MA 3972-MA-Book April 11, 2018 16:5
326 STEP 4. Review the Knowledge You Need to Score High
Step 3: Substitutek=
1
2
ln
(
8
5
)
into one of the equations.
500 =y 0 ek
500 =y 0 e
ln
(√ 8
5
)
500 =y 0
(√
8
5
)
y 0 =
500
√
8 / 5
= 125
√
10 ≈ 395. 285
Thus, there are 395 bacteria present initially.
(b) y 0 = 125
√
10,k=ln
√
8
5
y(t)=y 0 ekt
y(t)=
(
125
√
10
)
e
(
ln
√ 8
5
)t
=
(
125
√
10
)( 8
5
)(1/2)t
y( 4 )=
(
125
√
10
)( 8
5
)(1/2)4
=
(
125
√
10
)( 8
5
) 2
= 1011. 93
Thus, there are approximately 1012 bacteria present after 4 days.
TIP • Get a good night’s sleep the night before. Have a light breakfast before the exam.
Example 2 Radioactive Decay
Carbon-14 has a half-life of 5750 years. If initially there are 60 grams of carbon-14, how
many grams are left after 3000 years?
Step 1: y(t)=y 0 ekt= 60 ekt
Since half-life is 5750 years, 30= 60 ek(5750)⇒
1
2
=e^5750 k.
ln
(
1
2
)
=ln
(
e^5750 k
)
−ln 2= 5750 k
−ln 2
5750
=k