The half life of a radioactive substance is 12978 years. How log will it take until there is 52.9% of the substance remaining? Round your solution to the nearest tenth.

The decay constant is $LaTeX: \displaystyle k = \frac{\ln 2}{12978}$ . This gives the equation $LaTeX: \displaystyle 0.529 = e^{-\frac{\ln(2)}{12978}t}$ Taking the natural logarithm of both sides gives $LaTeX: \displaystyle \ln(0.529)= \frac{-t\ln(2)}{12978}$ . Solving for $LaTeX: \displaystyle t$ gives $LaTeX: \displaystyle t = -\frac{ 12978\ln(0.529) }{ \ln(2) }$ . It will take about about 11922.4 years.