Solve the exponential equation algebraically: -14 + 3e^x = 11
So I think I move the -14 to the other side, so now I'm left with
3e^x = 25
What do i do next?
$\displaystyle ln(x) = log_e(x)$
You don't need to unless you're making a decimal approximation. It is fine to say that $\displaystyle x = ln \left(\frac{25}{3}\right) = 2\ln\left(\frac{5}{3}\right) = 2ln5 - ln3$
Note that I used the laws of logs to simplify. If you're teacher wants you to make a decimal approximation there should be a button marked ln on your calculator