Question

ЛОГАРИФМЫ ВЫЧИСЛИТЬ ТОЛЬКО №1

Answer 1

$\frac{log _{2}24- \frac{1}{2} log _{2}72 }{log _{3}18- \frac{1}{3}log _{3}72 } = \frac{log _{2}24-log _{2} \sqrt{72} }{log _{3}18- log _{3} \sqrt[3]{72} } =$ $= \frac{log _{2} \frac{24}{6 \sqrt{2} } }{log _{3} \frac{18}{2 \sqrt[3]{9} } } = \frac{log _{2} \frac{4}{ \sqrt{2} } }{log _{3} \frac{9}{ \sqrt[3]{9} } }= \frac{log _{2} 2 ^{1,5} }{log _{3} 9^{ ^{ \frac{2}{3} } } } = \frac{1,5log _{2} 2}{ \frac{2}{3}log _{3}9 } = \frac{1,5}{ \frac{2}{3}*2 }= \frac{1,5}{ \frac{4}{3} } = \frac{9}{8}=1 \frac{1}{8}$

Answer 2

((2 ㏒₂24-㏒₂72)/2)/((3㏒₃18-㏒₃72)/3)=㏒₂512/(4㏒₃9)=9/8=1+1/8