The half life of a radioactive substance is 85507 seconds. How log will it take until there is 16.5% of the substance remaining? Round your solution to the nearest tenth.

The decay constant is $LaTeX: \displaystyle k = \frac{\ln 2}{85507}$ . This gives the equation $LaTeX: \displaystyle 0.165 = e^{-\frac{\ln(2)}{85507}t}$ Taking the natural logarithm of both sides gives $LaTeX: \displaystyle \ln(0.165)= \frac{-t\ln(2)}{85507}$ . Solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = -\frac{ 85507\ln(0.165) }{ \ln(2) }$ . It will take about about 222272.2 seconds.