Question

алгебра пожалуйста))))))​

Answer 1

$1) {4}^{ log_{2}(9) } = {2}^{2 log_{2}(9) } = {2}^{ log_{2}( {9}^{2} ) } = {9}^{2} = 81$

$3) {10}^{2 + lg8} = {10}^{2} \times {10}^{ log_{10}(8) } = 100 \times 8 = 800$

$5) {2}^{4 log_{2}(3) - 1} = {2}^{4 log_{2}(3) } \times {2}^{ - 1} = {2}^{4log_{2}(3) } \times \frac{1}{2} = \frac{ {2}^{ log_{2}( {3}^{4} ) } }{2} = \frac{81}{2} = 40.5$

$7) {8}^{1 - \frac{1}{3} log_{2}(12) } = 8 \times {2}^{ - 3 \times \frac{1}{3} log_{2}(12) } = 8 \times {2}^{ log_{2}( {12}^{ - 1} ) } = 8 \times \frac{1}{12} = \frac{8}{12} = \frac{2}{?3}$

$2) {( \frac{1}{9}) }^{ - 2 log_{3}(12) } = {3}^{ - 2 \times ( - 2 log_{3}(12)) } = {3}^{4 log_{3}(12) } = {3}^{ log_{3}( {12}^{4} ) } = {12}^{4} = 20736$

$4) {( \frac{1}{2} )}^{ log_{ \frac{1}{2} }(6) - 3} = { \frac{1}{2} }^{ log_{ \ \frac{1}{2} }(6) } \times { \frac{1}{2} }^{ - 3} = 6 \times 8 = 48$

$6) {( \frac{1}{5}) }^{ log_{25}(9) + 2} = {5}^{ - log_{25}(9) } \times {( \frac{1}{5} )}^{2} = {5}^{ - log_{ {5}^{2} }(9) } \times \frac{1}{25} = {5}^{ - \frac{1}{2} log_{5}(9) } \times \frac{1}{25} = {5}^{ log_{5}( {9}^{ - \frac{1}{2} } ) } \times \frac{1}{25} = \frac{1}{3} \times \frac{1}{25} = \frac{1}{75}$