Odpowiedź:
Szczegółowe wyjaśnienie:
[tex]\log_\frac12\dfrac{32}{\sqrt[4]8}=\log_\frac12\dfrac{2^5}{(2^3)^\frac14}=\log_\frac12\dfrac{2^5}{2^\frac34}=\log_\frac12\big2^{5-\frac34}=\log_\frac12\big2^{4\frac14}=\\\\{}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad=\log_\frac12\left(\frac12\right)^{-4\frac14}=-4\frac14[/tex]