Question 484241
Let z={{{e^x}}}
Note that {{{y^(-a)}}} = {{{1/y^a}}}
So we have z + 1/z = 4
*z: {{{z^2}}} + 1 = 4z
{{{z^2}}} - 4z +1 =0
z = {{{(-(-4)+-sqrt((-4)^2-4(1)(1)))/(2*1)}}}
z = {{{(4+-sqrt(16-4))/2}}}
{{{sqrt(12)}}} = {{{sqrt(4*3)}}} = 2 * sqrt(3)
z = {{{2+-sqrt(3))}}}
Ln(z) = x (natural log base e)
So x = {{{Ln(2 +- sqrt(3))}}}