Логарифмдік өрнектер (Сканави (А) 7.1-7.15)

 +/-  - Есептің жауабын көрсету/көрсетпеу.

▲/▼ - Жауап орнын жасыру/шығару

   ×    - Сұрақты алып тастау.

Логарифмдік өрнектерді ықшамдаңыз.

№ 7.1 Есептеңіз: $\sqrt {{{25}^{\frac{1}{{{{\log }_6}5}}}} + {{49}^{\frac{1}{{{{\log }_8}7}}}}} $

Шешуі: $${\sqrt {{{25}^{\frac{1}{{{{\log }_6}5}}}} + {{49}^{\frac{1}{{{{\log }_8}7}}}}} = \sqrt {{5^{2{{\log }_5}6}} + {7^{2{{\log }_7}8}}} = \sqrt {{5^{{{\log }_5}{6^2}}} + {7^{{{\log }_7}{8^2}}}} = \sqrt {{6^2} + {8^2}} = 10}$$

№ 7.2 Есептеңіз: ${81^{\frac{1}{{{{\log }_5}3}}}} + {27^{{{\log }_9}36}} + {3^{\frac{4}{{{{\log }_7}9}}}}$

Шешуі: $${{{81}^{\frac{1}{{{{\log }_5}3}}}} + {{27}^{{{\log }_9}36}} + {3^{\frac{4}{{{{\log }_7}9}}}} = {3^{4{{\log }_3}5}} + {3^{\frac{3}{2}{{\log }_3}36}} + {3^{\frac{4}{2}{{\log }_3}7}} = {5^4} + {{36}^{\frac{3}{2}}} + 49 = 625 + 216 + 49 = 890}$$

№ 7.3 Есептеңіз: $ - {\log _2}{\log _2}\sqrt {\sqrt[4]{2}} $

Шешуі: $$ - {\log _2}{\log _2}\sqrt {\sqrt[4]{2}} = - {\log _2}{\log _2}{2^{\frac{1}{8}}} = - {\log _2}\frac{1}{8}{\log _2}2 = - {\log _2}{2^{ - 3}} = 3.$$

№ 7.4 Есептеңіз: $ - {\log _3}{\log _3}\sqrt[3]{{\sqrt[3]{3}}}$

Шешуі: $$ - {\log _3}{\log _3}\sqrt[3]{{\sqrt[3]{3}}} = - {\log _3}{\log _3}{3^{\frac{1}{9}}} = - {\log _3}\frac{1}{9}{\log _3}3 = - {\log _3}{3^{ - 2}} = 2$$

№ 7.5 Есептеңіз: ${\dfrac{{\left( {{{27}^{\frac{1}{{{{\log }_2}3}}}} + {5^{{{\log }_{25}}49}}} \right)\left( {{{81}^{\frac{1}{{{{\log }_4}9}}}} - {8^{{{\log }_4}9}}} \right)}}{{3 + {5^{\frac{1}{{{{\log }_{16}}25}}}} \cdot {5^{{{\log }_5}3}}}}}$

Шешуі: $$ = \frac{{\left( {{{\left( {{3^3}} \right)}^{{{\log }_3}2}} + {5^{{{\log }_5}{{27}^2}}}} \right)\left( {{{\left( {{9^2}} \right)}^{{{\log }_9}4}} - {{\left( {{2^3}} \right)}^{{{\log }_2}{{23}^2}}}} \right)}}{{3 + {5^{{{\log }_5}{{24}^2}}} \cdot 3}} = $$ $$ = \frac{{\left( {{3^{{{\log }_3}{2^3}}} + {5^{{{\log }_5}7}}} \right)\left( {{9^{{{\log }_9}{4^2}}} - {2^{{{\log }_2}{3^3}}}} \right)}}{{3 + {5^{{{\log }_5}4}} \cdot 3}} = \frac{{\left( {{2^3} + 7} \right)\left( {{4^2} - {3^3}} \right)}}{{3 + 4 \cdot 3}} = $$ $$ = \frac{{15 \cdot ( - 11)}}{{15}} = - 11.$$

№ 7.6 Есептеңіз: ${36^{{{\log }_6}5}} + {10^{1 - \lg 2}} - {3^{{{\log }_9}36}}$

Шешуі: $${36^{{{\log }_6}5}} + {10^{1 - \lg 2}} - {3^{{{\log }_9}36}} = {6^{2{{\log }_6}5}} + \frac{{10}}{{{{10}^{\lg 2}}}} - {3^{{{\log }_3}{{26}^2}}} = $$ $$ = {6^{{{\log }_6}{5^2}}} + \frac{{10}}{2} - {3^{{{\log }_3}6}} = {5^2} + 5 - 6 = 24$$

№ 7.7 Есептеңіз: $\left( {{{81}^{\frac{1}{4} - \frac{1}{2}{{\log }_9}4}} + {{25}^{{{\log }_{125}}8}}} \right) \cdot {49^{{{\log }_7}2}}$

