It will take 71,947,270,950 years.
7-8 Using Exponential and Logarithmic Functions
c. Substitute 50 for a, 800,000,000 for t, and 1.42 ×
–11
10
for k , then evaluate f (x).
4. APPLY MATH The half-life of Rubidium-87 is
about 48.8 billion years.
a. Determine the value of k and the equation of
decay for Rubidium-87.
b. A specimen currently contains 50 milligrams of
Rubidium-87. How long will it take the specimen to
decay to only 18 milligrams of Rubidium-87?
c. How many milligrams of Rubidium-87 will be left
after 800 million years?
d. How long will it take Rubidium-87 to decay to
one-sixteenth its original amount?
SOLUTION: a. Exponential decay can be modeled by the function
After 800 million years, there will be about 49.4 mg
of Rubidium -87 remaining.
d. Substitute 50 for a,
–10
for f (x), and 1.42 × 10
for k , then solve for t.
–kt
f(x) = ae , where a is the initial value, t is time in
years, and k is a constant representing the rate of
continuous decay.
9
Substitute 48.8 × 10 for t, and 0.5a for f (x), then
solve for k.
It will take about 195.3 billion years.
5. BIOLOGY A certain bacteria is growing
kt
–11
b. Substitute 50 for a, 18 for f (x), and 1.42 × 10
for k , then solve for t.
exponentially according to the model y = 80e ,
where t is the time in minutes.
a. If there are 80 cells initially and 675 cells after 30
minutes, find the value of k for the bacteria.
b. When will the bacteria reach a population of 6000
cells?
c. If a second type of bacteria is growing
0.0978t
exponentially according to the model y = 35e
,
determine how long it will be before the number of
cells of this bacteria exceed the number of cells in
the other bacteria.
SOLUTION: a. Substitute 675 and 30 for y and t respectively, then
solve for k.
It will take 71,947,270,950 years.
c. Substitute 50 for a, 800,000,000 for t, and 1.42 ×
–11
10
for k , then evaluate f (x).
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After 800 million years, there will be about 49.4 mg
Page 1
the other bacteria.
SOLUTION: a. Substitute
675 andand
30 for
y and t respectively,
then
7-8 Using
Exponential
Logarithmic
Functions
solve for k.
It will take about 30.85 min.
7. PALEONTOLOGY A paleontologist finds a human
bone and determines that the Carbon-14 found in the
bone is 85% of that found in living bone tissue. How
old is the bone?
SOLUTION: –kt
b. Substitute 6000 and 0.071 for y and k respectively
then solve for t.
The formula for the decay of Carbon-14 is y = ae .
Let a be the initial amount of Carbon-14 in the
animal’s body. The amount y that remains after t
years is 85% of a or 0.85a. The rate of continuous
growth for Carbon-14 is k ≈ 0.00012.
Substitute 0.85a for y and 0.00012 for k, then solve
for t.
The bone is about 1354 years old.
It will take about 60.8 min.
19. Jada bought a used car for $6000. The value of the
car is expected to depreciate at a uniform rate of
30% per year. What will be the approximate value of
the car in 3 years? A $600
B $735
C $2060
D $2440
c. Equate the functions
, then solve for t.
SOLUTION: Substitute 6000 for a, 0.30 for k, and 3 for t in the
–kt
exponential decay formula y = ae , then evaluate
for y.
It will take about 30.85 min.
7. PALEONTOLOGY A paleontologist finds a human
bone and determines that the Carbon-14 found in the
bone is 85% of that found in living bone tissue. How
old is the bone?
SOLUTION: eSolutions
Manual - Powered by Cognero
The value of the car in 3 years is about $2440.
So, the correct answer is choice D.
Page 2
The formula for the decay of Carbon-14 is y = ae
Let a be the initial amount of Carbon-14 in the
–kt
.