Question 354497
{{{ln e^(sqrt(x-2))= 6}}}.  The ln and the base e will just cancel each other out, to give
{{{sqrt(x-2) = 6}}}.  Squaring both sides, we get
{{{x-2 = 36}}},
{{{x = 38}}}.  Checking this value against the original equation shows that indeed it is the solution to the equation.