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August 2022 1 17 Report

Вычислите:
1. [tex] log_{0,2}125: log_{16}64* log_{3}81 [/tex]
2. [tex] log_{ \frac{1}{2} }16* log_{5} \frac{1}{25} : 9^{ log_{3} 2} [/tex]
3. [tex] log_{ \frac{1}{3} }9* log_{2} \frac{1}{8} : 7^{ 2log_{49} 2} [/tex]
4. [tex](3 log_{7} 2-log _{7} 24):( log_{7}3+ log_{7}9) [/tex]

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arsenlevadniy

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$\log_{0,2}125: \log_{16}64\cdot\log_{3}81 = \log_{5^{-1}}5^3: \log_{2^4}2^6\cdot\log_{3}3^4 =\\= -3\log_{5}5: \frac{6}{4} \log_{2}2\cdot4\log_{3}3 = -3\cdot\frac{2}{3}\cdot4=-8;$

$\log_{\frac{1}{2} }16\cdot\log_{5}\frac{1}{25}:9^{\log_{3}2} = \log_{2^{-1}}2^4\cdot\log_{5}5^{-2}:(3^2)^{ log_{3}2} =\\= -4\log_{2}2\cdot(-2\log_{5}5):(3^{ log_{3}2})^2 = -4\cdot(-2):2^2=2;$

$\log_{\frac{1}{3} }9\cdot\log_{2}\frac{1}{8}: 7^{2\log_{49}2} = \log_{3^{-1}}3^2\cdot\log_{2}2^{-3}: 49^{\log_{49}2} =\\= -2\log_{3}3\cdot(-3\log_{2}2): 2= -2\cdot(-3): 2= 3;$

$(3\log_{7} 2-\log _{7} 24):(\log_{7}3+\ log_{7}9) = (\log_{7} 2^3-\log _{7} 24):\log_{7}(3\cdot9) =\\= \log_{7} \frac{8}{24}:\log_{7}27 = \log_{7} 3^{-1}:\log_{7}3^3 = -\log_{7} 3:(3\log_{7}3)=- \frac{1}{3} .$

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