Question

find the value of log root2 power8​

Answer 1

$\huge\star{\tt{\underline{\underline{\red{Answer\implies 6}}}}}\star$

$log_{\sqrt2}^{8}$

$\implies$ $\large{\log_{2^\frac{1}{2}}^{2^{3}}}$

$\implies$ $\large{\frac{3}{\frac{1}{2}}log_{2}^{2}}$

$\implies$ $6log_{2}^{2}$

$\implies$ $\large{\boxed{\boxed{6}}}$

Answer 2

$\huge \mathfrak{solution}$

To find

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

Formula used :$\boxed{ log_{a}(a) = 1}$

$\boxed{ log_{ {a}^{x} }( {b}^{y} ) } \: \\ = > \frac{y}{x} log_{a}(b)$