Көрсеткіштік теңдеулер (Рустюмова 3.1.3)

Шешуі: $${\left( {\frac{3}{2}} \right)^x} = m$$ $$4 \cdot {m^2} + 15m = 54$$ $$4{m^2} + 15m – 54 = 0$$ $$n = {15^2} + 4 \cdot 4 \cdot 54 = 1089 = {33^2}$$ $${m_{1,2}} = \frac{{ – 15 \pm 33}}{8} = \left\langle \begin{array}{l}\frac{{ – 48}}{8}\\\frac{{18}}{8}\end{array} \right.$$ $${\left( {\frac{3}{2}} \right)^x} \ne – \frac{{48}}{8};\quad {\left( {\frac{3}{2}} \right)^x} = \frac{{18}}{8}$$ $${\left( {\frac{3}{2}} \right)^x} = \frac{9}{4}$$ $${\left( {\frac{3}{2}} \right)^x} = {\left( {\frac{3}{2}} \right)^2}$$ $$x = 2$$

Шешуі: $${5^x} = m$$ $$0,2 \cdot {m^2} – 4,8 \cdot m – 5 = 0\quad \left| { \times 5} \right.$$ $${m^2} – 24m – 25 = 0$$ $$D = {( – 24)^2} + 4 \cdot 25 = 676 = {26^2}$$ $${m_{1,2}} = \frac{{24 \pm 26}}{2} = {\left\langle \begin{array}{l}25\\ – 1\end{array} \right.}$$ $${5^x} = 25;\quad {5^x} \ne – 1$$ $$x = 2$$

Шешуі: $${2^{2x}} – 10 \cdot {2^x} \cdot \dfrac{1}{2} – 24 = 0$$ $${2^{2x}} – 5 \cdot {2^x} – 24 = 0$$ $${2^x} = m$$ $${m^2} – 5m – 24 = 0$$ $$D = 25 + 4 \cdot 24 = 121 = {11^2}$$ $${m_{1,2}} = \frac{{5 \pm 11}}{2} = \left\langle \begin{array}{l}8\\ – 3\end{array} \right.$$ $${2^x} = 8;\quad {2^x} \ne – 3$$ $$x = 3$$

Шешуі: $${5^{2x}} + 24 \cdot {5^x} \cdot \frac{1}{5} – 1 = 0\quad \left| { \times 5} \right.$$ $$5 \cdot {5^{2x}} + 24 \cdot {5^x} – 5 = 0$$ $${5^x} = m$$ $$5 \cdot {m^2} + 24m – 5 = 0$$ $$D = {24^2} + 4 \cdot 5 \cdot 5 = 676 = {26^2}$$ $${m_{1,2}} = \frac{{ – 24 \pm 26}}{{5 \cdot 2}} = \left\langle \begin{array}{l} – 5\\\frac{1}{5}\end{array} \right.$$ $${5^x} = \frac{1}{5};\quad {5^x} \ne – 5$$ $$x = – 1$$

Шешуі: $${2^x} \cdot 2 + {2^{2x}} – 80 = 0$$ $${2^x} = m$$ $$2m + {m^2} – 80 = 0$$ $${m^2} + 2m – 80 = 0$$ $$D = 4 + 4 \cdot 80 = 324 = {18^2}$$ $${m_{1/2}} = \frac{{ – 2 \pm 18}}{2} = \left\langle \begin{array}{l} – 10\\8\end{array} \right.$$ $${2^x} \ne – 10;\quad {2^x} = 8$$ $$x = 3$$

Шешуі: $${2^{2x}} \cdot \frac{1}{8} – 3 \cdot {2^x} \cdot \frac{1}{4} + 1 = 0$$ $${2^{2x}} – 3 \cdot {2^x} \cdot \frac{1}{4} \cdot {8} + 8 = 0$$ $${2^{2x}} – 6 \cdot {2^x} + 8 = 0$$ $${2^x} = m$$ $${m^2} – 6m + 8 = 0$$ $$D = 36 – 4 \cdot 8 = 4 = {2^2}$$ $${m_{1,2}} = \frac{{6 \pm 2}}{2} = \left\langle \begin{array}{l}2\\4\end{array} \right.$$ $${2^x} = 2;\quad {2^x} = 4$$ $$x = 1;\quad x = 2$$

