Question

Função logarítmica
$log_{8} \: \sqrt[]{3} 2 = y$

Answer 1

$log_{8}( \sqrt{32} ) = y \\ {8}^{y} = \sqrt{32} \\ {8}^{y} = {32}^{ \frac{1}{2} } \\ {8}^{y} = {( {2}^{5} )}^{ \frac{1}{2} } \\ {( {2}^{3}) }^{y} = {2}^{ \frac{5}{2} } \\ {2}^{3y} = {2}^{ \frac{5}{2} } \\ \\ 3y = \frac{5}{2} \\ y = \frac{5}{2} \times \frac{1}{3} \\ y = \frac{5}{6}$


Logo, y = 5/6.

Answer 2

$log_8({\sqrt{32}})\\ =log_8{(32^{\frac{1}{2}}) \\ = \dfrac{1}{2} \times log_8({32}) \\ =\dfrac{1}{2} \times log_8({2^{5}}) \\ =\dfrac{1}{2} \times 5 \times log_8({2}) \\ =\dfrac{1}{2} \times 5 \times (\dfrac{1}{3}) \\ =\dfrac{5}{6}$

dessa forma

$log_8({\sqrt{32}}) = \dfrac{5}{6}\\ y = \dfrac{5}{6}$