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Exponential-Growth Word Problems
Log Probs Expo Growth Expo Decay
What is the formula for exponential growth and decay?
Exponential word problems almost always work off the growth / decay formula, A = Pe rt , where " A " is the ending amount of whatever you're dealing with (for example, money sitting in an investment, bacteria growing in a petri dish, or radioactive decay of an element highlighting your X-ray), " P " is the beginning amount of that same "whatever", " r " is the growth or decay rate, and " t " is time.
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Exponential Growth and Decay
The exponential-growth/-decay formula A = Pe rt is related to the compound-interest formula , and represents the case of the interest being compounded "continuously".
Note that the variables may change from one problem to another, or from one context to another, but that the structure of the equation is always the same. For instance, all of the following represent the same relationship:
A = P e r t A = P e k t Q = N e k t Q = Q 0 e k t
No matter the particular letters used, the green variable stands for the ending amount, the blue variable stands for the beginning amount, the red variable stands for the growth or decay constant, and the purple variable stands for time. Get comfortable with this formula; you'll be seeing a lot of it.
- A biologist is researching a newly-discovered species of bacteria. At time t = 0 hours, he puts one hundred bacteria into what he has determined to be a favorable growth medium. Six hours later, he measures 450 bacteria. Assuming exponential growth, what is the growth constant " k " for the bacteria? (Round k to two decimal places.)
For this exercise, the units on time t will be hours, because the growth is being measured in terms of hours.
The beginning amount P is the amount at time t = 0 , so, for this problem, P = 100 .
The ending amount is A = 450 at t = 6 hours.
The only variable I don't have a value for is the growth constant k , which also happens to be what I'm looking for. So I'll plug all the known values into the exponential-growth formula, and then solve for the growth constant:
A = Pe kt 450 = 100 e 6 k 4.5 = e 6 k ln(4.5) = 6 k ln(4.5) / 6 = k = 0.250679566129...
The growth constant is 0.25 /hour.
Many math classes, math books, and math instructors leave off the units for the growth and decay rates. However, if you see this topic again in chemistry or physics, you will probably be expected to use proper units ("growth-decay constant / time"), as I have displayed above. It's not a bad idea to get into the habit now of checking and reporting your units.
Note that the constant was positive, because it was a growth constant. If I had come up with a negative value for the growth constant, then I would have known to check my work to find my error(s).
- A certain type of bacteria, given a favorable growth medium, doubles in population every 6.5 hours. Given that there were approximately 100 bacteria to start with, how many bacteria will there be in a day and a half?
In this problem, I know that time t will be in hours, because they gave me growth in terms of hours. So I'll convert "a day and a half" to "thirty-six hours", so my units match.
I know that P = 100 , and I need to find A at t = 36 .
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But what is the growth constant k ? And why do they tell me what the doubling time is?
They gave me the doubling time because I can use this to find the growth constant k . Then, once I have this constant, I can go on to answer the actual question.
So this exercise actually has two unknowns, the growth constant k and the ending amount A . I can use the doubling time to find the growth constant, at which point the only remaining value will be the ending amount, which is what they actually asked for. So first I'll find the constant.
If the initial population is 100 , then, in 6.5 hours (being the specified doubling time), the population will be 200 . I'll set this up and solve for k :
A = Pe kt 200 = 100 e 6.5 k 2 = e 6.5 k
At this point, I need to use logs to solve:
ln(2) = 6.5 k ln(2) / 6.5 = k
I could simplify this to a decimal approximation, but I won't, because I don't want to introduce round-off error if I can avoid it. So, for now, the growth constant will remain this "exact" value. (I might want to check this value quickly in my calculator, to make sure that this growth constant is positive, as it should be. If I have a negative value at this stage, I need to go back and check my work.)
Now that I have the growth constant, I can answer the actual question, which was "How many bacteria will there be in thirty-six hours?" This means using 100 for P , 36 for t , and the above expression for k . I plug these values into the formula, and then I simplify to find A :
A = 100 e 36(ln(2)/6.5) = 4647.75313957...