Шешуі: $${\left( {{{81}^{\frac{1}{4} - \frac{1}{2}{{\log }_9}4}} + {{25}^{{{\log }_{125}}8}}} \right) \cdot {{49}^{{{\log }_7}2}} = \left( {\frac{{{{81}^{\frac{1}{4}}}}}{{{{\left( {{9^2}} \right)}^{\frac{1}{2}}}{{\log }_9}4}} + {5^{2{{\log }_5}{{32}^3}}}} \right) \cdot {7^{2{{\log }_7}2}} = }$$ $${ = \left( {\frac{3}{4} + 4} \right) \cdot 4 = 19.}$$

№ 7.8 Есептеңіз: ${\dfrac{{{{81}^{\frac{1}{{{{\log }_5}9}}}} + {3^{\frac{3}{{{{\log }_{\sqrt 6 }}3}}}}}}{{409}} \cdot \left( {{{(\sqrt 7 )}^{\frac{2}{{{{\log }_{25}}7}}}} - {{125}^{{{\log }_{25}}6}}} \right)}$

Шешуі: $${ = \frac{{{9^{2{{\log }_9}5}} + {3^{3{{\log }_3}\sqrt 6 }}}}{{409}} \cdot \left( {{{\left( {{7^{\frac{1}{2}}}} \right)}^{2{{\log }_7}25}} - {5^{3{{\log }_5}26}}} \right) = \frac{{{9^{{{\log }_9}{5^2}}} + {3^{{{\log }_3}{{(\sqrt 6 )}^3}}}}}{{409}} \times }$$ $${ \times \left( {{7^{{{\log }_7}25}} - {5^{{{\log }_5}{6^{\frac{3}{2}}}}}} \right) = \frac{{\left( {25 + {6^{\frac{3}{2}}}} \right)\left( {25 - {6^{\frac{3}{2}}}} \right)}}{{409}} = \frac{{625 - 216}}{{409}} = 1}$$

№ 7.9 Есептеңіз: ${\left( {{N^{\frac{1}{{{{\log }_2}N}}}} \cdot {N^{\frac{1}{{{{\log }_4}N}}}} \cdot {N^{\frac{1}{{{{\log }_8}N}}}} \cdots {N^{\frac{1}{{{{\log }_{512}}N}}}}} \right)^{\frac{1}{{15}}}}$

Шешуі: $${ = {{\left( {{N^{{{\log }_N}2}} \cdot {N^{{{\log }_N}4}} \cdot {N^{{{\log }_N}8}} \ldots {N^{{{\log }_N}512}}} \right)}^{\frac{1}{{15}}}} = }$$ $${ = {{(2 \cdot 4 \cdot 8 \cdots \cdot 512)}^{\frac{1}{{15}}}} = {{\left( {{2^1} \cdot {2^2} \cdot {2^3} \cdots {2^9}} \right)}^{\frac{1}{{15}}}} = {{\left( {{2^{1 + 2 + 3 + \ldots + 9}}} \right)}^{\frac{1}{{15}}}}}$$ $$S_n=1+2+3+…+9=45,$$ $${\left( {{2^{45}}} \right)^{\frac{1}{{15}}}} = {2^3} = 8$$

№ 7.10 Есептеңіз: ${\left( {{2^{{{\log }_{\sqrt[4]{2}}}a}} - {3^{{{\log }_{27}}{{\left( {{a^2} + 1} \right)}^3}}} - 2a} \right):\left( {{7^{4{{\log }_{49}}a}} - {5^{0,5{{\log }_{\sqrt 5 }}a}} - 1} \right) = }$

Шешуі: $${ = \left( {{2^{{{\log }_2}{a^4}}} - {3^{{{\log }_3}\left( {{a^2} + 1} \right)}} - 2a} \right):\left( {{7^{{{\log }_7}{a^2}}} - {5^{{{\log }_5}a}} - 1} \right) = }$$ $${ = \left( {{a^4} - \left( {{a^2} + 1} \right) - 2a} \right):\left( {{a^2} - a - 1} \right) = \frac{{{a^4} - {a^2} - 2a - 1}}{{{a^2} - a - 1}} = }$$ $${ = \frac{{\left( {{a^2} - a - 1} \right.}}{{{a^2} - a - 1}} \cdot \left( {{a^2} + a + 1} \right) = {a^2} + a + 1}$$

№ 7.11 Есептеңіз: $\dfrac{{{{\log }_a}\sqrt {{a^2} - 1} \cdot \log _{1/a}^2\sqrt {{a^2} - 1} }}{{{{\log }_{{a^2}}}\left( {{a^2} - 1} \right) \cdot {{\log }_{\sqrt[3]{a}}}\sqrt[6]{{{a^2} - 1}}}}$

Шешуі: $$ = \frac{{\frac{1}{2}{{\log }_a}\left( {{a^2} - 1} \right) \cdot \frac{1}{4}\log _a^2\left( {{a^2} - 1} \right)}}{{\frac{1}{2}{{\log }_a}\left( {{a^2} - 1} \right) \cdot \frac{1}{2}{{\log }_a}\left( {{a^2} - 1} \right)}} = $$ $$ = \frac{1}{2}{\log _a}\left( {{a^2} - 1} \right) = {\log _a}\sqrt {{a^2} - 1} $$

