Using logarithms to solve real world problems
Interest Compounded Annually
Suppose that $10,000 is invested at 6% interest compounded annually.
In
t years an investment will grow to the amount expressed by the function
, where
t is time (in years). (See the plot in
Figure 1).
How long will it take to accumulate $20,000 in the account?
Figure 1. Graph of the function 
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Solution
You should solve an equation S(t)=20000, which is  , for unknown t.
Divide both side of this equation by the initial amount of 10000. You get an equation  .
Take logarithm base 10 from both sides. You get an equation  .
Apply the Power Rule to the logarithm. You get an equation  .
Therefore,  (approximately 12 years).
Note that this result is in agreement with the plot in Figure 1.
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Population Growth
An initial number of bacteria presented in a culture is 10000. This number doubles every 30 minutes.
1) Write a function expressed the number of bacteria in time.
2) How long it will take to get the bacteria number 100000?
Figure 2. Graph of the function 
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Solution
1) The function expressing the bacteria growth is  , where  is an initial number of bacteria in the culture, t is time in minutes. The plot for the ratio  is shown in Figure 2.
2) To estimate the time for bacteria to get the number 100000 in the culture,
you should solve an equation  for unknown t.
Divide both sides by N(0), the initial number of bacteria.
You get an equation  .
Take logarithm base 10 from both sides. You get an equation  .
Apply the Power Rule to the logarithm. You get an equation  .
Therefore,  (approximately 100 minutes).
Note that this result is in agreement with the plot in Figure 2.
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Radioactive decay
Polonium Po-210 has a half-life of 138 days.
1) Write the decay function for the amount of Polonium Po-210 that remains in a sample after t days.
2) Estimate time for Polonium Po-210 to get 0.1 of its initial amount in the sample.
Figure 3. Graph of the function 
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Solution
1) The decay function for Polonium P-210 is  , where  is an initial amount of Polonium P-210 in the sample, t is time in days. The plot for the ratio  (percentage) is shown in Figure 3.
2) To estimate time for Polonium Po-210 to get 0.1 of its initial amount in the sample,
you should solve an equation  for unknown t.
Take logarithm base 10 from both sides. You get an equation  .
Apply the Power Rule to the logarithm. You get an equation  .
Therefore,  (approximately 458.5 days).
Note that this result is in agreement with the plot in Figure 3.
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Carbon Dating
The following introductory info is from the Internet
NDT Resource Center article
Carbon-14 Dating .
Carbon (C) has three naturally occurring isotopes. Both C-12 and C-13 are stable, but C-14 is radioactive
and decays to nitrogen-14 with a half-life of approximately 5,730 years. Naturally occurring radiocarbon is
produced as a secondary effect of cosmic-ray bombardment of the upper atmosphere. Plants transpire
to take in atmospheric carbon, which is the beginning of absorption of carbon into the food chain.
Animals eat the plants and this action introduces carbon into their bodies.
As long as the animal is alive, the ratio of C-14 to C-12 in its body remains constant.
After the animal dies, carbon-14 continues to decay without being replaced.
The current amount of radioactive Carbon-14 present in the remains of animal bones can be measured,
and the ratio of the current amount of Carbon-14 to its initial amount can be used to determine age.
The last ratio is described by the function
, where
t is time (in years). (See the percentage plot in
Figure 4).
How old is an animal bone that has lost 40% of its initial Carbon-14?
Figure 4. Graph of the function
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Solution
Since 40% of Carbon-14 is lost, 60% is remained, or 0.6 of its initial amount.
Thus, you should solve an equation C(t)=0.6, which is  , for unknown t.
Take logarithm base 10 from both sides. You get an equation  .
Apply the Power Rule to the logarithm. You get an equation  .
Therefore,  (approximately 4200 years).
Note that this result is in agreement with the plot in Figure 4.
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