In 36 hours, there will be about 4648 bacteria.
You can do a rough check of this answer, using the fact that exponential processes involve doubling (or halving) times. The doubling time in this case is 6.5 hours, or between 6 and 7 hours.
If the bacteria doubled every six hours, then there would be 200 in six hours, 400 in twelve hours, 800 in eighteen hours, 1600 in twenty-four hours, 3200 in thirty hours, and 6400 in thirty-six hours.
If the bacteria doubled every seven hours, then there would be 200 in seven hours, 400 in fourteen hours, 800 in twenty-one hours, 1600 in twenty-eight hours, and 3200 in thirty-five hours. The answer we got above, 4678 in thirty-six hours, fits nicely between these two estimates.
Warning: When doing the above simplification of 100 e 36(ln(2)/6.5) , try to do the calculations completely within your calculator in order to avoid round-off error. It is best to work from the inside out, starting with the exponent, then the exponential, and finally the multiplication, like this:
Note: When you are given a nice, neat doubling time, another method for solving the exercise is to use a base of 2 . First, figure out how many doubling-times that you've been given. In the above case, this would start by noting that "a day and a half" is 36 hours, so we have:
36 ÷ 6.5 = 72/13
Use this as the power on 2 :
100 × 2 (72/13) = 4647.75314...
Not all algebra classes cover this method. If you're required to use the first method for every exercise of this type, then do so (in order to get the full points). Otherwise, this base- 2 trick can be a time-saver. And, yes, you'd use a base of 3 if you'd been given a tripling-time, a base of 4 for a quadrupling-time, etc.
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Exponential Growth – Examples and Practice Problems
Exponential functions can be used to model population growth scenarios or other situations that follow patterns with growth at fixed rates. There are formulas that can be used to find solutions to most problems related to exponential growth.
Here, we will look at a summary of exponential growth and the formulas that can be used to solve these types of problems. In addition, we will look at several examples with answers of exponential growth in order to learn how to apply these formulas.
Relevant for …
Exploring examples of exponential growth.
See examples
Summary of exponential growth
Exponential growth – examples with answers, exponential growth – practice problems.
Exponential growth is a pattern of data that shows larger increases over time, creating the curve of an exponential function. For example, if a bacteria population starts with 2 in the first month, then with 4 in the second month, 16 in the third month, 256 in the fourth month, and so on, it means that the population grows exponentially with a power of 2 every month.
The following formula is used to model exponential growth. If a quantity grows by a fixed percentage at regular intervals, the pattern can be described by this function:
We recall that the original exponential function has the form $latex y = a{{b}^x}$. In the original growth formula, we have replaced b with $latex 1+ r$. So, in this formula we have:
- $latex a=$ initial value. This is the starting amount before growth.
- $latex r=$ growth rate. This is represented as a decimal.
- $latex x=$ time interval. This is the time that has passed.
Most naturally occurring events continually grow. For example, bacteria continue to grow over a 24-hour period. Bacteria don’t wait until the end of 24 hours to reproduce all at the same time.
To model the continuous growth that occurs naturally such as populations, bacteria, etc., we use the exponential e . e can be thought of as a universal constant that represents growth possibilities using a continuous process. Furthermore, using e we can also represent growth measured periodically over time.
Therefore, if a quantity is continually growing with a fixed percentage, we can use the following formula to model this pattern:
In this formula we have:
- $latex A=$ final value. This is the amount after growth.
- $latex A_{0}=$ initial value. This is the amount before growth.
- $latex e=$ exponential. e is approximately equal to 2718…
- $latex k =$ continuous growth rate. It is also called the constant of proportionality.
- $latex t =$ elapsed time.
The following examples use the formulas detailed above and some variations to find the solution. It is recommended that you try to solve the exercises yourself before looking at the answer.
A population of bacteria grows according to the function $latex f(x)=100{{e}^{0.02t}}$, where t is measured in minutes. How many bacteria will there be after 4 hours (240 minutes)?