Шешуі: $${3^{2x – 2}} – {3^x} \cdot 3 + {3^x} \cdot \frac{1}{{27}} = 1$$ $${3^{2x}} \cdot \frac{1}{9} – {3^x} \cdot 3 + {3^x} \cdot \frac{1}{{27}} = 1;\quad \left| { \times 27} \right.$$ $${3^{2x}} \cdot 3 – 81 \cdot {3^x} + {3^x} = 27$$ $$3 \cdot {3^{2x}} – 80 \cdot {3^x} – 27 = 0$$ $${3^x} = m$$ $$3{m^2} – 80m – 27 = 0$$ $$D = 6400 + 4 \cdot 3 \cdot 27 = 6724 = {82^2}$$ $${m_{1,2}} = \frac{{80 \pm 82}}{{3 \cdot 2}} = \left\langle \begin{array}{l} – \frac{1}{3}\\27\end{array} \right.$$ $${3^x} \ne – \frac{1}{3};\quad {3^x} = 27$$ $$x = 3.$$

Шешуі: $$5 \cdot {7^{6x}} – 2 \cdot {7^{3x}} – 3 = 0$$ $${7^{3x}} = m$$ $$5 \cdot {m^2} – 2m – 3 = 0$$ $$D = 4 + 4 \cdot 5 \cdot 3 = 64 = {8^2}$$ $${m_{1,2}} = \frac{{2 \pm 8}}{{5 \cdot 2}} = \left\langle \begin{array}{l}1\\ – \frac{2}{5}\end{array} \right.$$ $${7^{3x}} = 1;\quad {7^{3x}} \ne – \frac{2}{5}$$ $${7^{3x}} = {7^0}$$ $$3x = 0$$ $$x = 0$$

Шешуі: $${3^{2x}} + 8 \cdot {3^x} \cdot \frac{1}{3} – 1 = 0;\quad \left| { \times 3} \right.$$ $$3 \cdot {3^{2x}} + 8 \cdot {3^x} – 3 = 0$$ $${3^x} = m$$ $$3{m^2} + 8m – 3 = 0$$ $$D = 64 + 4 \cdot 3 \cdot 3 = 100 = {10^2}$$ $${m_{1,2}} = \frac{{ – 8 \pm 10}}{{3 \cdot 2}} = \left\langle \begin{array}{l} – 3\\\frac{1}{3}\end{array} \right.$$ $${3^x} \ne – 3;\quad {3^x} = \frac{1}{3}$$ $$x = – 1$$

Шешуі: $$33 \cdot {2^x} \cdot \frac{1}{2} – {2^{2x + 2}} – 2 = 0$$ $$\frac{{33}}{2} \cdot {2^x} – {2^{2x}} \cdot 4 – 2 = 0,\quad \left| { \times 2} \right.$$ $$33 \cdot {2^x} – 8 \cdot {2^{2x}} – 4 = 0$$ $${2^x} = m$$ $$33m – 8{m^2} – 4 = 0\quad \left| { \times ( – 1)} \right.$$ $$8{m^2} – 33m + 4 = 0$$ $$D = {33^2} – 4 \cdot 8 \cdot 4 = 961 = {31^2}$$ $${m_{1,2}} = \frac{{33 \pm 31}}{{8 \cdot 2}} = \left\langle \begin{array}{l}4\\\frac{1}{8}\end{array} \right.$$ $${2^x} = 4;\quad {2^x} = \frac{1}{8}$$ $$x = 2;\quad x = – 3$$

