Applications of Logarithmic Functions

8/6/2019 Applications of Logarithmic Functions

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Applications of Logarithmic

Functions-- Section 4.4 --

Annual Growth

Rate

An investment is growing continuously for t years.P = current value, P0 = original amount

A $10,000 stock investment in 1993 was worth $22,400 in 2000. Find theaverage annual growth rate of the stock.

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Annual GrowthRate

Find your ln key. (Natural log)

Identify each part HINT: use these key strokes

P= 22,400 current amount =22,400

1 7 ln (2.24) [enter or =]

Po= 10,000initial amount =10,000

Some calculators do notrequire the

parentheses

t= 7t = 2000 - 1993 =7

Annual GrowthRate

Interpret The investment earn about 12% a year for 7 years.

A $5,000 stock investment in 1992 was worth $32,500 in 1998. Find theaverage annual growth rate of the stock.

[Solution]

If $10,000 is invested each year in an annuity earning 10% annual interest,when will the account be worth $1,000,000?


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*Annuities:

Pis amount invested each year.r is the annual interest rate.A is the amount aftern years.

Solve Step Calculator Help

annuity formula Find your log key. (common log)

Identify each part HINT: use these key strokes

P= 10,000invested eachyear

log(1000000.110000 + 1)log(1 + .1) [enteror =]

r= 0.1

annual interest

rate pare

ntheses

are

nee

d here

A = 1,000,000amount after nyears

plug in theparts

and simplify

Interpret It will take about 25 years to have 1,000,000 by investing 10,000 a year at10% annual interest.

*An annuity is an invested or loan with a fixed payment and interest rate. There can be

variations.

If $5,000 is invested each year in an annuity earning 8% annual interest,when will the account be worth $500,000?

Calculate pH of a solution: pH=-log[H+].


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Exponential and

Logarithmic Models

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Contents: This page corresponds to 4.5 (p. 359) of the text.

Suggested problems from text:

p. 366 #7,9,15,17,25,33,35,69

Exponential Growth and Decay

Other Models

Exponential Growth and Decay

Two common types of mathematical models are

Exponential Growth: y = a e bx, b > 0.

Exponential Decay: y = a e -bx, b > 0.

Example 1.

During the 1980s the population of a certain city went from 100,000 to 205,000. Populations byyear are listed in the table below.

Year 1980 1981 1982 1983 1984 1985 1986 1987 1988 1989

Population in thousands 100 108 117 127 138 149 162 175 190 205

This data is approximated well by the exponential growth model P = 100 e0.08t, where t is thenumber of years since 1980. In other words, the year 1980 corresponds to t = 0, 1981

corresponds to t = 1, etc. The data points and model are graphed below.


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Population data points and model P = 100 e0.08t

where t is number of years since 1980.

Problem 1: Use the model to predict the population of the city in 1994.

1991 corresponds to t = 11, so our model predicts that the population will be

P = 100 e0.08*11 = 241 thousand.

Problem 2: According to our model, when will the population reach 300 thousand?

To solve this problem we set 100 e0.08t

equal to 300 and solve for t.

100 e0.08t = 300

e0.08t = 3 Take the natural logarithm of both sides.

ln e0.08t = ln 3

0.08t = ln 3

t = (ln 3)/0.08 = 13.73, approximately.

Therefore, the population is expected to reach 300 thousand about three fourths of the waythrough the year 1993.

It is important to recognize the limitations of this model. While it is obvious from the graph thatfor t between 0 and 9, the model values are very close to the actual population values, we should


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not assume that our model will give an accurate prediction of population for values of t muchlarger than 9. For instance, the model predicts that in the year 2080 (t = 100), the population ofthe city will be almost 300 million! That is not likely.

Suppose we know that a variable y can be expressed in the form aebx, but we don't know the

values a and b. If we are given any two points on the graph of y, then it is possible to find thenumbers a and b. The simplest case, and one that is often encountered in applications, is wherewe know the value of y when x = 0 and one other point on the graph of y.

Example 2.

Assume that a population P is growing exponentially, so P = aebt, where t is measured in years.

If P = 15000 in 1990, and P has grown to 17000 in 1993, find the formula for P.

Let t be the number of years since 1990. Then a = 15000, the value of P when t = 0.

Note that t could have been chosen differently. For instance, we could let t be the number ofyears since 1900. But then we would not know the value of P when t = 0, so we would not knowthe value of a immediately.

We still need to find b. All we know about b is that it is positive, since the population is growing.Using the value we have found for a, we have

P = 15000 ebt.

The year 1993 corresponds to t = 3, so we substitute P = 17000 and t = 3 in the equation above

and solve for b.

17000 = 15000 e3b

We have solved equations of this form several times. The first step is to isolate the exponentialterm. Then take the natural logarithm of both sides.

17/15 = e3b

ln (17/15) = ln e3b

ln (17/15) = 3b

b = (ln(17/15))/3 = 0.0417 (approximately)

Therefore P = 15000 e0.0417 t

Based on this model, when will the population reach 20000?


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Set P equal to 20000 and solve for t.

20000 = 15000 e0.0417 t

4/3 = e0.0417 t

ln (4/3) = ln e0.0417 t

ln (4/3) = 0.0417 t

t = (ln (4/3))/0.0417 = 6.9 years

So, we expect the population to reach 20000 toward the end of 1996.

