Question 715949
With an equation in the form:
log(expression) = number
the next step is to rewrite the equation in exponential form. In general, {{{log(a, (p)) = n}}} is equivalent to {{{p = a^n}}}. Using this pattern on your equation we get:
{{{2+x = 10^2.1}}}
Subtracting 2 we get:
{{{x = 10^2.1-2}}}
This is an exact expression for the solution to your equation. For a decimal approximation we get out our calculators:
{{{x = 125.89254117941672104239541063958-2}}}
which simplifies to:
{{{x = 123.89254117941672104239541063958}}}
Rounded to two places this is:
{{{x = 123.89}}}