

$${{2^{{x^2} – 6x – 2,5}} = {2^4} \cdot {2^{\frac{1}{2}}}}$$ $${{2^{{x^2} – 6x – 2,5}} = {2^{4\frac{1}{2}}}}$$

$${{x^2} – 6x – 2,5 = 4,5}$$ $${{x^2} – 6x – 7 = 0}$$ $${{x_1} = 7,\quad {x_2} = – 1}$$

Жауабы: ${ – 1;7}$

$${2^{3x + 1}} = {2^{ – 5}}$$ $$3x + 1 = – 5$$

$$3x = – 6$$ $$x = – 2$$

Жауабы: $ – 2$

$${{{10}^{(5 – x)(6 – x)}} = {{10}^2}}$$ $${(5 – x)(6 – x) = 2}$$ $${30 – 5x – 6x + {x^2} – 2 = 0}$$

$${{x^2} – 11x + 28 = 0}$$ $${{x_1} = 7,\quad {x_2} = 4}$$

Жауабы: ${4;7}$

$${{{(2,5)}^{2x – 3}} = {{\left( {\frac{5}{2}} \right)}^3}}$$ $${{{\left( {\frac{5}{2}} \right)}^{2x – 3}} = {{\left( {\frac{5}{2}} \right)}^3}}$$

$${2x – 3 = 3}$$ $${2x = 6}$$ $${x = 3}$$

Жауабы: $3$

$${{3^{{x^2} – 7,2x}} = {3^{ – 1}} \cdot {3^{ – \frac{2}{5}}}}$$ $${{3^{{x^2} – 7,2x}} = {3^{ – 1 – \frac{2}{5}}}}$$ $${{x^2} – 7,2x + 1\frac{2}{5} = 0}$$

$${{x^2} – 7,2x + 1,4 = 0}$$ $${D = {{3,6}^2} – 1,4 = 11,56}$$ $${{x_{1,2}} = 3,6 \pm \sqrt {11,56} = 3,6 \pm 3,4}$$ $${x_1} = 7;\quad {x_2} = 0,2$$

Жауабы: ${\quad 7; 0,2}$

$${{{(2 \cdot 3)}^{\frac{x}{2}}} = {6^2}}$$ $${{6^{\frac{x}{2}}} = {6^2}}$$

$${\frac{x}{2} = 2}$$ $${x = 4}$$

Жауабы: $4$

$${{{\left( {\frac{4}{9}} \right)}^x} \cdot {{\left( {\frac{{27}}{8}} \right)}^x} \cdot {{\left( {\frac{{27}}{8}} \right)}^{ – 1}} = \frac{2}{3}}$$ $${{{\left( {\frac{4}{9} \cdot \frac{{27}}{8}} \right)}^x} = \frac{2}{3}:\frac{8}{{27}}}$$

$${{{\left( {\frac{3}{2}} \right)}^x} = \frac{2}{3} \cdot \frac{{27}}{8}}$$ $${{{\left( {\frac{3}{2}} \right)}^x} = {{\left( {\frac{3}{2}} \right)}^2}}$$ $${x = 2}$$

Жауабы: $2$

$${{{20}^x} \cdot 5 = 5 \cdot {{20}^{2 – x}}}$$ $${{{20}^x} = {{20}^{2 – x}}}$$

$${x = 2 – x}$$ $${2x = 2}$$ $${x = 1}$$

Жауабы: $1$

$${\left( {\frac{1}{4}} \right)^x} = \frac{{{2^7}}}{{{2^{x – 1}}}}$$ $${2^{ – 2x}} = {2^{7 – x + 1}}$$

$$ – 2x = 8 – x$$ $$ – x = 8$$ $$x = – 8$$

Жауабы: $ – 8$

$${{2^{ – {x^2} + 20x – 61,5}} = {2^{3 – \frac{1}{2}}}}$$ $${ – {x^2} + 20x – 61,5 = 2,5}$$

$${{x^2} – 20x + 64 = 0}$$ $${D = 100 – 64 = 36}$$ $${x_1} = 10 + 6 = 16$$ $${x_2} = 10 – 6 = 4$$

