Көрсеткіштік теңдеулер (Рустюмова 3.1.2)

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Теңдеулерді көбейткіштерге жіктеу әдісімен шешіңіз.

№ 1 Теңдеуді шешіңіз: ${5^{x + 1}} - {5^x} = 100$

Шешуі: $${5^x} \cdot 5 - {5^x} = 100$$ $${5^x}(5 - 1) = 100$$ $${5^x} \cdot 4 = 100$$ $${5^x} = 100:4$$ $${5^x} = 25$$ $${5^x} = {5^2}$$ $$x = 2$$

№ 2 Теңдеуді шешіңіз: $4 \cdot {3^{x + 2}} + 5 \cdot {3^x} - 7 \cdot {3^{x + 1}} = 20$

Шешуі: $${4 \cdot {3^x} \cdot {3^2} + 5 \cdot {3^x} - 7 \cdot {3^x} \cdot 3 = 20}$$ $${{3^x}(36 + 5 - 21) = 20}$$ $${{3^x} \cdot 20 = 20}$$ $${{3^x} = 20:20}$$ $${{3^x} = 1.}$$ $${{3^x} = {3^0}}$$ $${x = 0.}$$

№ 3 Теңдеуді шешіңіз: ${{3^{x + 1}} - 2 \cdot {3^{x - 2}} = 25}$

Шешуі: $${{3^x} \cdot 3 - 2 \cdot {3^x} \cdot {3^{ - 2}} = 25}$$ $${{3^x}\left( {3 - 2 \cdot \frac{1}{9}} \right) = 25}$$ $${{3^x} \cdot \frac{{25}}{9} = 25}$$ $${{3^x} = 25:\frac{{25}}{9}}$$ $${{3^x} = 9}$$ $${x = 2}$$

№ 4 Теңдеуді шешіңіз: ${{{10}^x} - {5^{x - 1}} \cdot {2^{x - 2}} = 950}$

Шешуі: $${{{10}^x} - {5^x} \cdot \frac{1}{5} \cdot {2^x} \cdot \frac{1}{4} = 950}$$ $${{{10}^x} - {{(5 \cdot 2)}^x} \cdot \frac{1}{{20}} = 950}$$ $${{{10}^x} - {{10}^x} \cdot \frac{1}{{20}} = 950}$$ $${{{10}^x}\left( {1 - \frac{1}{{20}}} \right) = 950}$$ $${{{10}^x} \cdot \frac{{19}}{{20}} = 950}$$ $${{{10}^x} = 950:\frac{{19}}{{20}}}$$ $${{{10}^x} = 1000}$$ $${{{10}^x} = {{10}^3}}$$ $${x = 3}$$

№ 5 Теңдеуді шешіңіз: ${{7^{x + 2}} - \frac{1}{7} \cdot {7^{x + 1}} - 14 \cdot {7^{x - 1}} + 2 \cdot {7^x} = 48}$

Шешуі: $${{7^x} \cdot {7^2} - \frac{1}{7} \cdot {7^x} \cdot {7^x} - 14 \cdot {7^x} \cdot \frac{1}{7} + 2 \cdot {7^x} = 48}$$ $${{7^x}(49 - 1 - 2 + 2) = 48}$$ $${{7^x} \cdot 48 = 48}$$ $${{7^x} = 48:48}$$ $${{7^x} = 1}$$ $${{7^x} = {7^0}}$$ $${x = 0}$$

№ 6 Теңдеуді шешіңіз: ${2 \cdot {{16}^x} - {2^{4x}} - {4^{2x - 2}} = 15}$

Шешуі: $${2 \cdot {{16}^x} - {{16}^x} - {{16}^x} \cdot \frac{1}{{16}} = 15}$$ $${{{16}^x}\left( {2 - 1 - \frac{1}{{16}}} \right) = 15}$$ $${{{16}^x}\left( {\frac{{15}}{{16}}} \right) = 15}$$ $${{{16}^x} = 15:\frac{{15}}{{16}}}$$ $${{{16}^x} = 16}$$ $${x = 1.}$$

