The half life of a radioactive substance is 42279 days. How log will it take until there is 10.4% of the substance remaining? Round your solution to the nearest tenth.

The decay constant is $LaTeX: \displaystyle k = \frac{\ln 2}{42279}$ . This gives the equation $LaTeX: \displaystyle 0.104 = e^{-\frac{\ln(2)}{42279}t}$ Taking the natural logarithm of both sides gives $LaTeX: \displaystyle \ln(0.104)= \frac{-t\ln(2)}{42279}$ . Solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = -\frac{ 42279\ln(0.104) }{ \ln(2) }$ . It will take about about 138055.5 days.