Exercise 1:

Find a model of the type P = ae

bt

, where t is the number of years since 1970, if P = 30000 in1970 and P = 36000 in 1977. Use this model to predict the value of P in 1980.

Answer

When the coefficient of x (or whatever the independent variable is named) is negative, then weare modeling a decreasingvariable. This is called exponential decay. We will illustrateexponential decay by considering a radioactive substance. A sample of a radioactive substancedecays with time.

Example 3.

The mass (in grams) of radioactive material in a sample is given by N = 100e-0.0017t, where t ismeasured in years.

Find the half-life of this radioactive substance.

The half-life of a radioactive substance is the amount of time required for half of a give sampleto decay. Note that half-life is independent of the size of the sample. If the half-life of a certainradioactive material is 700 years, then if the initial mass of the sample is 1000 grams, in 700years there will be 500 grams. If the initial mass of the sample is only 8 grams, in 700 yeas therewill be 4 grams.

In this example, the mass of the radioactive material is 100 grams at time t = 0. Therefore thehalf-life is the amount of time necessary for the sample to decay to 50 grams. So we can find thehalf-life by setting N equal to 50 and solving for t.

100e-0.0017t = 50

e-0.0017t = 0.5


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Note: If the initial amount had been 800 grams, then we would have to solve the equation 800e-0.0017t=400, and after dividing both sides by 800 we would have e-0.0017t = 0.5, which is the sameas the equation above. That is why half-life is independent of initial quantity.

ln e-0.0017t = ln 0.5

-0.0017t = ln 0.5

t = (ln 0.5)/ -0.0017 = 408 years (approximately)

Question: Why is (ln 0.5)/-0.0017 equal to (ln 2)/0.0017 ?

In an earlier section we discussed an important example of exponential growth, namelycontinuous compounding of interest. Computing the half-life of a radioactive substance is verysimilar to computing the doubling time for an investment.

Example 4.

If $1000 is invested at 9% annual interest compounded continuously, how long will it take forthe investment to double? Using the compound interest formula A = Pert, we have

A = 1000e0.09t

We want to find the amount of time it will take for the investment to double, that is grow to$2000, so we set A equal to 2000 and solve for t.

1000e0.09t = 2000 Divide both sides by 1000.

e0.09t = 2

Note: Just like half-life, doubling time is independent of the initial investment P. If we hadstarted with $50 and asked how long it will be before we have $100, then we would have solvedthe equation

50e0.09t = 100, and after dividing both sides by 50 we would again have e0.09t = 2.

ln e0.09t = ln 2

0.09 t = ln 2

t = (ln 2)/0.09 = 7.7 years (approximately).

Question: If you invest $3000 at 9% compounded continuously, about how much will you havein 15 and a half years?


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Answer: Around $12000. This is easy to approximate because doubling time is 7.7 years, so 15.5years corresponds to just over two doublings. 3000 -> 6000 -> 12000.

Exercise 2:

How long does it take an investment to triple at %11 compounded continuously? Answer

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Other Models

Example 5.

Consider the following data points.

x 1 2 3 4 5 6 7

y 0.5 2.6 3.6 4.5 4.9 5.4 5.7

When we plot these points we see that they do not seem to lie on any exponential curve, but theshape is very much like a logarithmic graph. This suggests that a logarithmic model isreasonable. The graph below shows the data points and the function y = 0.63 + 2.7 ln x whichfits the data points quite well.

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Line & Area: Logarithmic scale chart

Logarithmic scaleYou can have the Y-axis on a logarithmic scale instead of a linear one. On a logarithmic axis,equal distance along it represents an equal percentage change. It is very useful if you want topresent data with large differences in scale on the same chart. A logarithmic scale is also widelyused for showing stock price changes over a longer period of time. Column & Bar and Scatter &Bubble charts can also have a logarithmic axis.

Rotated axis values

Both X and Y-axis values can be rotated by a particular degree, so that more values can fit.

SAO Maths Primer

Mathematics Primer: The Logarithmic Scale

LogarithmUse of the logarithmic scale

Logarithm is the exponent or power to which a base (number) must be raised to yield agiven number.

An example of a logarithm is as follows. In the expression above, if the base

bis equalto 10 and N is a number equal to 100, thenxis equal to 2 and is said to be the

logarithm of 100 to the base 10.This is written: log 100 = 2, in which it is understood that log means logarithm to thebase 10. Logarithms to the base 10 are also called a common logarithm.


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The problem is, that when we draw a graphof Money versus Age, the detail near thebeginning isn't clear.

This is because the vertical scale has to goup to $100,000, so amounts like $1, $5 and$10 are too tiny to show.

The solution is to draw the graphic againbut this time using a logarithmic verticalscale, where you increment the y-axis byfactors of ten (i.e. axis units are 10, 100,1,000 ...)

This is just like expressing the vertical scaleas the logarithm in base 10 of the moneyowned (i.e. y = log(Money)).

Any scale that increases by multiplyingrather than by adding is called alogarithmic scale.

Using logarithmic scales is particularly useful when drawing graphs in astronomy, whenamounts that need to be plotted can greatly vary in size.


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For example, this graphic shows theluminosity-period relation for 1500

variable stars in the Large MagellanicCloud.

As you can see the horizontal axisgives the period (P) using alogarithmic scale. If you didn't draw itthis way you wouldn't be able toappreciate variations for small valuesof P, just like in our example aboveyou couldn't see the differencebetween amounts like $1, $5 and

$10.

Applications of Logarithmic Functions

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