Question 133119
{{{ln(x + 2) = ln(e^ln(2)) - ln(x)}}}
We need to break this down:
{{{y = ln(e^ln(2))}}}
{{{e^y = e^ln(2)}}}
{{{y = ln(2)}}}
{{{ln(e^ln(2)) = ln(2)}}}
So:
{{{ln(x + 2) = ln(2) - ln(x)}}}
{{{ln(x + 2) = ln(2/x)}}}
{{{x + 2 = 2/x}}}
{{{x^2 + 2x = 2}}}
{{{x^2 + 2x + 1 = 2 + 1}}}
{{{(x + 1)^2 = 3}}}
{{{x + 1 = 0+-sqrt(3)}}}
{{{x = -1+-sqrt(3)}}}
The negative will result with a natural log of a negative, so only the positive term works: {{{x = -1+sqrt(3)}}}