This is continuous growth, so we have the formula $latex A=A_{0}{{e}^{kt}}$. We can recognize the following data:
- $latex A_{0}=100$
- $latex k=0.02$
- $latex t=240$
Therefore, we have:
$latex f(240)=100{{e}^{0.02(240)}}\approx 12151$
Therefore, there will be 12 151 bacteria after 4 hours.
A population of bacteria grows according to the function $latex f(x)=100{{e}^{0.02t}}$, where t is measured in minutes. When will the population reach 50 000?
Here, we have the same formula as the previous exercise, but now we have to find the time knowing the final quantity. We can recognize the following data:
- $latex A=50000$
$latex 50 000=100{{e}^{0.02t}}$
$latex 500={{e}^{0.02t}}$
$latex \ln(500)=0.02t$
$latex t=\frac{\ln(500)}{0.02}$
$latex t\approx 310.73$
Therefore, the population of bacteria will become 50 000 after 310.73 minutes.
We can model the population of a community with the formula $latex A=10000({{e}^{0.005t}})$. Here, A represents population and t represents time in years. What is the population after 10 years?
We already have a given formula: $latex A=10000({{e}^{0.005t}})$. We have to calculate the population using time $latex t=10$. Therefore, we substitute $latex t=10$ to get:
$latex A=10000({{e}^{0.005(10)}})$
$latex =10000({{e}^{0.05}})$
$latex =10000(1.0513)$
$latex =10513$
Therefore, the population in the community after 10 years will be 10 513.
The population of a certain community was 10 000 in 1980. In 2000, it was found to have grown to 20 000. Form an exponential function to model the population of community P that changes through time t .
When we have continuous population growth, we can model the population with the general formula $latex P=P_{0}({{e}^{\lambda t}})$, where $latex P_{0}$ represents the initial population, λ is the exponential growth constant and t is time.
Using the given information, we have to find the constant λ to complete the formula. Therefore, we have:
$latex P=P_{0}({{e}^{\lambda t}})$
$latex 20000=10000({{e}^{20 \lambda}})$
$latex \frac{20000}{10000}={{e}^{20 \lambda}}$
$latex 2={{e}^{20 \lambda}}$
$latex \ln(2)=20 \lambda$
$latex \frac{\ln(2)}{20}=\lambda$
$latex 0.0347=\lambda$
Thus, we can model the population growth of the community with the formula $latex P=10000({{e}^{0.0347 t}})$.
The population growth of a small city is modeled with the function $latex P= P_{0}({{e}^{0.1234t}})$. When did the population reach 37 500 if in 1980 the population was 12 500?
We can substitute the values in the formula with the given information:
$latex P=P_{0}({{e}^{0.1234t}})$
$latex 37500=12500({{e}^{0.1234t}})$
Now, we have to solve for time:
$latex \frac{37500}{12500}=({{e}^{0.1234t}})$
$latex 3=({{e}^{0.1234t}})$
$latex \ln(3)=0.1234t$
$latex \frac{\ln(3)}{0.1234}=t$
$latex 8.9=t$
Thus, the city’s population reached 37 500 in 1989.
One type of bacteria doubles every 5 minutes. Assuming we start with one bacterium, how many bacteria will we have at the end of 96 minutes?
We know that bacteria grow continuously, so we have to use the formula:
$latex A=A_{0}({{e}^{kt}})$
The bacteria doubles every 5 minutes, so after 5 minutes, we will have 2. We use this to find the value of k :
$latex 2=1({{e}^{5k}})$
$latex \ln(2)=\ln({{e}^{5k}})$
$latex \ln(2)=5k$
$latex k=\frac{\ln(2)}{5}$
$latex k=0.13863$
Now, we form the equation using this value of k and solve using the time of 96 minutes:
$latex A=A_{0}({{e}^{0.13863t}})$
$latex A=1({{e}^{0.13863(96)}})$
$latex A\approx602 248$
→ Exponential Equations Calculator
Practice using the exponential growth formulas with the following exercises. Solve the problems and select an answer. Check your answer to verify that you selected the correct one.
