The half life of a radioactive substance is 18296 seconds. How log will it take until there is 27.6% of the substance remaining? Round your solution to the nearest tenth.

The decay constant is $LaTeX: \displaystyle k = \frac{\ln 2}{18296}$ . This gives the equation $LaTeX: \displaystyle 0.276 = e^{-\frac{\ln(2)}{18296}t}$ Taking the natural logarithm of both sides gives $LaTeX: \displaystyle \ln(0.276)= \frac{-t\ln(2)}{18296}$ . Solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = -\frac{ 18296\ln(0.276) }{ \ln(2) }$ . It will take about about 33980.4 seconds.