Шешуі: $${6^x} = m$$ $${m^2} – 8m + 12 = 0$$ $$D = 64 – 4 \cdot 12 = 16 = {4^2}$$ $${m_{1,2}} = \frac{{8 \pm 4}}{2} = \left\langle \begin{array}{l}6\\2\end{array} \right.$$ $${6^x} = 6;\quad {6^x} = 2$$ $$x = 1;\quad x = {\log _6}2$$

Шешуі: $${3^{2\sqrt x }} = m$$ $${m^2} – 4m + 3 = 0$$ $$D = 16 – 4 \cdot 3 = 4 = {2^2}$$ $${m_{1,2}} = \frac{{4 \pm 2}}{2} = \left\langle \begin{array}{l}3\\1\end{array} \right.$$ $${3^{2\sqrt x }} = 3,\quad \quad {3^{2\sqrt x }} = 1$$ $$2\sqrt x = 1,\quad \quad 2\sqrt x = 0$$ $$\sqrt x = \frac{1}{2},\quad \quad x = 0$$ $$x = \frac{1}{4}$$

Шешуі: $$17 \cdot {2^{\sqrt {{x^2} – 8x} }} – 8 – 2 \cdot {2^{2\sqrt {{x^2} – 8x} }} = 0$$ $${2^{\sqrt {{x^2} – 8x} }} = m$$ $$17m – 8 – 2{m^2} = 0$$ $$2{m^2} – 17m + 8 = 0$$ $$D = 289 – 4 \cdot 8 \cdot 2 = 225 = {15^2}$$ $${m_{1,2}} = \frac{{17 \pm 15}}{{2 \cdot 2}} = \left\langle \begin{array}{l}\frac{1}{2}\\8\end{array} \right.$$ $${m_1} = \frac{1}{2}\quad {m_2} = 8$$ $${2^{\sqrt {{x^2} – 8x} }} = \frac{1}{2}$$ $$\sqrt {{x^2} – 8x} \ne – 1$$ $$x \in \emptyset $$ $$\text{ММЖ}:( – \infty ;0] \cup [8; + \infty )$$ $${2^{\sqrt {{x^2} – 8x} }} = {2^3}$$ $${x^2} – 8x = 9$$ $${x^2} – 8x – 9 = 0$$ $$D = 64 + 36 = 100$$ $${x_{1,2}} = \frac{{8 \pm 10}}{2} = \left\langle \begin{array}{l}9\\ – 1\end{array} \right.$$ $$x = 9,\quad \,x = – 1$$

Шешуі: $${5^{2x}} – 24 \cdot {5^x} = 25$$ $${5^x} = m$$ $${m^2} – 24m = 25$$ $${m^2} – 24m – 25 = 0$$ $$D = {24^2} + 4 \cdot 25 = 676 = {26^2}$$ $${m_{1,2}} = \frac{{24 \pm 26}}{2} = \left\langle \begin{array}{l}25\\ – 1\end{array} \right.$$ $${5^x} = 25;\quad {5^x} \ne – 1$$ $$x = 2$$

Шешуі: $$10 \cdot {5^x} \cdot \frac{1}{5} – {5^x} \cdot 5 + {5^{2x}} – 10 = 0$$ $$2 \cdot {5^x} – 5 \cdot {5^x} + {5^{2x}} – 10 = 0$$ $${5^{2x}} – 3 \cdot {5^x} – 10 = 0$$ $${5^x} = m$$ $${m^2} – 3m – 10 = 0$$ $$D = 9 + 40 = 49 = {7^2}$$ $${m_{1,2}} = \frac{{3 \pm 7}}{2} = \left\langle \begin{array}{l} – 2\\5\end{array} \right.$$ $${5^x} \ne – 2;\quad {5^x} = 5$$ $$x = 1.$$