The population of a community was 12 500. After 20 years it was found that the population grew to 16 000. What will the population be after 50 years?
Choose an answer
One type of bacteria triples every 8 hours. Starting with 100 bacteria, how many will there be after 18 hours?
One type of bacteria doubles every 6.5 hours. if there were 100 bacteria at the beginning, how many will there be after 1 and a half days.
Interested in learning more about exponential functions? Take a look at these pages:
- Exponential Equations Calculator
- Exponential Decay – Formulas and Examples
- Domain and Range of Exponential Functions
- Examples of Exponential Function Problems
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EXPONENTIAL GROWTH AND DECAY WORD PROBLEMS
In this section, we are going to see how to solve word problems on exponential growth and decay.
Before look at the problems, if you like to learn about exponential growth and decay,
please click here
Problem 1 :
David owns a chain of fast food restaurants that operated 200 stores in 1999. If the rate of increase is 8% annually, how many stores does the restaurant operate in 2007 ?
Number of years between 1999 and 2007 is
n = 2007 - 1999
n = 8
No. of stores in the year 2007 = P(1+r)ⁿ
Substitute P = 200, r = 8% or 0.08 and n = 8.
No. of stores in the year 2007 = 200(1 + 0.08) 8
No. of stores in the year 2007 = 200(1.08 ) 8
No. of stores in the year 2007 = 200(1.8509)
No. of stores in the year 2007 = 370.18
So, the number of stores in the year 2007 is about 370.
Problem 2 :
You invest $2500 in bank which pays 10% interest per year compounded continuously. What will be the value of the investment after 10 years ?
We have to use the formula given below to know the value of the investment after 3 years.
A = Pe rt
Substitute
P = 2500
r = 10% or 0.1
t = 10
e = 2.71828
Then, we have
A = 2500(2.71828) (0.1)10
A = 6795.70
So, the value of the investment after 10 years is $6795.70.
Problem 3 :
Suppose a radio active substance decays at a rate of 3.5% per hour. What percent of substance will be left after 6 hours ?
Since the initial amount of substance is not given and the problem is based on percentage, we have to assume that the initial amount of substance is 100.
We have to use the formula given below to find the percent of substance after 6 hours.
A = P(1 + r) n
P = 100
r = -3.5% or -0.035
t = 6
(Here, the value of "r" is taken in negative sign. because the substance decays)
A = 100(1-0.035) 6
A = 100(0.935 ) 6
A = 100(0.8075)
A = 80.75
Because the initial amount of substance is assumed as 100, the percent of substance left after 6 hours is 80.75%
Problem 4 :
The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture initially, how many bacteria will be present at the end of 8th hour?
Note that the number of bacteria present in the culture doubles at the end of successive hours.
Since it grows at the constant ratio "2", the growth is based is on geometric progression.
We have to use the formula given below to find the no. of bacteria present at the end of 8th hour.
A = ab x
a = 30
b = 2
x = 8
Then, we have
A = 30(2 8 )
A = 30(256 )
A = 7680
So, the number of bacteria at the end of 8th hour is 7680.
Problem 5 :
A sum of money placed at compound interest doubles itself in 3 years. If interest is being compounded annually, in how many years will it amount to four times itself ?
Let "P" be the amount invested initially. From the given information, P becomes 2P in 3 years.
Since the investment is in compound interest, for the 4th year, the principal will be 2P. And 2P becomes 4P (it doubles itself) in the next 3 years.
Therefore, at the end of 6 years accumulated value will be 4P. So, the amount deposited will amount to 4 times itself in 6 years.
Related Topics
Doubling-Time Growth Formula
Half-Life Decay Formula
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Exponential Growth – Examples and Practice Problems
Exponential functions can be used to model population growth scenarios or other situations that follow patterns with growth at fixed rates. There are formulas that can be used to find solutions to most problems related to exponential growth.