№ 7 Теңдеуді шешіңіз: $5 \cdot {2^{\sqrt x }} - 3 \cdot {2^{\sqrt x - 1}} = 56$

Шешуі: $${{2^{\sqrt x }}\left( {5 - 3 \cdot \frac{1}{2}} \right) = 56}$$ $${{2^{\sqrt x }}\left( {\frac{7}{2}} \right) = 56}$$ $${{2^{\sqrt x }} = 56:\frac{7}{2}}$$ $${{2^{\sqrt x }} = 16}$$ $${{2^{\sqrt x }} = {2^4}}$$ $${\sqrt x = 4}$$ $${x = 16}$$

№ 8 Теңдеуді шешіңіз: ${{2^{x - 1}} + {2^{x - 2}} + {2^{x - 3}} = 448}$

Шешуі: $${{2^{x - 3}}\left( {{2^2} + {2^1} + 1} \right) = 448}$$ $${{2^{x - 3}} \cdot (7) = 448}$$ $${{2^{x - 3}} = 448:7}$$ $${{2^{x - 3}} = 64}$$ $${{2^{x - 3}} = {2^6}}$$ $${x - 3 = 6}$$ $${x = 9}$$

№ 9 Теңдеуді шешіңіз: ${{2^{x + 1}} + 3 \cdot {2^{x - 1}} - 5 \cdot {2^x} + 6 = 0}$

Шешуі: $${{2^x}\left( {2 + 3 \cdot \frac{1}{2} - 5} \right) + 6 = 0}$$ $${{2^x}\left( { - 3 + \frac{3}{2}} \right) = - 6}$$ $${{2^x}\left( { - \frac{3}{2}} \right) = - 6}$$ $${{2^x} = - 6:\left( { - \frac{3}{2}} \right)}$$ $${{2^x} = 4}$$ $${x = 2}$$

№ 10 Теңдеуді шешіңіз: ${{4^{x - 1}} + 11 \cdot {4^{x - 2}} = 15 \cdot {2^{ - 4}}}$

Шешуі: $${{4^{x - 2}}(4 + 11) = 15 \cdot {2^{ - 4}}}$$ $${{4^{x - 2}} \cdot 15 = 15 \cdot {2^{ - 4}}}$$ $${{2^{2x - 4}} = {2^{ - 4}}}$$ $${2x - 4 = - 4}$$ $${2x = 0}$$ $${x = 0}$$

№ 11 Теңдеуді шешіңіз: ${{3^{x - 1}} \cdot {2^{x + 1}} + {2^{x - 1}} \cdot {3^x} = \frac{7}{{36}}}$

Шешуі: $${{3^x} \cdot \frac{1}{3} \cdot {2^x} \cdot 2 + {2^x} \cdot \frac{1}{2} \cdot {3^x} = \frac{7}{{36}}}$$ $${{{(3 \cdot 2)}^x} \cdot \frac{2}{3} + {{(2 \cdot 3)}^x} \cdot \frac{1}{2} = \frac{7}{{36}}}$$ $${{6^x}\left( {\frac{2}{3} + \frac{1}{2}} \right) = \frac{7}{{36}}}$$ $${{6^x} \cdot \frac{7}{6} = \frac{7}{{36}}}$$ $${{6^x} = \frac{7}{{36}}:\frac{7}{6}}$$ $${{6^x} = \frac{1}{6}}$$ $${x = - 1.}$$

№ 12 Теңдеуді шешіңіз: ${{3^{2x - 3}} - {9^{x - 1}} + {{27}^{\frac{{2x}}{3}}} = 675}$

Шешуі: $${{3^{2x}} \cdot \frac{1}{{27}} - {3^{2x - 2}} + {3^{3 \cdot \frac{{2x}}{3}}} = 675}$$ $${{3^{2x}} \cdot \frac{1}{{27}} - {3^{2x}} \cdot \frac{1}{9} + {3^{2x}} = 675}$$ $${{3^{2x}}\left( {\frac{{{1}}}{{27}} - \frac{1}{9} + 1} \right) = 675}$$ $${{3^{2x}} \cdot \frac{{25}}{{27}} = 675}$$ $${{3^{2x}} = 675:\frac{{25}}{{27}}}$$ $${{3^{2x}} = 27 \cdot 27}$$ $${2x = 6}$$ $${x = 3}$$