№ 7.12 Есептеңіз: ${{a^{\frac{2}{{{{\log }_h}a}} + 1}} \cdot b - 2{a^{{{\log }_a}b + 1}} \cdot {b^{{{\log }_b}a + 1}} + a{b^{\frac{2}{{{{\log }_a}b}} + 1}}}$

Шешуі: $$ = a \cdot {a^{2{{\log }_a}b}} \cdot b - 2a \cdot {a^{{{\log }_a}b}} \cdot b \cdot {b^{{{\log }_b}a}} + a \cdot b \cdot {b^{2{{\log }_h}a}} = $$ $${ = a \cdot {a^{{{\log }_a}{b^2}}} \cdot b - 2a \cdot b \cdot b \cdot a + a \cdot b \cdot {b^{{{\log }_h}{a^2}}} = a{b^2}b - 2{a^2}{b^2} + ab{a^2} = }$$ $${ = a{b^3} - 2{a^2}{b^2} + ab{a^2} = a{b^3} - 2{a^2}{b^2} + {a^3}b = }$$ $${ = ab\left( {{b^2} - 2ab + {a^2}} \right) = ab{{(b - a)}^2} = ab{{(a - b)}^2}}$$

№ 7.13 Есептеңіз: ${\dfrac{{\left( {{{25}^{\frac{1}{{2{{\log }_9}25}}}} + 2{{\log }_2}{{\log }_2}{{\log }_2}{a^{2{{\log }_a}4}}} \right) \cdot {4^{\frac{2}{{{{\log }_3}4}}}} - {a^2}}}{{1 - a}}}$

Шешуі: $${ = \frac{{\left( {\left( {{{25}^{{{\log }_{25}}49}}\frac{1}{2} + 2{{\log }_2}{{\log }_2}4} \right) \cdot {{\left( {{4^{2{{\log }_4}3}}} \right)}^{ - 1}} - {a^2}} \right.}}{{1 - a}} = }$$ $${ = \frac{{\left( {(49)\frac{1}{2} + 2{{\log }_2}2} \right) \cdot {9^{ - 1}} - {a^2}}}{{1 - a}} = \frac{{(7 + 2) \cdot \frac{1}{9} - {a^2}}}{{1 - a}} = \frac{{1 - {a^2}}}{{1 - a}} = 1 + a.}$$

№ 7.14 Есептеңіз: ${\left( {{{\log }_a}b + {{\log }_b}a + 2} \right)\left( {{{\log }_a}b - {{\log }_{ab}}b} \right){{\log }_b}a - 1}$

Шешуі: $$ = \left( {{{\log }_a}b + \frac{1}{{{{\log }_a}b}} + 2} \right) \times \left( {{{\log }_a}b - \frac{{{{\log }_a}b}}{{{{\log }_a}ab}}} \right)\frac{1}{{{{\log }_a}b}} - 1$$ $${ = \frac{{\log _a^2b + 2{{\log }_a}b + 1}}{{{{\log }_a}b}} \times \left( {{{\log }_a}b - \frac{{{{\log }_a}b}}{{{{\log }_a}a + {{\log }_a}b}}} \right) \cdot \frac{1}{{{{\log }_a}b}} - 1 = }$$ $${ = \frac{{{{\left( {{{\log }_a}b + 1} \right)}^2}}}{{{{\log }_a}b}} \cdot \left( {{{\log }_a}b - \frac{{{{\log }_a}b}}{{1 + {{\log }_a}b}}} \right)\frac{1}{{{{\log }_a}b}} - 1 = }$$ $${ = \frac{{{{\left( {{{\log }_a}b + 1} \right)}^2}}}{{{{\log }_a}b}}{{\log }_a}b\left( {1 - \frac{1}{{1 + {{\log }_a}b}}} \right)\frac{1}{{{{\log }_a}b}} - 1 = }$$ $${ = \frac{{{{\left( {{{\log }_a}b + 1} \right)}^2}\left( {1 + {{\log }_a}b - 1} \right)}}{{\left( {1 + {{\log }_a}b} \right){{\log }_a}b}} - 1 = {{\log }_a}b + 1 - 1 = {{\log }_a}b}$$

№ 7.15 Есептеңіз: ${\dfrac{{1 - \log _a^3b}}{{\left( {{{\log }_a}b + {{\log }_b}a + 1} \right) \cdot {{\log }_a}\frac{a}{b}}}}$

Шешуі: $$ = \frac{{\left( {1 - {{\log }_a}b} \right)\left( {1 + {{\log }_a}b + \log _a^2b} \right)}}{{\left( {{{\log }_a}b + \frac{1}{{{{\log }_a}b}} + 1} \right)\left( {{{\log }_a}a - {{\log }_a}b} \right)}} = $$ $${ = \frac{{\left( {1 - {{\log }_a}b} \right)\left( {1 + {{\log }_a}b + \log _a^2b} \right){{\log }_a}b}}{{\left( {\log _a^2b + 1 + {{\log }_a}b} \right)\left( {1 - {{\log }_a}b} \right)}} = {{\log }_a}b.}$$

 
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