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Showing posts with label logarithm. Show all posts
Showing posts with label logarithm. Show all posts

## Saturday, July 27, 2013

### Exponential Growth and Decay

In this section we will solve typical word problems that involve exponential growth or decay.
If k is positive then we will have a growth model and if k is negative then we will have a decay model.

Use the exponential growth/decay model to answer the questions.

Example: A certain bacterium has an exponential growth rate of 25% per day.  If we start with 0.5 gram and provide unlimited resources how much bacteria can we grow in 2 weeks?

Answer: We can grow 16.56 grams of bacteria in 2 weeks time.

Typically the exponential growth rate will not be given.  In this case, we must determine that before we can use the model to answer the question.

Step 1: Use the given information to calculate the growth/decay rate k.
Step 2: Substitute the initial amount and k to formulate a model.
Step 3: Us the model to answer the question.

Example: During its exponential growth phase, a certain bacterium can grow from 5,000 cells to 12,000 cells in 10 hours.  At this rate how many cells will be present after 36 hours?

Answer: About 116,877 cells in 36 hours time.

Tip: Use the exact value for k and avoid round-off error.  If we use the approximate rounded-off value for k we will compound the error by rounding off again at the end when calculating the final result.

Example: During its exponential growth phase, a certain bacterium can grow from 5,000 cells to 12,000 cells in 10 hours.  At this rate how long will it take to grow to 50,000 cells?

Answer: It will take approximately 26.3 hours.

Example: A certain animal species can double its population every 30 years. Assuming exponential growth, how long will it take the population to grow from 40 specimens to 500?

Answer: it will take about 109.3 years.

Up to this point, we have seen only exponential growth.  We will conclude this section with some exponential decay applications.  Often exponential rate of decay can be determined from the half-life information.  Half-life is the amount of time it takes for a substance to decay to half of the original amount.

Example: A certain isotope has a half-life of 4.2 days. How long will it take a 150-milligram sample to decay so that only 10 milligrams are left?

Answer: It will take about 16.4 days.

Example: The half-life of carbon-14 is 5730 years. If it is determined that an old bone contains 85% of it original carbon-14 how old is the bone?

Answer: It takes about 1,343.5 years for a bone to lose 15% of its carbon-14.

To summarize, first find the growth/decay rate. Put together a mathematical model using the initial amount and the exponential rate of growth/decay.  Then use the model to answer the question.

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## Friday, July 26, 2013

### Interest Problems

In this section we cover compound interest and continuously compounded interest.
Use the compound interest formula to solve the following.

Example: If a \$500 certificate of deposit earns 4 1/4% compounded monthly then how much will be accumulated at the end of a 3 year period?

Answer: At the end of 3 years the amount is \$576.86.

Example: A certain investment earns 8 3/4% compounded quarterly.  If \$10,000 is invested for 5 years, how much will be in the account at the end of that time period?

Answer: At the end of 5 years the account have \$15,415.42 in it.

The basic idea is to first determine the given information then substitute the appropriate values into the formula and evaluate.  To avoid round-off error, use the calculator and round-off only once as the last step.

• Annual  n = 1
• Semiannual n = 2
• Quarterly n = 4
• Monthly n = 12
• Daily n = 365

One important application is to determine the doubling time.  How long does it take for the principal in an account earning compound interest to double?

Example: How long does it take to double \$1000 at an annual interest rate of 6.35% compounded monthly?

Answer:  The account will double in approximately 10.9 years.

The key step in this process is to apply the common logarithm to both sides so that we can apply the power rule and solve for time t.  Use the calculator in the last step and round-off only once.

Example: How long will it take \$30,000 to accumulate to \$110,000 in a trust that earns a 10% return compounded semiannually?

Answer: Approximately 13.3 years.

Example: How long will it take our money to triple in a bank account with an annual interest rate of 8.45% compounded annually?

Answer: Approximately 13.5 years to triple.

Make a note that doubling or tripling time is independent of the principal. In the previous problem, notice that the principal was not given and that the variable P cancelled.
Use the continuously compounding interest formula to solve the following.

Example: If a \$500 certificate of deposit earns 4 1/4% annual interest compounded continuously then how much will be accumulated at the end of a 3 year period?

Answer: the amount at the end of 3 years will be \$576.99.

Example: A certain investment earns 8 3/4% compounded continuously.  If \$10,000 dollars is invested, how much will be in the account after 5 years?

Answer: The amount at the end of five years will be \$15,488.30.

The previous two examples are the same examples that we started this chapter with.  This allows us to compare the accumulated amounts to that of regular compound interest.

As we can see, continuous compounding is better, but not by much.  Instead of buying a new car for say \$20,000, let us invest in the future of our family.  If we invest the \$20,000 at 6% annual interest compounded continuously for say, two generations or 100 years, then how much will our family have accumulated in that time?
The answer is over 8 million dollars. One can only wonder actually how much that would be worth in a century.

Given continuously compounding interest, we are often asked to find the doubling time.  Instead of taking the common log of both sides it will be easier take the natural log of both sides, otherwise the steps are the same.

Example: How long does it take to double \$1000 at an annual interest rate of 6.35% compounded continuously?

Answer: The account will double in about 10.9 years.

The key step in this process is to apply the natural logarithm to both sides so that we can apply the power rule and solve for t.  Use the calculator in the last step and round-off only once.

Example: How long will it take \$30,000 to accumulate to \$110,000 in a trust that earns a 10% annual return compounded continuously?

Answer: Approximately 13 years.

Example: How long will it take our money to triple in a bank account with an annual interest rate of 8.45% compounded continuously?

Answer: Approximately 13 years.

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### Graphing Logarithmic Functions

One way to graph logarithmic functions is to first graph its inverse exponential. Then use your knowledge about the symmetry of inverses to graph the logarithm. Recall that inverses are symmetric about the line y = x. For example,
Step 1: Find some points on the exponential f(x). The more points we plot the better the graph will look.

Step 2: Switch the x and y values to obtain points on the inverse.

Step 3: Determine the asymptote.

In practice, we use a combination of techniques to graph logarithms.  We can use our knowledge of transformations, techniques for finding intercepts, and symmetry to find as many points as we can to make these graphs. General guidelines follow:

1. Graph the vertical asymptote. All logarithmic functions of the form
have a vertical asymptote at x = h.

2. Find the x- and y-intercepts if they exist. To find x-intercepts set y = f(x) to zero and to find y-intercepts set x = 0.

3. Plot a few more points and graph it.

Graph the following logarithmic functions. State the domain and range.

In the previous solved problem, make a note of the rigid transformations.  If we start with the basic graph y = log(x) then the first translation is a shift to the left 3 units y = log(x+3).  Next we see a vertical shift up 2 units y = log(x+3)+2 .

In the above problem there was a reflection about the x-axis as well as a shift to the left 3 units.

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