Question

me ajudem
log√2 na base 1/8

Answer 1

boa noite!

$Log_{\frac {1}{8}}\sqrt{2} = x$
$(\frac {1}{8})^x=\sqrt{2}$
$(8^{-1})^x=2^{\frac {1}{2}}$
$2^{-3x}=2^{\frac {1}{2}$
corta as bases
$-3x=\frac {1}{2}$
$x=\frac{1}{2}.-\frac{1}{3}$
$x=-\frac {1}{6}$

espero ter ajudado.
bons estudos!