The half life of a radioactive substance is 80746 hours. How log will it take until there is 44.0% of the substance remaining? Round your solution to the nearest tenth.

The decay constant is $LaTeX: \displaystyle k = \frac{\ln 2}{80746}$ . This gives the equation $LaTeX: \displaystyle 0.44 = e^{-\frac{\ln(2)}{80746}t}$ Taking the natural logarithm of both sides gives $LaTeX: \displaystyle \ln(0.44)= \frac{-t\ln(2)}{80746}$ . Solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = -\frac{ 80746\ln(0.44) }{ \ln(2) }$ . It will take about about 95637.5 hours.