Question

Como resolve a E e a F?

Answer 1

As respostas das questões são elaboradas a seguir.

Explicação passo-a-passo:

Em ambas as respostas, vamos usar as seguintes propriedades dos logaritmos:

$log_{\;b}\;a^c=c\,.\,log_{\;b}\;a\\\\log_{\;b^c}\;a=\frac{1}{c}\,.\,log_{\;b}\;a$

letra e)

$x=\;log_{\;0,5}\;8\sqrt{2}\\\\x=\;log_{\;\frac{1}{2}}\;8\sqrt{2}\\\\x=\;log_{\;2^{-1}}\;8\sqrt{2}\\\\x=\;log_{\;2^{-1}}\;2^3\,.\,\sqrt{2}\\\\x=\;log_{\;2^{-1}}\;2^3\,.\,2^{\frac{1}{2}}\\\\x=\;log_{\;2^{-1}}\;2^{(3+\frac{1}{2})}\\\\x=\;log_{\;2^{-1}}\;2^{(\frac{6}{2}+\frac{1}{2})}\\\\x=\;log_{\;2^{-1}}\;2^{\frac{7}{2}}\\\\x=-1\,.\,\;log_{\;2}\;2^{\frac{7}{2}}\\\\x=-1\,.\,\;\frac{7}{2}\,.\,log_{\;2}\;2\\\\x=-1\,.\,\;\frac{7}{2}\,.\,1\\\\x=-\frac{7}{2}$

letra f)

$x=log_{\,\sqrt{2}}\;\frac{\sqrt{2}}{2}\\\\x=log_{\,{2^{\frac{1}{2}}}}\;\frac{\sqrt{2}}{2}\\\\x=2\,.\,log_{\;{2}}}\;\frac{\sqrt{2}}{2}\\\\x=2\,.\,log_{\;{2}}}\;(\sqrt{2}\,.\,\frac{1}{2})\\\\x=2\,.\,log_{\;{2}}}\;(2^{\frac{1}{2}}\,.\,2^{-1})\\\\x=2\,.\,log_{\;{2}}}\;2^{(\frac{1}{2}-1})\\\\x=2\,.\,log_{\;{2}}}\;2^{(\frac{1}{2}-\frac{2}{2}})\\\\x=2\,.\,log_{\;{2}}}\;2^{-\frac{1}{2}}\\\\x=2\,.\,-\frac{1}{2}\,.\,log_{\;{2}}}\;2\\\\x=2\,.\,-\frac{1}{2}\,.\,1\\\\x=-1$