Question

Me ajudem valendo 24 pontos.

Answer 1

Aplicando propriedades de logaritmos

       $log( \sqrt{2}) \frac{ \sqrt{2} }{2} = x \\ \\ \frac{ \sqrt{2} }{2} =( \sqrt{2})^x \\ \\ 2^{ \frac{1}{2} -1} = 2^{ \frac{1}{2} x}$

        Base iguais , expoentes iguais

       $\frac{1}{2} -1= \frac{1}{2} x \\ \\ \frac{1}{2} - \frac{2}{2} =\frac{1}{2}x \\ \\ - \frac{1}{2} = \frac{1}{2} x \\ \\ x= \frac{- \frac{1}{2} }{ \frac{1}{2} } \\ \\ x=-1$

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        $log(0,5)8 \sqrt{2} =y \\ \\ 8 \sqrt{2} =(0,5)^y \\ \\ 2^3. 2^{ \frac{1}{2} } =( \frac{1}{2} )^y \\ \\ 2^{3+ \frac{1}{2} } = 2^{-y} \\ \\ 3+ \frac{1}{2} =-y \\ \\ \frac{7}{2} =-y \\ \\ y=- \frac{7}{2}$