The half-life of cesium-137 is 30 years. Suppose we have a 180 mg sample. Find the mass that remains after t years. Let y(t) be the mass (in mg) remaining after t years. Then we know the following. y ( t ) = y ( 0 ) e k t = ⋅ e k t Since the half-life is 30 years, then y ( 30 ) = mg. Thus, e 30 k = . Therefore, k = . Now, remembering that ln x n = n ln x and that e ln z = z , we have y ( t ) = 180 e ( t / 30 ) ( − ln 2 ) = mg. How much of the sample remains after 70 years? After 70 years we have the following. y ( 70 ) = 180 ⋅ 2 = mg (Round your answer to two decimal places.) After how long will only 1 mg remain? To find the time at which only 1 mg remains, we must solve 1 = y ( t ) = 180 ( 2 − t / 30 ) , and so we get the following. t = − 30 log 2 ( ) Hence, we conclude the following. t = yr (Round your final answer to one decimal place.)