The half life of a radioactive substance is 57315 hours. How log will it take until there is 84.9% of the substance remaining? Round your solution to the nearest tenth.

The decay constant is $LaTeX: \displaystyle k = \frac{\ln 2}{57315}$ . This gives the equation $LaTeX: \displaystyle 0.849 = e^{-\frac{\ln(2)}{57315}t}$ Taking the natural logarithm of both sides gives $LaTeX: \displaystyle \ln(0.849)= \frac{-t\ln(2)}{57315}$ . Solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = -\frac{ 57315\ln(0.849) }{ \ln(2) }$ . It will take about about 13535.7 hours.