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5.2
LOGARITHMS AND
EXPONENTIAL MODELS
1
Using Logarithms to Undo Exponents
Example 2
The US population, P, in millions, is currently growing according
to the formula P = 299e0.009t, where t is in years since 2006.
When is the population predicted to reach 350 million?
Solution
We want to solve the following equation for t: 299e0.009t = 350.
299 e
0.009 t
 350
e
0.009 t
 350 / 299
ln e
0.009 t
 ln(350 / 299 )
0.009 t  ln(350 / 299 )
t
ln(350 / 299 )
 17.5 years
0.009
The US population is predicted to reach 350 million during the
year 2024.
2
Doubling Time
Example 4 (a)
Find the time needed for the turtle population described
by the function P = 175(1.145)t to double its initial size.
Solution (a)
The initial size is 175 turtles; doubling this gives 350 turtles.
We need to solve the following equation for t:
175(1.145)t = 350
1.145t = 2
log 1.145t = log 2
t · log 1.145 = log 2
t = log 2/log 1.145 ≈ 5.119 years.
Note that
175(1.145)5.119 ≈ 350
3
Since the growth rate is 0.000121, the growth rate factor is b = 1 – 0.000121
= 0.999879. Thus after t years the amount left will be
4
5
Half-Life
Example 8
The quantity, Q, of a substance decays according to the
formula Q = Q0e−kt, where t is in minutes. The half-life of
the substance is 11 minutes. What is the value of k?
Solution
We know that after 11 minutes, Q = ½ Q0. Thus, solving for k,
we get
Q0e−k·11 = ½ Q0
e−11k = ½
−11k = ln ½
k = ln ½ / (−11) ≈ 0.06301,
so k = 0.063 per minute. This substance decays at the
continuous rate of 6.301% per minute.
6
Converting Between Q = a bt and Q = a ekt
Any exponential function can be
written in either of the two forms:
Q = a bt
or
Q = a ekt.
If b = ek, so k = ln b, the two formulas
represent the same function.
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Converting Between Q = a bt and Q = a ekt
Example 9
Convert the exponential function P = 175(1.145)t to the form P = aekt.
Solution
The parameter a in both functions represents the initial population. For
all t,
175(1.145)t = 175(ek)t,
so we must find k such that
ek = 1.145.
Therefore k is the power of e which gives 1.145. By the definition of ln,
we have
k = ln 1.145 ≈ 0.1354.
Therefore,
P = 175e0.1354t.
Example 10
Convert the formula Q = 7e0.3t to the form Q = a bt.
Solution
Using the properties of exponents,
Q = 7e0.3t = 7(e0.3)t.
Using a calculator, we find e0.3 ≈ 1.3499, so Q = 7(1.3499)t.
8
Exponential Growth Problems That Cannot Be
Solved By Logarithms
Example 13
With t in years, the population of a country (in millions) is given by
P = 2(1.02)t, while the food supply (in millions of people that can be fed)
is given by N = 4+0.5t. Determine the year in which the country first
experiences food shortages.
Finding the intersection of
linear and exponential graphs
Solution
Setting P = N, we get 2(1.02)t = 4+0.5t, Population
which, after dividing by 2 and taking
(millions)
the log of both sides can simplify to
t log 1.02 = log(2 + 0.25t).
We cannot solve this equation
algebraically, but we can estimate
its solution numerically or graphically:
t ≈ 199.381. So it will be nearly 200 years
before shortages are experienced
Shortages start →
N = 4+0.5t
P = 2(1.02)t
t, years
199.381
9
Q  ab
t
The growth factor b and the doubling time T are
related by:
t
Q  ab
2a = ab
2= b
t
t
ln 2  ln b
t
t ln b  ln 2
D oubling T im e T 
ln 2
ln b
10
Q  ae
kt
The growth factor k and the doubling time T are
kt
related by:
Q  ae
2a = ae
2= e
kt
kt
ln 2  ln e
kt
kt  ln 2
D oubling T im e T 
ln 2
k
11

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