Here, we will look at a summary of exponential growth and the formulas that can be used to solve these types of problems. In addition, we will look at several examples with answers of exponential growth in order to learn how to apply these formulas.
Relevant for …
Exploring examples of exponential growth.
See examples
Summary of exponential growth
Exponential growth – examples with answers, exponential growth – practice problems.
Exponential growth is a pattern of data that shows larger increases over time, creating the curve of an exponential function. For example, if a bacteria population starts with 2 in the first month, then with 4 in the second month, 16 in the third month, 256 in the fourth month, and so on, it means that the population grows exponentially with a power of 2 every month.
The following formula is used to model exponential growth. If a quantity grows by a fixed percentage at regular intervals, the pattern can be described by this function:
We recall that the original exponential function has the form $latex y = a{{b}^x}$. In the original growth formula, we have replaced b with $latex 1+ r$. So, in this formula we have:
- $latex a=$ initial value. This is the starting amount before growth.
- $latex r=$ growth rate. This is represented as a decimal.
- $latex x=$ time interval. This is the time that has passed.
Most naturally occurring events continually grow. For example, bacteria continue to grow over a 24-hour period. Bacteria don’t wait until the end of 24 hours to reproduce all at the same time.
To model the continuous growth that occurs naturally such as populations, bacteria, etc., we use the exponential e . e can be thought of as a universal constant that represents growth possibilities using a continuous process. Furthermore, using e we can also represent growth measured periodically over time.
Therefore, if a quantity is continually growing with a fixed percentage, we can use the following formula to model this pattern:
In this formula we have:
- $latex A=$ final value. This is the amount after growth.
- $latex A_{0}=$ initial value. This is the amount before growth.
- $latex e=$ exponential. e is approximately equal to 2718…
- $latex k =$ continuous growth rate. It is also called the constant of proportionality.
- $latex t =$ elapsed time.
The following examples use the formulas detailed above and some variations to find the solution. It is recommended that you try to solve the exercises yourself before looking at the answer.
A population of bacteria grows according to the function $latex f(x)=100{{e}^{0.02t}}$, where t is measured in minutes. How many bacteria will there be after 4 hours (240 minutes)?
This is continuous growth, so we have the formula $latex A=A_{0}{{e}^{kt}}$. We can recognize the following data:
- $latex A_{0}=100$
- $latex k=0.02$
- $latex t=240$
Therefore, we have:
$latex f(240)=100{{e}^{0.02(240)}}\approx 12151$
Therefore, there will be 12 151 bacteria after 4 hours.
A population of bacteria grows according to the function $latex f(x)=100{{e}^{0.02t}}$, where t is measured in minutes. When will the population reach 50 000?
Here, we have the same formula as the previous exercise, but now we have to find the time knowing the final quantity. We can recognize the following data:
- $latex A=50000$
$latex 50 000=100{{e}^{0.02t}}$
$latex 500={{e}^{0.02t}}$
$latex \ln(500)=0.02t$
$latex t=\frac{\ln(500)}{0.02}$
$latex t\approx 310.73$
Therefore, the population of bacteria will become 50 000 after 310.73 minutes.
We can model the population of a community with the formula $latex A=10000({{e}^{0.005t}})$. Here, A represents population and t represents time in years. What is the population after 10 years?
We already have a given formula: $latex A=10000({{e}^{0.005t}})$. We have to calculate the population using time $latex t=10$. Therefore, we substitute $latex t=10$ to get:
$latex A=10000({{e}^{0.005(10)}})$
$latex =10000({{e}^{0.05}})$
$latex =10000(1.0513)$
$latex =10513$
Therefore, the population in the community after 10 years will be 10 513.
The population of a certain community was 10 000 in 1980. In 2000, it was found to have grown to 20 000. Form an exponential function to model the population of community P that changes through time t .
When we have continuous population growth, we can model the population with the general formula $latex P=P_{0}({{e}^{\lambda t}})$, where $latex P_{0}$ represents the initial population, λ is the exponential growth constant and t is time.
