Independent Practice with Common Log Word Problems

1)Use algebraic reasoning with logarithms to solve the following equations for x.

  1. log x = 2
  2. 15 = 10x
  3. 5(10)2x = 60
  4. 103x – 1 = 100,000

2)When archaeologists discover remains of an ancient civilization, they use a technique called carbon dating to estimate the time when the person or animal died or when the artifact was made from living material.

The amount of radioactive Carbon-14 in such an artifact decreases exponentially according to the function C(t) = 100(10-0.00005255t), where t is time in years and C(t) is the percent of the Carbon 14 present in the artifact when it was last living material.

  1. What is the half-life of Carbon-14, the time when only 50% remains from an original amount?
  1. Suppose that a skeleton is discovered that has only 10% of the Carbon-14 that one finds in living animals. When was that skeleton part of a living animal?

3)The Washington Nationals baseball team was purchased in 2006 for 450 million dollars. If the value of this investment grows at a rate of 5% compounded yearly, the purchase price of the team in 2006 + t will be given by V(t) = 450(100.021t).

  1. Explain how you can be sure that this function gives the correct value of the investment in 2006.
  1. Use the function to estimate the value of the investment in 2010.
  1. Use logarithms and other algebraic reasoning to estimate the time when the value of the investment will be $1 billion ($1,000 million).

4)Suppose that the average rent for a two-bedroom apartment in Indianapolis is currently $750 per month and increasing at a rate of 8% per year. The function R(t) = 750(100.033t) provides a model for the pattern of expected increase in monthly rent after t more years.

  1. Explain how you know that R(0) = 750.
  1. Find the time when the average rent for a two-bedroom apartment will be $1,000 per month, if inflation continues at the current 8% rate. Show how to find that time in two ways – one by estimation using a table or graph of R(t) and another using logarithms and algebraic reasoning.

5)If an athlete tries to improve his or her performance by taking an illegal drug, the amount of that drug in his or her blood will decline exponentially over time, but tests are quite sensitive to small amounts.

For example, a 200 mg dose of a steroid might decay so that the amount remaining after t days is given by the rule s(t) = 200(10-0.046t).

  1. Explain how you know that s(0) = 200.
  1. Estimate the time when only 5 mg of the steroid remains active in the athlete’s blood. Show how to find that time in two ways – one by estimation using a table or graph of s(t) and another using logarithms and algebraic reasoning.

6)Suppose that 500 mg of a medicine enters a hospital patient’s bloodstream at noon and decays exponentially at a rate of 15% per hour.

  1. Use the exponential function D(t) = 500(10-0.07t) to predict the amount of medicine active in the patient’s blood at a time 5 hours later, where t is time in hours.
  1. Find the time when only 5% (25 mg) of the original amount of medicine will be active in the patient’s body. Show how to find that time in two ways – one by estimation by using a table or graph of D(t) and another using logarithms and algebraic reasoning.

Adapted from Core Plus Mathematics 2, Unit 5, Investigation 2, pp. 381 – 386.

Solutions:

1a. x = 100

  1. Log 15 = x, so x  1.18
  2. 102x = 12

2x = log 12

X  0.54

  1. 3x – 1 = 5

X = 2

2a. 50 = 100(10-0.00005255t) when t  5,728 years.

  1. 10 = 100(10-0.00005255t) when t  19,029 years.

3. V(0) = 450(100.021(0)) = 450(100) = 450 because 100 = 1

  1. V(4) = 450(100.021(4)), so V(4) = 546.02
  2. 1,000 = 450(100.021t) when t = 16.5 or 6 months into the year 2022.

4a. R(0) = 750(100.033(0)) = 750(100) = 750 because 100 = 1

  1. 1,000 = 750(100.033t) when t  4 years. This value can be estimated by scanning a table or graph of the function R(t) = 750(100.033t) to find values of t when s(t) = 1,000 or by reasoning algebraically.

If 1,000 = 750(100.033t)

Then 1.33 = 100.033t

Log 1.33 = 0.033t

3.75  t

5a. s(0) = 200(10-0.046(0)) = 200(100). Thus, because 100 = 1, s(0) = 200.

  1. 5 = 200(10-0.046t) when t  35 days. This value can be estimated by scanning a table or graph of the function s(t) = 200(10-0.046t) to find values of t when s(t) = 5 or by reasoning algebraically.

If 5 = 200(10-0.046t),

Then 0.025 = 10-0.046t

Log 0.025 = -0.046t

34.8  t

6a. D(5) = 500(10-0.07(5)) = 500(10-0.35)  223. So, only about 223 mg of the drug remain active 5 hours after the initial injection.

  1. 5% of 500 is 25, so we need to find t when 25 = 500(10-0.07t). This occurs when t  18 hours. This value can be estimated by scanning a table or graph of the function D(t) = 500(10-0.07t) to find values of t when D(t) = 500 or by reasoning algebraically.

If 25 = 500(10-0.07t),

Then 0.05 = 10-0.07t

Log 0.05 = -0.07t

18.6  t

Adapted from Core Plus Mathematics 2, Unit 5, Investigation 2, pp. 381 – 386.