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APPLICATIONS OF SOLVING EXPONENTIAL EQUATIONS
USING LOGARITHMS – ANSWER KEY
1)
Depreciation (the decline in cash value) on a car can be determined by the
formula
V = C (1 − r )t , where:
V = the value of the car after t years
C = the original cost
r = the rate of depreciation
If the car's cost, when new, is $15,000, the rate of depreciation is 30%, and the
value of the car now is $3,000, how old is the car to the nearest tenth of a year?
3000 = 15000(1 − 0.3)t →
→t =
2)
1
= 0.7t → 0.2 = 0.7t → log 0.2 = log 0.7t → log 0.2 = t log 0.7
5
log 0.2
→ t ≈ 4.5 years
log 0.7
Growth of a certain strain of bacteria is modeled by the equation
G = A(2.7)0.584 t
where:
G = final number of bacteria
A = initial number of bacteria
t = time (in hours)
In approximately how many hours will 4 bacteria first increase to 2,500 bacteria?
Round your answer to the nearest hour.
2500 = 4(2.7)0.584 t → 625 = (2.7)0.584 t → log 625 = (0.584t ) log 2.7
→
3)
log 625
= 0.584t → 6.481 = 0.584t → t ≈ 11 hours
log 2.7
The amount A, in milligrams, of a 10-milligram dose of a drug remaining in the
t
body after t hours is given the formula A = 10(0.8) . Find, to the nearest tenth of
an hour, how long it takes for half of the drug dose to be left in the body.
5 = 10(0.8)t → 0.5 = (0.8)t → log 0.5 = t log 0.8 →
log 0.5
= t → t ≈ 3.1 hours
log 0.8
4)
Sean invests $10,000 at an annual rate of 5% compounded continuously,
according to the formula
A = amount
P = principal
e = 2.718
A = Pe rt , where:
r = rate of interest
t = time (in years)
a) Determine, to the nearest dollar, the amount of money he will have after 2 years.
A = 10000e0.05(2) → A = 10000e0.1 → A ≈ $11, 052
b) Determine how many years, to the nearest year, it will take for his initial
investment to double.
20000 = 10000(2.718)0.05t → 2 = (2.718)0.05t → 0.05t =
5)
The equation for radioactive decay is
log 2
→ 0.05t = 0.693 → t ≈ 14 years
log 2.718
p = (0.5)
t
H
, where:
p = the part of the substance
H = half-life of substance
t = period of time that the substance remains radioactive
A given substance has a half-life of 6,000 years. After t years, one-fifth of the
original sample remains radioactive. Find t, to the nearest thousand years.
0.2 = (0.5)
t
6000
log 0.2
t
t
⎛ t ⎞
→ log 0.2 = ⎜
=
→ 2.322 =
→ t ≈ 14, 000 years
⎟ log(0.5) →
log 0.5 6000
6000
⎝ 6000 ⎠