Шешуі: $${2^{ – 2x}} – {2^{ – (x – 1)}} – 8 = 0$$ $${2^{ – 2x}} – {2^{ – x + 1}} – 8 = 0$$ $${2^{ – 2x}} – 2 \cdot {2^{ – x}} – 8 = 0$$ $${2^{ – x}} = m$$ $${m^2} – 2m – 8 = 0$$ $$D = 4 + 32 = 36 = {6^2}$$ $${m_{1,2}} = \frac{{2 \pm 6}}{2} = \left\langle \begin{array}{l} – 2\\4\end{array} \right.$$ $${2^{ – x}} \ne – 2;\quad {2^{ – x}} = 4$$ $$ – x = 2,\quad x = – 2$$

Шешуі: $${2^{2x}} – {2^x} \cdot {2^2} + 3 + \left( {{2^{\frac{x}{2}}} + 1} \right)\left( {{2^{\frac{x}{2}}} – 1} \right) = 0;\quad {2^{\frac{x}{2}}} \ne 1$$ $${2^{2x}} – 4 \cdot {2^x} + 3 + {2^x} – 1 = 0;\quad \boxed{x \ne 0}$$ $${2^{2x}} – 3 \cdot {2^x} + 2 = 0$$ $$m = {2^x}$$ $${m^2} – 3m + 2 = 0$$ $$D = 9 – 8 = 1$$ $${m_{1,2}} = \frac{{3 \pm 1}}{2} = \left\langle \begin{array}{l}2\\1\end{array} \right.$$ $${2^x} = 2;\quad {2^x} = 1$$ $$x = 1;\quad {2^x} = {2^0}$$ $$\quad \quad \quad x = 0\quad \left( {бөгде \, түбір} \right)$$ Жауабы: $x=1$

Шешуі: $${2^x} – 7 = {2^{3 – x}}$$ $${2^x} – 7 = {2^3} \cdot \frac{1}{{{2^x}}}\quad \left| { \times {2^x}} \right.$$ $${2^{2x}} – 7 \cdot {2^x} = 8$$ $${2^{2x}} – 7 \cdot {2^x} – 8 = 0$$ $${2^x} = m$$ $${m^2} – 7m – 8 = 0$$ $$D = 49 + 32 = 81 = {9^2}$$ $${m_{1,2}} = \frac{{7 \pm 9}}{2} = \left\langle \begin{array}{l}8\\ – 1\end{array} \right.$$ $${2^x} = 8;\quad {2^x} \ne – 1$$ $$x = 3\quad $$

Шешуі: $${2^x} \cdot {2^2} – {2^2} \cdot \frac{1}{{{2^x}}} – 15 = 0,\quad \left| { \times {2^x}} \right.$$ $${2^{2x}} \cdot 4 – 4 – 15 \cdot {2^x} = 0.$$ $${2^x} = m$$ $$4{m^2} – 15m – 4 = 0.$$ $$D = 225 + 4 \cdot 4 \cdot 4 = 289 = {17^2}$$ $${m_{1,2}} = \frac{{15 \pm 17}}{{4 \cdot 2}} = \left\langle \begin{array}{l}4\\ – \frac{1}{4}\end{array} \right.$$ $${2^x} = 4;\quad {2^x} \ne – \frac{1}{4}$$ $$x = 2$$

Шешуі: $${2^{3x}} – {2^{2x}} – {2^x} = 0$$ $${2^x}\left( {{2^{2x}} – {2^x} – 1} \right) = 0$$ $${2^x} = 0;\quad \quad {2^{2x}} – {2^x} – 1 = 0$$ $$x \in \emptyset \quad \quad m = {2^x}$$ $${m^2} – m – 1 = 0$$ $$n = 1 + 4 = 5$$ $${m_{1,2}} = \frac{{1 \pm \sqrt 5 }}{2} = \left\langle \begin{array}{l}\frac{{1 + \sqrt 5 }}{2}\\\frac{{1 – \sqrt 5 }}{2}\end{array} \right.$$ $${2^x} = \frac{{1 + \sqrt 5 }}{2};\quad {2^x} \ne \frac{{1 – \sqrt 5 }}{2}$$ $$x = {\log _2}\left( {\frac{{1 + \sqrt 5 }}{2}} \right)$$

