1. ## Logarithms - Half-life

A 105-milligram sample of a radioactive substance decays according to the equation

N=105 * e^-0.038t
where N is the number of milligrams present after t years.

(a) Find the half-life of the substance to the nearest tenth of a year.

The half-life is _________ years.

(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.

2. Originally Posted by lemontea
A 105-milligram sample of a radioactive substance decays according to the equation

N=105 * e^-0.038t
where N is the number of milligrams present after t years.

(a) Find the half-life of the substance to the nearest tenth of a year.

The half-life is _________ years.
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 $P = P_0e^{-rt}$, the following formula holds:

$rt_h = \ln 2$

where $t_h$ is the half-life and $r$ is the rate of decay

(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.
Solve the equation 105/4 = 105e^{-0.038t} for t

can you continue?