Using the given information, we have to find the constant λ to complete the formula. Therefore, we have:
$latex P=P_{0}({{e}^{\lambda t}})$
$latex 20000=10000({{e}^{20 \lambda}})$
$latex \frac{20000}{10000}={{e}^{20 \lambda}}$
$latex 2={{e}^{20 \lambda}}$
$latex \ln(2)=20 \lambda$
$latex \frac{\ln(2)}{20}=\lambda$
$latex 0.0347=\lambda$
Thus, we can model the population growth of the community with the formula $latex P=10000({{e}^{0.0347 t}})$.
The population growth of a small city is modeled with the function $latex P= P_{0}({{e}^{0.1234t}})$. When did the population reach 37 500 if in 1980 the population was 12 500?
We can substitute the values in the formula with the given information:
$latex P=P_{0}({{e}^{0.1234t}})$
$latex 37500=12500({{e}^{0.1234t}})$
Now, we have to solve for time:
$latex \frac{37500}{12500}=({{e}^{0.1234t}})$
$latex 3=({{e}^{0.1234t}})$
$latex \ln(3)=0.1234t$
$latex \frac{\ln(3)}{0.1234}=t$
$latex 8.9=t$
Thus, the city’s population reached 37 500 in 1989.
One type of bacteria doubles every 5 minutes. Assuming we start with one bacterium, how many bacteria will we have at the end of 96 minutes?
We know that bacteria grow continuously, so we have to use the formula:
$latex A=A_{0}({{e}^{kt}})$
The bacteria doubles every 5 minutes, so after 5 minutes, we will have 2. We use this to find the value of k :
$latex 2=1({{e}^{5k}})$
$latex \ln(2)=\ln({{e}^{5k}})$
$latex \ln(2)=5k$
$latex k=\frac{\ln(2)}{5}$
$latex k=0.13863$
Now, we form the equation using this value of k and solve using the time of 96 minutes:
$latex A=A_{0}({{e}^{0.13863t}})$
$latex A=1({{e}^{0.13863(96)}})$
$latex A\approx602 248$
→ Exponential Equations Calculator
Practice using the exponential growth formulas with the following exercises. Solve the problems and select an answer. Check your answer to verify that you selected the correct one.
The population of a community was 12 500. After 20 years it was found that the population grew to 16 000. What will the population be after 50 years?
Choose an answer
One type of bacteria triples every 8 hours. Starting with 100 bacteria, how many will there be after 18 hours?
One type of bacteria doubles every 6.5 hours. if there were 100 bacteria at the beginning, how many will there be after 1 and a half days.
Interested in learning more about exponential functions? Take a look at these pages:
- Exponential Equations Calculator
- Exponential Decay – Formulas and Examples
- Domain and Range of Exponential Functions
- Examples of Exponential Function Problems
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Exponential growth population word problems
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Exponential Growth and Decay Word Problems
Exponential Growth
This algebra and precalculus video tutorial explains how to solve exponential growth and decay word problems. It provides the formulas and
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POPULATION GROWTH AND DECAY WORD PROBLEMS
Exponential Word Problems: Growth & Decay
Growth Formula: y = a (1 + r) t
Decay Formula: y = a (1 – r) t
where a = original number
r = rate (% in decimal form)
t = time periods
Write an exponential function to model each situation. Find each amount at the end of the specified time. Round your answers to the nearest whole number.
Problem 1 :
A town with a population of 5,000 grows 3% per year. Find the population at the end of 10 years.
y = a(1 + r) t
Here a = initial population, r = increasing rate and t = number of years
Initial population = 5000
Increasing rate = 3% and
number of years = 10
y = 5000(1 + 3%) 10
y = 5000(1 + 0.03) 10
y = 5000(1.03) 10
y = 5000(1.344)
Problem 2 :
The population of Boomtown is 475,000 and is increasing at a rate of 3.75% each year. When will the population exceed 1 million people (to the nearest year)?
Initial population = 475,000
Increasing rate = 3.75%
475000(1 + 3.75%) t > 1000000
475000(1 + 0.0375) t > 1000000
Dividing by 475000 on both sides.