Шешуі: $${3^{2{x^2} – 2}} – 36 \cdot {3^{{x^2}}} \cdot \frac{1}{{27}} + 3 = 0$$ $${3^{2{x^2}}} \cdot \frac{1}{9} – \frac{4}{3} \cdot {3^{{x^2}}} + 3 = 0,\quad \left| { \times 9} \right.$$ $${3^{2{x^2}}} – 12 \cdot {3^{{x^2}}} + 27 = 0$$ $$m = {3^{{x^2}}}$$ $${m^2} – 12m + 27 = 0$$ $$D = 144 – 4 \cdot 27 = 36 = {6^2}$$ $${m_{1,2}} = \frac{{12 \pm 6}}{2} = \left\langle \begin{array}{l}9\\3\end{array} \right.$$ $${3^{{x^2}}} = 9; \quad {3^{{x^2}}} = 3$$ $${x^2} = 2; \quad {x^2} = 1$$ $$x = \pm \sqrt 2 ; \quad x = \pm 1$$

Шешуі: $${4^x} \cdot 4 + 4 \cdot \frac{1}{{{4^x}}} – 10 = 0,\quad \left| { \times {4^x}} \right.$$ $${4^{2x}} \cdot 4 + 4 – 10 \cdot {4^x} = 0$$ $$m = {4^x}$$ $${m^2} \cdot 4 – 10m + 4 = 0$$ $$4{m^2} – 10m + 4 = 0$$ $$D = 100 – 4 \cdot 4 \cdot 4 = 36 = {6^2}$$ $${m_{1,2}} = \frac{{10 \pm 6}}{{4 \cdot 2}} = \left\langle \begin{array}{l}2\\\frac{1}{2}\end{array} \right.$$ $${4^x} = 2;\quad \quad {4^x} = \frac{1}{2}$$ $$2x = 1;\quad \quad {2^{2x}} = {2^{ – 1}}$$ $$x = \frac{1}{2};\quad \quad x = – \frac{1}{2}$$

Шешуі: $$3 \cdot {9^{2 \cdot \left( {\frac{1}{x}} \right)}} – 10 \cdot {9^{\frac{1}{x}}} + 3 = 0$$ $${9^{\frac{1}{x}}} = m$$ $$3 \cdot {m^2} – 10m + 3 = 0$$ $$D = 100 – 4 \cdot 3 \cdot 3 = 64 = {8^2}$$ $${m_{1,2}} = \frac{{10 \pm 8}}{{3 \cdot 2}} = \left\langle \begin{array}{l}3\\\frac{1}{3}\end{array} \right.$$ $${9^{\frac{1}{x}}} = 3;\quad \quad {9^{\frac{1}{x}}} = \frac{1}{3}$$ $$\frac{2}{x} = 1;\quad \quad \frac{2}{x} = – 1$$ $$x = 2;\quad \quad x = – 2$$

Шешуі: $${2^x} – {\left( {\frac{1}{2}} \right)^x} = 3\frac{3}{4}$$ $${2^x} – \frac{1}{{{2^x}}} = \frac{{15}}{4},\quad \left| { \times \left( {4 \cdot {2^x}} \right)} \right.$$ $$4 \cdot {2^{2x}} – 4 = 15 \cdot {2^x}$$ $$4 \cdot {2^{2x}} – 15 \cdot {2^x} – 4 = 0$$ $${2^x} = m$$ $$4{m^2} – 15m – 4 = 0$$ $$D = 225 + 4 \cdot 4 \cdot 4 = 289 = {17^2}$$ $${m_{1,2}} = \frac{{15 \pm 17}}{{4 \cdot 2}} = \left\langle \begin{array}{l}4\\ – \frac{1}{4}\end{array} \right.$$ $${2^x} = 4;\quad {2^x} \ne – \frac{1}{4}$$ $$x = 2$$