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Wiemy, że \(2^a=3\). Oblicz \(24^{ \frac{2}{3+a} }\)

\(2^a=3 \So a=\log_23\)
\(24^{ \frac{2}{3+a} }=24^{\frac{2}{\log_28+\log_23}}=(24^{\frac{1}{\log_{2}24}})^2=(24^{\log_{24}2})^2=2^2=4 \)

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