1

5.2

LOGARITHMS AND EXPONENTIAL MODELS

2

Using Logarithms to Undo Exponents

Example 2The 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?SolutionWe want to solve the following equation for t: 299e0.009t = 350.

The US population is predicted to reach 350 million during the year 2024.

0.009299 350te

0.009ln ln(350 / 299)te 0.009 ln(350 / 299)t

ln(350 / 299)17.5

0.009t years

0.009 350 / 299te

3

Doubling TimeExample 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

4

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

5

6

Half-LifeExample 8The 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?SolutionWe 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.

7

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.

8

Converting Between Q = a bt and Q = a ekt

Example 9Convert 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 10Convert 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.

9

199.381

Exponential Growth Problems That Cannot Be Solved By Logarithms

Example 13With 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.

SolutionSetting P = N, we get 2(1.02)t = 4+0.5t,which, after dividing by 2 and taking the log of both sides can simplify to t log 1.02 = log(2 + 0.25t).

We cannot solve this equationalgebraically, but we can estimateits solution numerically or graphically:t ≈ 199.381. So it will be nearly 200 years before shortages are experienced

Finding the intersection of linear and exponential graphs

P = 2(1.02)t

N = 4+0.5t

Shortages start →

t, years

Population(millions)

10

The growth factor b and the doubling time T arerelated by:

2a =

2 =

ln 2 ln

ln ln 2

t

t

t

t

Q ab

ab

b

b

t b

tQ ab

ln 2

lnDoubling Time T

b

11

The growth factor k and the doubling time T arerelated by:

2a =

2 = e

ln 2 ln

ln 2

kt

kt

kt

kt

Q ae

ae

e

kt

ktQ ae

ln 2Doubling Time T

k