Жауабы: $4;16$

$${{5^{\frac{{5 – 3x}}{{x – 2}}}} = {5^{ – 2}}}$$ $${\frac{{5 – 3x}}{{x – 2}} = – 2}$$

$${5 – 3x = – 2x + 4}, \quad x \ne 2$$ $${x = 1}$$

Жауабы: $1$

$${\left( {\frac{5}{{12}}} \right)^{2x – 3}} = {\left( {\frac{{24}}{{10}}} \right)^{3x – 2}}$$ $${\left( {\frac{5}{{12}}} \right)^{2x – 3}} = {\left( {\frac{5}{{12}}} \right)^{2 – 3x}}$$

$$2x – 3 = 2 – 3x$$ $$5x = 5$$ $$x = 1$$

Жауабы: $1$

$${{4^{\frac{4}{x}}} = {4^x},\quad x \in N}$$ $${\frac{4}{x} = x}$$

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

Жауабы: $2$

$$\boxed{\sqrt[q]{a^p} = a^{\frac{p}{q}},\quad a \ge 0,\,q \ne 1}$$ $${{{10}^3} \cdot {{10}^{ – \frac{1}{x}}} = {{10}^{2x}}}$$ $${{{10}^{3 – \frac{1}{x}}} = {{10}^2x}}$$ $${3 – \frac{1}{x} = 2x}$$

$${2x + \frac{1}{x} = 3}$$ $${2{x^2} – 3x + 1 = 0}$$ $${D = 9 – 8 = 1}$$ $${{x_{1,2}} = \frac{{3 \pm 1}}{4} = \left\langle \begin{array}{l}\frac{1}{2}\\1\end{array} \right.}$$ $${x \in N,\quad x \ne 1}$$

Жауабы: $\emptyset $

$${{2^{\cos 2x}} = {2^{\frac{1}{2}}}}$$ $${\cos 2x = \frac{1}{2}}$$

$${2x = \pm \frac{\pi }{3} + 2\pi n}$$ $${x = \pm \frac{\pi }{6} + \pi n,\quad n \in Z}$$

Жауабы: ${ \pm \frac{\pi }{6} + \pi n,\quad n \in Z}$

$${{{15}^{\frac{1}{x}}} = {{15}^{ 2}}}$$ $${\frac{1}{x} = 2}$$

$$x = \frac{1}{2}$$ $$x \in N$$

Жауабы: $\emptyset $

$${{{10}^{{x^2} – 3}} = {{10}^{ – 2}} \cdot {{10}^{3x – 3}}}$$ $${{x^2} – 3 = 3x – 5}$$

$${{x^2} – 3x + 2 = 0}$$ $${{x_1} = 2,\quad {x_2} = 1}$$

Жауабы: ${1;2}$

$${27 = {{\left( {\frac{1}{3}} \right)}^{6 – x}}}$$ $${{3^3} = {3^{x – 6}}}$$

$${3 = x – 6}$$ $${x = 9}$$

Жауабы: $9$

$${5^2}^{|1 – 2x|} = {5^{4 – 6x}}$$ $$2|1 – 2x| = 4 – 6x$$ $$|1 – 2x| = 2 – 3x$$

$$\left\{ \begin{array}{l}\left[ {\begin{array}{*{20}{l}}{1 – 2x = 2 – 3x}\\{1 – 2x = 3x – 2}\end{array}} \right.\\2 – 3x \ge 0\end{array} \right. \Rightarrow $$ $$ \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = 1\\x = \frac{3}{5}\end{array} \right.\\x \le \frac{2}{3}\end{array} \right.$$

Жауабы: $\frac{3}{5}$

$${{2^{|x + 1|}} = {2^{\frac{1}{2}( – 2x + 3)}}}$$ $${|x + 1| = – x + \frac{3}{2}}$$

$$\left\{ \begin{array}{l}\left[ \begin{array}{l}x + 1 = – x + \frac{3}{2}\\x + 1 = x – \frac{3}{2}\end{array} \right.\\ – x + \frac{3}{2} \ge 0\end{array} \right. \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x = \frac{1}{4}\\\emptyset \end{array} \right.\\x \le \frac{3}{2}\end{array} \right.$$

