The amount remaining of a radioactive substance after t hour

The amount remaining of a radioactive substance after t hours is given by A(t) = 100ekt. After 12 hours, the initial amount has decreased by 7%.

How much remains after 48 hours? (Round your answer to one decimal place.) %

What is the half-life of the substance? (Round your answer to one decimal place.)

A(t)=100e^(kt)

After 12 hours the initial amount decreased by 7%

therefore after 12 hours A(t)= 93

93=100e^(12k)

.93=e^(12k)

Taking ln on both sides

ln .93= 12k

k=-.00605

after 48 hours

N(t)=100e^(-.00605*48)

N(t)=74.8

For the half life,initially it is 100 therefore at half life it becomes half of 100=50

50=100e^(-.00605t)

.5=e^(-.00605t)

taking ln on both sides

t=114.6 hours

ln .5= -.00605k