№ 13 Теңдеуді шешіңіз: ${{4^{x - 1}} + {4^x} + {4^{x + 1}} = 84.}$

Шешуі: $${{4^x} \cdot \frac{1}{4} + {4^x} + {4^x} \cdot 4 = 84.}$$ $${{4^x}\left( {\frac{1}{4} + 1 + 4} \right) = 84}$$ $${{4^x}\left( {\frac{{21}}{4}} \right) = 84.}$$ $${{4^x} = 84:\frac{{21}}{4}}$$ $${{4^x} = 16}$$ $${x = 2}$$

№ 14 Теңдеуді шешіңіз: ${{5^{x + 1}} + {5^x} + {5^{x - 1}} = 155}$

Шешуі: $${{5^x} \cdot 5 + {5^x} + {5^x} \cdot \frac{1}{5} = 155}$$ $${{5^x}\left( {5 + 1 + \frac{1}{5}} \right) = 155}$$ $${{5^x} \cdot \frac{{31}}{5} = 155}$$ $${{5^x} = 155:\frac{{31}}{5}}$$ $${{5^x} = {5^2}}$$ $${x = 2}$$

№ 15 Теңдеуді шешіңіз: ${{3^{3x + 1}} - 4 \cdot {{27}^{x - 1}} + {9^{1,5x - 1}} = 80}$

Шешуі: $${{3^{3x}} \cdot 3 - 4 \cdot {3^{3x - 3}} + {3^{3x - 2}} = 80}$$ $${{3^{3x}}\left( {3 - 4 \cdot \frac{1}{{27}} + \frac{1}{9}} \right) = 80}$$ $${{3^{3x}}\left( {{{3}} - \frac{4}{{27}} + \frac{1}{9}} \right) = 80}$$ $${{3^{3x}} \cdot \frac{{80}}{{27}} = 80}$$ $${{3^{3x}} = 80:\frac{{80}}{{27}}}$$ $${{3^{3x}} = 27}$$ $${3x = 3}$$ $${x = 1}$$

№ 16 Теңдеуді шешіңіз: ${5^{x + 6}} - {3^{x + 7}} = 43 \cdot {5^{x + 4}} - 19 \cdot {3^{x + 5}}$

Шешуі: $${5^{x + 4}} \cdot {5^2} - {3^{x + 5}} \cdot {3^2} = 43 \cdot {5^{x + 4}} - 19 \cdot {3^{x + 5}}$$ $$19 \cdot {3^{x + 5}} - 9 \cdot {3^{x + 5}} = 43 \cdot {5^{x + 4}} - 25 \cdot {5^{x + 4}}$$ $$10 \cdot {3^{x + 4}} \cdot 3 = 18 \cdot {5^{x + 4}}$$ $$30 \cdot {3^{x + 4}} = 18 \cdot {5^{x + 4}}\quad | \div {3^{x + 4}}$$ $${30 = 18 \cdot \frac{{{5^{x + 4}}}}{{{3^{x + 4}}}}}$$ $${\frac{{30}}{{18}} = {{\left( {\frac{5}{3}} \right)}^{x + 4}}}$$ $${\frac{5}{3} = {{\left( {\frac{5}{3}} \right)}^{x + 4}}}$$ $${1 = x + 4}$$ $${x = - 3}$$

№ 17 Теңдеуді шешіңіз: ${{2^{{x^2} - 1}} - {3^{{x^2}}} = {3^{{x^2} - 1}} - {2^{{x^2} + 2}}}$

