Question

ВЫЧИСЛИТЬ!
$log _{8} 5^{2log _{25}32 }$
$log _{3} (7 sqrt{3})-log_{3} (7)$
$frac{1}{4}^{log _{ frac{1}{4} }7-3 }$

Answer 1

$1)log _{8} 5^{2log _{25}32 }=log _{8} (5^2)^{log _{25}32 }= log _{8} (25^{log _{25}32 })=log_8{32}=log_{2^3}{2^5}= \ \ = frac{5}{3} log_22= frac{5}{3}$

$2)log _{3} (7 sqrt{3})-log_{3} (7)=log_3 frac{7 sqrt{3} }{7}=log_3 sqrt{3}= frac{1}{2}$

$3)frac{1}{4}^{log _{ frac{1}{4} }7-3 } =frac{1}{4}^{log _{ frac{1}{4} }7} cdot frac{1}{4}^{-3} =7cdot 4^3=7cdot 64=448$