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August 2021 1 13 Report

Вычислите:
1. [tex](3lg2+lg0,25):(lg14-lg7)[/tex]
2. [tex]( log_{2}12- log_{2}3+ 3^{log _{3}8 } ) ^{lg5} [/tex]
3. [tex]( log_{6}2+ log_{6}3+ 2^{ log_{2}4 }) ^{ log_{5}7 } [/tex]
4. [tex] \frac{ log_{3}63 }{2+ log_{3}7 } [/tex]

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arsenlevadniy

Verified answer

$(3\lg2+\lg0,25):(\lg14-\lg7) = (\lg2^3+\lg0,25):\lg\frac{14}{7} =\\= \lg(8\cdot0,25):\lg2 = \lg2:\lg2 = 1;$

$(\log_{2}12- \log_{2}3+ 3^{\log _{3}8 } )^{\lg5} = (\log_{2}\frac{12}{3} + 8)^{\lg5} =\\= (\log_{2}4+ 8)^{\lg5} = (2\log_{2}2+ 8)^{\lg5} = (2+ 8)^{\lg5} = 10^{\lg5} = 5;$

$( \log_{6}2+ \log_{6}3+ 2^{ \log_{2}4 }) ^{ \log_{5}7 } =( \log_{6}(2\cdot3)+4) ^{\log_{5}7} =(1+4) ^{\log_{5}7} =\\= 5^{\log_{5}7} = 7;$

$\frac{\log_{3}63 }{2+\log_{3}7} = \frac{\log_{3}63 }{2\log_{3}3+\log_{3}7} = \frac{\log_{3}63 }{\log_{3}3^2+\log_{3}7} = \frac{\log_{3}63 }{\log_{3}(9\cdot7)} = \frac{\log_{3}63 }{\log_{3}63} = 1.$

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