The half life of a radioactive substance is 81129 minutes. How log will it take until there is 85.6% of the substance remaining? Round your solution to the nearest tenth.

The decay constant is $LaTeX: \displaystyle k = \frac{\ln 2}{81129}$ . This gives the equation $LaTeX: \displaystyle 0.856 = e^{-\frac{\ln(2)}{81129}t}$ Taking the natural logarithm of both sides gives $LaTeX: \displaystyle \ln(0.856)= \frac{-t\ln(2)}{81129}$ . Solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = -\frac{ 81129\ln(0.856) }{ \ln(2) }$ . It will take about about 18198.6 minutes.