Question 353706
{{{ln(x) = -0.123}}}
Just rewrite your equation in exponential form. In general, {{{log(a, (p)) = q}}} is equivalent to {{{a^q = p}}}. Using this on your equation we get:
{{{x = e^(-0.123)}}}
For a decimal approximation of you answer, use your calculator on the above. If your calculator doesn't have a button for "e", then use 2.718281828459045 (or a rounded off version of this number).