Шешуі: $${{2^{{x^2}}} \cdot \frac{1}{2} + {2^{{x^2}}} \cdot 4 = {3^{{x^2}}} \cdot \frac{1}{3} + {3^{{x^2}}}}$$ $${{2^{{x^2}}}\left( {\frac{1}{2} + 4} \right) = {3^{{x^2}}}\left( {\frac{1}{3} + 1} \right)}$$ $${2^{{x^2}}} \cdot \frac{9}{2} = {3^{{x^2}}} \cdot \frac{4}{3}\quad \left| { \div \left( {{2^{{x^2}}} \cdot \frac{4}{3}} \right)} \right.$$ $${\frac{9}{2}:\frac{4}{3} = \frac{{{3^{{x^2}}}}}{{{2^{{x^2}}}}}}$$ $${\frac{{27}}{8} = {{\left( {\frac{3}{2}} \right)}^{{x^2}}}}$$ $${{{\left( {\frac{3}{2}} \right)}^3} = {{\left( {\frac{3}{2}} \right)}^{{x^2}}}}$$ $${{x^2} = 3}$$ $${x = \pm \sqrt 3 }$$

№ 18 Теңдеуді шешіңіз: ${{5^{2x}} - {7^x} - {5^{2x}} \cdot 35 + {7^x} \cdot 35 = 0}$

Шешуі:$${{5^{2x}}(1 - 35) - {7^x}(1 - 35) = 0}$$ $${{5^{2x}} \cdot ( - 34) + 34 \cdot {7^x} = 0}$$ $${34 \cdot {7^x} = 34 \cdot {{25}^x}}$$ $${{7^x} = {{25}^x}\quad | \div {7^x}}$$ $${1 = {{\left( {\frac{{25}}{7}} \right)}^x}}$$ $${{{\left( {\frac{{25}}{7}} \right)}^0} = {{\left( {\frac{{25}}{7}} \right)}^x}}$$ $${x = 0}$$

№ 19 Теңдеуді шешіңіз: $${{x^2} \cdot {2^x} + {2^x} = x \cdot {2^{x + 1,5}}}$$

Шешуі: $${{x^2} \cdot {2^x} + {2^x} - x \cdot {2^x} \cdot {2^{1,5}} = 0}$$ $${{2^x}\left( {{x^2} + 1 - x \cdot {2^{1,5}}} \right) = 0.}$$ $${{2^x} = 0.}$$ $${x \in \emptyset }$$ $${x^2} + 1 - x\sqrt {{2^3}} = 0$$ $${x^2} - 2\sqrt 2 x + 1 = 0$$ $${D = {{(2\sqrt 2 )}^2} - 4 = 4}$$ $${{x_{1,2}} = \frac{{2\sqrt 2 \pm 2}}{2} = \sqrt 2 \pm 1}$$

№ 20 Теңдеуді шешіңіз: ${{3^{x + 3}} + {3^3} \cdot {3^{x - 1}} - 3 = 0}$

Шешуі: $${{3^x} \cdot {3^3} + {3^3} \cdot {3^x} \cdot \frac{1}{3} = 3}$$ $${{3^x}(27 + 9) = 3}$$ $${{3^x} \cdot 36 = 3}$$ $${{3^x} = 3:36}$$ $${3^x} = \frac{1}{{12}}$$ $$x = {\log _3}\frac{1}{{12}}$$ $${x = - {{\log }_3}12}$$ $${x = - \left( {{{\log }_3}3 + {{\log }_3}4} \right)}$$ $${x = - 1 - {{\log }_3}2}$$

