The half life of a radioactive substance is 39258 minutes. How log will it take until there is 77.6% of the substance remaining? Round your solution to the nearest tenth.

The decay constant is $LaTeX: \displaystyle k = \frac{\ln 2}{39258}$ . This gives the equation $LaTeX: \displaystyle 0.776 = e^{-\frac{\ln(2)}{39258}t}$ Taking the natural logarithm of both sides gives $LaTeX: \displaystyle \ln(0.776)= \frac{-t\ln(2)}{39258}$ . Solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = -\frac{ 39258\ln(0.776) }{ \ln(2) }$ . It will take about about 14363.4 minutes.