Question

Ребят, кто шарит
С решением

Answer 1

$log_{0.3}( \frac{1}{0.09} ) = - 2$

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

$log_{6}(8) - log_{6}(2) + log_{6}(9) = log_{6}( \frac{8 \times 9}{2} ) = log_{6}(36) = 2$

$log_{2}( log_{2}( \sqrt[4]{2} ) ) = log_{2}( log_{2}( {2}^{ \frac{1}{4} } ) ) = log_{2}( \frac{1}{4} ) = - 2 \\ log_{5}(8) - log_{5}(2) + log_{5}( \frac{25}{4} ) = log_{5}( \frac{8 \times 25}{2 \times 4} ) = log_{5}(25) = 2$

$log_{35}(7) + \frac{1}{ log_{5}(35) } = log_{35}(7) + log_{35}(5) = log_{35}(7 \times 5) = log_{35}(35) = 1$

$log_{ \sqrt{7} }( \frac{1}{7} ) + {3}^{ log_{3}(7) } = log_{ \sqrt{7} }( \frac{1}{7} ) + 7 = - 2 + 7 = 5$

${ \sqrt{5} }^{ log_{5}(16) } = { ({5}^{ \frac{1}{2} }) }^{ log_{5}(16) } = { ({5}^{ log_{5}(16) } )}^{ \frac{1}{2} } = {16}^{ \frac{1}{2} } = 4$

${10}^{1 - log(5) } = {10}^{1} \div {10}^{ log(5) } = 10 \div 5 = 2$

$( log_{5}(36) + log_{5}(2) - log_{5}(8) ) log_{9}( \frac{1}{25} ) = log_{5}( \frac{36 \times 2}{8} ) log_{9}( \frac{1}{25} ) = log_{5}(9) log_{9}( \frac{1}{25} ) = log_{5}(9) log_{9}( {( 5)}^{ - 2} ) = - 2 log_{5}(9) log_{9}(5) = \frac{ - 2 log_{5}(9) }{ log_{5}(9) } = - 2$

${( \frac{1}{3} )}^{4 log_{ \frac{1}{3} }(2) } = {( ({ \frac{1}{3} )}^{ log_{ \frac{1}{3} }(2) }) }^{4} = {2}^{4} = 16$

$log_{2}( \frac{2}{3} ) + log_{4}( \frac{9}{4} ) = log_{2}( \frac{2}{3} ) + log_{ {2}^{2} }( { \frac{3}{2} )}^{2} ) = log_{2}( \frac{2}{3} ) + log_{2}( \frac{3}{2} ) = log_{2}( \frac{2}{3} \times \frac{3}{2} ) = log_{2}(1) = 0$

$\sqrt{ {25}^{ \frac{1}{ log_{6}(5) } } + {49}^{ \frac{1}{ log_{8}(7) } } } = \sqrt{ {25}^{ log_{5}(6) } + {49}^{ log_{7}(8) } } = \sqrt{ { ({5}^{2}) }^{ log_{5}(6) } + { ({7}^{2} )}^{ log_{7}(8) } } = \sqrt{ { ({5}^{ log_{5}(6) }) }^{2} + { ({7}^{ log_{7}(8) } )}^{2}} = \sqrt{ {6}^{2} + {8}^{2} } = \sqrt{36 + 64} = \sqrt{100} = 10$