Помогите пожалуйста срочно зарание спасибо)!!!

Решите задачу:

$log_{ \frac{1}{ \sqrt{2} } }( \frac{1}{8} ) = x \\ log_{ \frac{1}{ \sqrt{2} } }( \frac{1}{2} ) ^{3} = x \\ x = log_{ \frac{1}{ \sqrt{2} } }( \frac{1}{ \sqrt{2} } ) ^{6} \\ x = 6$
$log_{3 \sqrt{3} }( \frac{1}{27} ) = x \\ x = log_{( \sqrt{3}) ^{3} }(3) ^{ - 3} \\ x = log_{( \sqrt{3}) ^{3} }( \sqrt{3}) ^{ - 6} \\ x = \frac{ - 6}{3} \\ x = - 2$
$log_{x}(0.125) = - 3 \\ log_{x}( {0.5}^{3} ) = - 3 \\ 3 log_{x}(0.5) = - 3 \\ log_{x}( \frac{1}{2} ) = -1 \\ x = 2 \\$
$log_{x}(4) = - \frac{1}{2} \\ {x}^{ - \frac{1}{2} } = 4 \\ \frac{1}{ \sqrt{x} } = 4 \\ x = \frac{1}{16}$
$log_{16}(x) = \frac{3}{4} \\ x = {16}^{ \frac{3}{4} } \\ x = {2}^{4 \times \frac{3}{4} } \\ x = {2}^{3} \\ x = 8$
$log_{5}(x) = - 3 \\ x = {5}^{ - 3} \\ x = \frac{1}{ {5}^{3} } \\ x = \frac{1}{125}$
$log_{6}(x) = - {2} \\ x = {6}^{ - 2} \\ x = \frac{1}{36}$
$log_{x}(8) = - \frac{1}{2} \\ {x}^{ - \frac{1}{2} } = 8 \\ \frac{1}{ \sqrt{x} } = 8 \\ x = \frac{1}{64}$