№ 21 Теңдеуді шешіңіз: ${{9^x} - {2^{x + 0,5}} = {2^{x + 3,5}} - {3^{2x - 1}}}$

Шешуі: $${{9^x} + {3^{2x - 1}} = {2^{x + 3,5}} + {2^{x + 0,5}}}$$ $${{3^{2x}} + {3^{2x}} \cdot \frac{1}{3} = {2^x}\left( {{2^{3,5}} + {2^{0,5}}} \right){\rm{. }}}$$ $${{3^{2x}}\left( {1 + \frac{1}{3}} \right) = {2^x}\left( {{2^{\frac{7}{2}}} + {2^{\frac{1}{2}}}} \right)}$$ $${{3^{2x}}\left( {\frac{4}{3}} \right) = {2^x}\left( {\sqrt {{2^7}} + \sqrt 2 } \right)}$$ $${{9^x} \cdot \frac{4}{3} = {2^x} \cdot \left( {{2^{\sqrt[3]{2}}} + \sqrt 2 } \right)}$$ $${9^x} \cdot \frac{4}{3} = {2^x} \cdot 9\sqrt 2 \quad \left| { \div \left( {{2^x} \cdot \frac{4}{3}} \right)} \right.$$ $${\frac{{{9^x}}}{{{2^x}}} = \frac{{9\sqrt 2 }}{{\frac{4}{3}}}}$$ $${{{\left( {\frac{9}{2}} \right)}^x} = \frac{{27}}{{2\sqrt 2 }}}$$ $${{{\left( {\frac{9}{2}} \right)}^x} = \frac{{{9^{\frac{3}{2}}}}}{{{2^{\frac{3}{2}}}}}}$$ $${{{\left( {\frac{9}{2}} \right)}^x} = {{\left( {\frac{9}{2}} \right)}^{\frac{3}{2}}}}$$ $${x = \frac{3}{2}}$$

№ 22 Теңдеуді шешіңіз: ${{3^x} + {3^{x + 1}} + {3^{x + 2}} = {5^x} + {5^{x + 1}} + {5^{x + 2}}}$

Шешуі: $${{3^x} + {3^x} \cdot 3 + {3^x} \cdot 9 = {5^x} + {5^x} \cdot 5 + {5^x} \cdot 25}$$ $${{3^x}(1 + 3 + 9) = {5^x}(1 + 5 + 25)}$$ $${{3^x} \cdot 13 = {5^x} \cdot 31\quad \quad | \div \left( {{3^x} \cdot 31} \right)}$$ $${\frac{{13}}{{31}} = \frac{{{5^x}}}{{{3^x}}}}$$ $${\frac{{13}}{{31}} = {{\left( {\frac{5}{3}} \right)}^x}}$$ $${x = {{\log }_{\frac{5}{3}}}\frac{{13}}{{31}}}$$ $${x = {{\log }_{\frac{3}{5}}}\frac{{31}}{{13}}}$$

№ 23 Теңдеуді шешіңіз: ${{{\left( {\frac{1}{3}} \right)}^{2 - x}} - 6 \cdot {9^{\frac{{x - 4}}{2}}} + 2 \cdot {3^{x - 6}} = 29}$

Шешуі: $${{{\left( {\frac{1}{3}} \right)}^2} \cdot {3^x} - 6 \cdot {3^{x - 4}} + 2 \cdot {3^x} \cdot \frac{1}{{{3^6}}} = 29}$$ $${{3^x}\left( {\frac{1}{9} - \frac{6}{{{3^4}}} + \frac{2}{{{3^6}}}} \right) = 29}$$ $${{3^x}\left( {\frac{{81 - 54 + 2}}{{729}}} \right) = 29}$$ $${{3^x} \cdot \frac{{29}}{{729}} = 29}$$ $${{3^x} = 29:\frac{{29}}{{729}}}$$ $${{3^x} = 729}$$ $${x = 6}$$

№ 24 Теңдеуді шешіңіз: ${\left( {\frac{1}{3}} \right)^{2 - x}} + {3^{x - 3}} = 99 + \sqrt {{{\left( {\frac{1}{9}} \right)}^{4 - x}}} $

Шешуі: $${{3^{x - 2}} + {3^{x - 3}} = 99 + \sqrt {{3^{ - 8 + 2x}}} }$$ $${{3^{x - 2}} + {3^{x - 3}} - {3^{\frac{{ - 8 + 2x}}{2}}} = 99}$$ $${{3^{x - 2}} + {3^{x - 3}} + {3^{x - 4}} = 99}$$ $${{3^{x - 4}}\left( {{3^2} + {3^1} + 1} \right) = 99}$$ $${{3^{x - 4}} \cdot 11 = 99}$$ $${{3^{x - 4}} = 9}$$ $${x - 4 = 2}$$ $$x = 6.$$

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