Жауабы: $\frac{1}{4}$

$${{x^2} – 7,2x + 3,9 = 2,5}$$ $$\left[ \begin{array}{l}{3^{{x^2} – 7,2x + 3,9}} – 9\sqrt 3 = 0,\\\sqrt {5 – x} = 0\\5 – x \ge 0 \Rightarrow x \le 5\end{array} \right. \Rightarrow $$ $$\left[ \begin{array}{l}{x^2} – 7,2x + 3,9 = 2,5\\x = 5\end{array} \right.$$

$${x^2} – 7,2x + 1,4 = 0$$ $$D = {3,6^2} – 1,4 = 11,56$$ $${{x_{1,2}} = 3,6 \pm 3,4 = \left\langle \begin{array}{l}7 -\text{бөгде түбір} \\0,2\end{array} \right.}$$

Жауабы: $0,2; 5$

$${{2^{\frac{{4x + 5}}{4}}} = {{\left( {\frac{{\sqrt 2 }}{2}} \right)}^{\frac{{2x}}{3}}}}$$ $${{2^{\frac{{4x + 5}}{4}}} = {2^{ – \frac{1}{2} \cdot \frac{{2x}}{3}}}}$$ $${x + \frac{5}{4} = – \frac{x}{3}}$$

$${\frac{4}{3}x = – \frac{5}{4}}$$ $${x = – \frac{5}{4} \cdot \frac{3}{4} = – \frac{{15}}{{16}}}$$

Жауабы: ${ – \frac{{15}}{{16}}}$

$${7^{\frac{1}{3}}} \cdot {7^{2,5}} = {7^{\frac{1}{2} – \frac{2}{3}}} \cdot {7^2} \cdot {x^{0,5}}$$ $${\frac{{{7^{2,5 + \frac{1}{3}}}}}{{{7^{\frac{1}{2} – \frac{2}{3} + 2}}}} = {x^{\frac{1}{2}}}}$$

$${{7^{\frac{5}{2} + \frac{1}{3} – \frac{1}{2} + \frac{2}{3} – 2}} = {x^{\frac{1}{2}}}}$$ $${{7^{2 + 1 – 2}} = {x^{\frac{1}{2}}}}$$ $${x = 49}$$

Жауабы: ${49}$

$${{2^{4 + \frac{1}{8} + \frac{3}{{40}}}} \cdot x = {2^{6 + \frac{6}{5} – 4}}}$$ $${x = {2^{2 + \frac{6}{5} – 4 – \frac{1}{8} – \frac{3}{{40}}}}}$$

$${x = {2^{ – 2 + 1}} = {2^{ – 1}} = \frac{1}{2}}$$

Жауабы: ${\frac{1}{2}}$

$$\left| {{x^2} – 3x + 2} \right| = 2(x + 1)$$ $$\left\{ \begin{array}{l}\left[ \begin{array}{l}{x^2} – 3x + 2 = 2x + 2\\{x^2} – 3x + 2 = – 2x – 2\end{array} \right.\\2x + 2 \ge 0\end{array} \right.$$

$$ \Rightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}{x_1} = 0,\quad {x_2} = 5\\{x^2} – x + 4 = 0,\quad x \in \emptyset \end{array} \right.\\x \ge – 1\end{array} \right.$$

Жауабы: $0;\,5$

$${{{1,8}^{4\sqrt x – 10}} = {{\left( {\frac{5}{9}} \right)}^{5\sqrt x + 1}}}$$ $${{{\left( {\frac{5}{9}} \right)}^{10 – 4\sqrt x }} = {{\left( {\frac{5}{9}} \right)}^{5\sqrt x + 1}}}$$

$${10 – 4\sqrt x = 5\sqrt x + 1}$$ $${9 = 9\sqrt x }$$ $${x = 1}$$

Жауабы: $1$

$${{3^{ – 3}} \cdot {3^{\frac{{6x – 2}}{4}}} = {3^{ – 2}}}$$ $${{3^{ – 3 + \frac{{3x – 1}}{2}}} = {3^{ – 2}}}$$ $${\frac{{ – 6 + 3x – 1}}{2} = – 2}$$

$${ – 6 + 3x – 1 = – 4}$$ $${3x = 3}$$ $${x = 1}$$

Жауабы: $1$