Question

математика логарифм на фото

Answer 1

${( \sqrt{2}) }^{ \log_{4}(49)+7 } = {( \sqrt{2}) }^{\log_4 49} \cdot {( \sqrt{2}) }^{7} = \\ = {( \sqrt{2}) }^{\log_{2^2}{7^2}} \cdot {( \sqrt{2}) }^{6} \cdot \sqrt{2} =( {2}^{ \frac{1}{2} } )^{\log_2 7}\cdot {( {2}^{ \frac{1}{2} } )}^{6} \cdot \sqrt{2} = \\ = {2}^{ \frac{1}{2} \log_{2}7 } \cdot {2}^{ \frac{1}{2}\cdot6} \cdot \sqrt{2} = {2}^{\log_{2}{7}^{ \frac{1}{2} } }\cdot {2}^{3} \cdot \sqrt{2} = \\ = {2}^{\log_{2} \sqrt{7} }\cdot 8\cdot \sqrt{2} = \sqrt{7} \cdot 8\cdot \sqrt{2} = 8 \sqrt{7\cdot2} = 8 \sqrt{14}$

Answer 2

$\Big (\sqrt2\Big )^{log_449+7}=\Big (2^{\frac{1}{2}}\Big )^{log_{2^2}7^2}\cdot \Big (2^{\frac{1}{2}}\Big )^7=2^{\frac{1}{2}\cdot log_27}\cdot 2^{\frac{7}{2}}=\\\\=2^{log_2\sqrt7}\cdot \sqrt{2^7}=\sqrt7\cdot \sqrt{2^6\cdot 2}=\sqrt7\cdot 2^3\sqrt2=8\sqrt{7\cdot 2}=8\sqrt{14}$