(1 + 0.0375) t > 2.105
(1.0375) t > 2.105
To solve use equal sign, we get
t log(1.0375) = log(2.105)
t(0.015) = 0.323
t = 0.323/0.015
So, it will take 21 years tp reach the population of 1 million.
Problem 3 :
The population of Leave town is 123,000 and is decreasing at a rate of 2.375% each year.
• When will the population of Leave town drop below 50,000 (to the nearest year)?
• What will the population of Leave town be 100 years from now?
(i) y = a(1 - r) t
Initial population = 123000
When will the population become below 50000.
123000(1 - 2.375%) t < 50000
Divide by 123000 on both sides.
(1 - 0.02375) t < 0.4065
(0.97625) t < 0.4065
Take log on both sides, we get
t log (0.97625) < log (0.4065)
t(-0.0104) < -0.3909
t < 0.3909/0.0104
t < 37.58
So, it will take 38 years.
(ii) y = a(1 - r) t
After 100 years the population will be.
y = 123000(1 - 2.375%) 100
y = 123000 (0.97625) 100
y = 123000(0.09038)
Problem 4 :
Problem The 1989 population of Mexico was estimated at 87,000,000. The annual growth rate is 2.4%. When will the population reach 100,000,000 (to the nearest year)?
Population is increasing, so we will use the formula
Initial population = 87000000
Growth rate = 2.4%
After how many year the population will become 100,000,000.
87000000(1 + 2.4%) t = 100,000,000.
(1 + 0.024) t = 100,000,000/ 87000000
(1 + 0.024) t = 100,000,000/ 87000000
(1.024) t = 1.149
Taking log on both sides, we get
log (1.024) t = log (1.149)
t log(1.024) = log(1.149)
t = log(1.149)/log(1.024)
t = 0.0603/0.0102
So, the population will become 100,000,000 after 6 years.
Problem 5 :
The population of Small town in the year 1890 was 6,250. Since then, it has increased at a rate of 3.75% each year.
a) What was the population of Small town in the year 1915?
a) In 1940?
c) What will the population of Small town be in the year 2003?
d) When will the population reach 1,000,000 (to the nearest year)?
Population is increasing :
(a) Initial population a = 6250, increasing rate = 3.75%
Population will be at 1915 :
Difference in years = 1915 - 1890 ==> 25 years
y = 6250(1 + 3.75%) 25
y = 6250(1 + 0.0375) 25
y = 6250(1.0375) 25
y = 6250(2.510)
(b) In 1940 :
Difference between 1940 and 1890
= 1940-1890
y = 6250(1 + 3.75%) 50
y = 6250( 1.0375) 50
y = 6250(6.3)
So, at 1914 the population will be 39380.
c) Population at 2003 :
= 2003 - 1890
y = 6250( 1.0375) 113
y = 6250(64.07)
y = 400438.
(d) After how many the population will be become 1,000,000
1,000,000 = 6250(1 + 3.75%) t
Dividing by 6250 on both sides.
160 = ( 1.0375 ) t
log 160 = t log ( 1.0375 )
2.204 = t(0.015)
t = 2.204/0.015
t = 147
After 147 years, the population will be 1,000,000.
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Exponential growth - population problem.
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Exponential growth y=a { { (1+r)}^x} y = a(1 + r)x We recall that the original exponential function has the form y = a { {b}^x} y = abx. In the original growth formula, we have replaced b with 1+ r 1 + r. So, in this formula we have: a= a = initial value. This is the starting amount before growth. r= r = growth rate.
-- [ (√x)^2]^k can be simplified by multiply the exponents. This creates: (√x)^2k -- So, your problem becomes: log√x (√x)^2k -- The logarithm is asking you what exponent would you apply to √x to create (√x)^2k. The answer is 2k. Thus: log√x (x^2) = log√x (√x)^2k = 2k Hope this helps. Comment ( 2 votes) Upvote Downvote Flag more sanangelo9250
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