The hard way: solve the equation 105/2 = 105e^{-0.038t} for t. why?

can you continue?

The easy way: half-life = (ln(2))/(0.038)

this is going by the formula: For the radioactive decay equation of the form , the following formula holds:

where is the half-life and is the rate of decay

Solve the equation 105/4 = 105e^{-0.038t} for t(b):After how many years, to the nearest tenth of a year, there will be 105/4 milligrams present?

After _______ years there will be 26.25=105/4 milligrams of the substance left.

can you continue?