Question

задача с логарифмами во вложении!

Answer 1

$(36^{log_{6} 5} +10^{1- lg_{} 2} -3^{log_{9}36 } )*(2^{-3log_{a}2-1 })^{(log_{3} 2)}^{-1} = ((6^{2})^{log_{6} 5}+10^{1}*10^{-lg_{2} } -3^{log_{3^{2} }36 } )*(2^{-3log_{a}1 })^{(log_{3} 2)}^{-1} = (6^{2log_{6} 5} +10 * 2^{-1} -3^{log_{3} 6} )*(2^{0})^{log_{3}\frac{1}{2} } = 5^{2} +10*\frac{1}{2} -6 = 25 + 10*\frac{1}{2} -6 = 24$