Question

Логорифмы logloglogl

Answer 1

Ответ:

a)

$log_{ \frac{1}{3} }(1) = 0 \\ log_{2}( \sqrt[4]{2} ) = log_{2}( {2}^{ \frac{1}{4} } ) = \frac{1}{4} = 0.25$

$log_{3}( \frac{1}{3 \sqrt{3} } ) = log_{3}( {3}^{ - \frac{3}{2} } ) = - \frac{3}{2} = - 1.5$

$log_{7}( \frac{ \sqrt[3]{7} }{49} ) = log_{7}( \frac{ {7}^{ \frac{1}{3} } }{ {7}^{2} } ) = log_{7}( {7}^{ \frac{1}{3} - 2} ) = \frac{1}{3} - 2 = - 1 \frac{2}{3}$

b)

${7}^{ \frac{1}{2} log_{7}(9) } = {7}^{ log_{7}( {9}^{ \frac{1}{2} } ) } = \sqrt{9} = 3$

${( \frac{1}{9}) }^{ \frac{1}{2} log_{3}(4) } = {3}^{ - 2 \times \frac{1}{2} log_{3}(4) } = {3}^{ log_{3}( {4}^{ - 1} ) } = \frac{1}{4} = 0.25$

${( \frac{1}{6}) }^{1 + 2 log_{ \frac{1}{6} }(3) } = \frac{1}{6} \times {( \frac{1}{6} )}^{ log_{ \frac{1}{6} }( {3}^{2} ) } = \frac{1}{6} \times {3}^{2} = \frac{9}{6} = \frac{3}{2} = 1.5$

c)

$log_{2}( log_{4}(16) ) + log_{ \frac{1}{2} }(2) = log_{2}(2) - 1 = 1 - 1 = 0$

$log_{4}(12) - log_{4}(15) + log_{4}(20) = log_{4}( \frac{12 \times 20}{15} ) = log_{4}(4 \times 4) = 2$

$\frac{ log_{5}(36) - log_{5}(12) }{ log_{5}(9) } = \frac{ log_{5}( \frac{36}{12} ) }{ log_{5}(9) } = \frac{ log_{5}(3) }{ log_{5}(9) } = log_{9}(3) = \frac{1}{2} = 0.5$