Question 309530
Well {{{ x=e^(1/2)^"" }}} is the answer.



{{{ log( 2, (ln (x)) )= -1 }}} Start with the given equation.



{{{ log( 2, (ln (e^(1/2)^"")) )= -1 }}} Plug in {{{ x=e^(1/2)^"" }}}



{{{ log( 2, ((1/2)ln (e)) )= -1 }}} Use the identity  {{{ln(x^y)=y*ln(x))}}} to pull down the exponent.



{{{ log( 2, ((1/2)*1) )= -1 }}} Take the natural log of 'e' to get 1.



{{{ log( 2, (1/2) )= -1 }}} Multiply and simplify



{{{ log( 2, (2^(-1)) )= -1 }}} Rewrite {{{1/2}}} as {{{2^(-1)}}}



{{{ -1*log( 2, (2)) )= -1 }}} Pull down the exponent using the identity  {{{log(b,(x^y))=y*log(b,(x))}}}



{{{ -1*1= -1 }}} Evaluate the log base 2 of 2 to get 1.



{{{ -1= -1 }}} Multiply



Since the final equation is an identity (ie is always true), this verifies the answer.