Question

Hallar "k"
$2^{k} = \sqrt[\sqrt{2}-1 ]{\sqrt{2}^{\sqrt{2}+1 } } \sqrt{2}$

Answer 1

$k = 2 + \sqrt{2}$

Explicación paso a paso:

${2}^{k} = \sqrt[ \sqrt{2} - 1 ]{ { \sqrt{2} }^{ \sqrt{2} + 1 } } \sqrt{2}$

${2}^{k} = \sqrt[2 \sqrt{2} - 2 ]{ {2}^{ \sqrt{2} + 1 } } \sqrt{2}$

${2}^{k} = {2}^{ \sqrt{2} + \frac{3}{2} } \times {2}^{ \frac{1}{2} }$

${2}^{k} = {2}^{( \sqrt{2} + \frac{3}{2} ) + \frac{1}{2} }$

$k = \sqrt{2} + \frac{3}{2} + \frac{1}{2}$

$k = \sqrt{2} + 2$

Answer 2

Respuesta:

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