infinity

$x=\sqrt { 2+\sqrt { 2+\sqrt { 2+\sqrt { 2..... } } } }$

$x=\sqrt { 2+x }$

${ x }^{ 2 }=2+x$

${ x }^{ 2 }-x-2=0$

$(x-2)(x+1)=0$

$x=2$ or $x=-1$

The value of $x$ cannot be negative (because of the definition of the square root function), so the solution is 2.

$\log _{ \frac { 1 }{ 2 } }{ 2 }$

${ \frac { 1 }{ 2 } }^{ x }=2$

answer = $-1$