Алгебра: Найдите значения логарифмических выражений.

Найдите значения логарифмических выражений.

Ответ:

fcnhff7

25.12.2020 19:46

$2)100^{2-\frac{1}{2}lg12} =(10^{2})^{2-\frac{1}{2} lg12}=10^{4-lg12}=10^{4}*10^{-lg12}=\\\\=10000*(10^{lg12})^{-1}=10000*12^{-1}=10000*\frac{1}{12}=833\frac{1}{3} \\\\\\3)7^{\frac{3}{log_{8}7}} =(7^{\frac{1}{log_{8}7}})^{3} =(7^{log_{7}8 })^{3}=8^{3}=\boxed{512}$

$4)log_{\frac{1}{6}}-log_{\frac{1}{6}}=log_{\frac{1}{6}}\frac{3}{18}=log_{\frac{1}{6}}\frac{1}{6}=\boxed1\\\\\\5)\frac{log_{5}12-2log_{5} 2}{log_{5}18+log_{5}0,5}=\frac{log_{5}12-log_{5} 2^{2} }{log_{5}(18*0,5)}=\frac{log_{5}12-log_{5}4}{log_{5}9}=\frac{log_{5}(\frac{12}{4})}{log_{5}3^{2}} =\frac{log_{5}3 }{2log_{5}3}=\frac{1}{2}=\boxed{0,5}$

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Ответ:

KitKat321456

25.12.2020 19:46

$2) {100}^{2 - \frac{1}{2}lg(12) } = {100}^{2} \times {100}^{ - \frac{1}{2} log_{10}(12) } = {100}^{2} \times {10}^{ - 2 \times \frac{1}{2} log_{10}(12) } = {100}^{2} \times {10}^{ log_{10}( {12}^{ - 1} ) } = {100}^{2} \times \frac{1}{12} = \frac{10000}{12} = \frac{2500}{3}$

$3) {7}^{ \frac{3}{ log_{8}(7) } } = {7}^{3 log_{7}(8) } = {7}^{ log_{7}( {8}^{3} ) } = {8}^{3} = 512$

$4) log_{ \frac{1}{6} }(3) - log_{ \frac{1}{6} }(18) = log_{ \frac{1}{6} }( \frac{3}{18} ) = log_{ \frac{1}{6} }( \frac{1}{6} ) = 1$

$5) \frac{ log_{5}(12) - 2 log_{2}(5) }{ log_{5}(18) + log_{5}(0.5) } = \frac{ log_{5}(12) - log_{5}( {2}^{2} ) }{ log_{5}(18 \times 0.5) } = \frac{ log_{5}( \frac{12}{4} ) }{ log_{5}(9) } = log_{9}( 3 ) = \frac{1}{2}$