Question

математика логарифм на фото

Answer 1

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

Answer 2

$Big (sqrt2Big )^{log_449+7}=Big (2^{frac{1}{2}}Big )^{log_{2^2}7^2}cdot Big (2^{frac{1}{2}}Big )^7=2^{frac{1}{2}cdot log_27}cdot 2^{frac{7}{2}}=\\=2^{log_2sqrt7}cdot sqrt{2^7}=sqrt7cdot sqrt{2^6cdot 2}=sqrt7cdot 2^3sqrt2=8sqrt{7cdot 2}=8sqrt{14}$