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  • Elementary Functions Part 3, Exponential Functions & Logarithms

    Lecture 3.6a, Applications of Logarithms

    Dr. Ken W. Smith

    Sam Houston State University

    2013

    Smith (SHSU) Elementary Functions 2013 1 / 21

    Using logarithms to solve Newton’s Law of Cooling

    Recall Newton’s Law of Cooling in which the ratio of differences in temperature decays exponentially:

    T (t)−Ta T0−Ta = e

    −kt

    Here k is a constant that depends on the liquid and the environment. We solve for the temperature T at time t.

    T (t) = Ta + (T0 − Ta)e−kt

    (Here we use the notation of the textbook by Stitz and Zeager.) In most applications of this law, we need to solve for k and/or t. Since both k and t are exponents, we must use logarithms.

    Smith (SHSU) Elementary Functions 2013 2 / 21

    Applications of logarithms

    A worked example. Suppose the water in a hot tub is heated to 150◦. After the heater is turned off, the hot tub takes an hour to cool to 120◦. The temperature of the surrounding air is 80◦. Use Newton’s Law of Cooling:

    T (t) = Ta + (T0 − Ta)e−kt

    to find the temperature of the hot tub after 3 hours.

    Solution. At time t = 1 we have

    120 = 80 + (150− 80)e−k

    Subtracting 80 from both sides and then dividing by 70 gives

    40 = 70e−k

    4 7 = e

    −k.

    Taking the natural log of both sides we have

    ln(47) = −k

    k = − ln(4/7) = ln(7/4) .Smith (SHSU) Elementary Functions 2013 3 / 21

    Applications of logarithms

    So the equation for T at time t is

    T (t) = 80 + (150− 80)et ln(4/7).

    We may simplify this by replacing (150− 80) by 70. We can also (if we wish) replace et ln(4/7) by (eln(

    4 7 ))t = (47)

    t so that

    T (t) = 80 + 70et ln(4/7) = 80 + 70(47) t.

    At time t = 3,

    T (3) = 80 + 70(4/7)3 ≈ 80 + 13.0612 = 93.0612

    Smith (SHSU) Elementary Functions 2013 4 / 21

  • Applications of logarithms

    Now let us suppose that we want to know the time t this hot tub will cool down to 85◦.

    Solution. Here we need to solve for t in the equation

    85 = 80 + 70(47) t.

    Subtract 80 from both sides and divide by 70 and simplify 570 = 1 14 so that

    1 14 = (4/7)

    t

    Taking the natural log of both sides we have

    ln( 114) = ln(( 4 7)

    t).

    Rewriting the lefthand side as ln( 114) = ln 1− ln 14 = − ln 14 and rewriting the righthand side as ln(47

    t ) = t ln 47 gives us

    − ln 14 = t ln 47 .

    So t = − ln 14 ln 4

    7

    = ln 14 ln 7

    4

    ≈ 4.7158 hours. Smith (SHSU) Elementary Functions 2013 5 / 21

    Radioactive half-life

    A radioactive isotope decays at random, but with a probability computable over an interval of time. The proportion of radioactive material that decays in a particular interval depends on the isotope. Since the amount of material that decays is proportional to the total amount of material, radioactive decay is an example of exponential decay. Here is an example. A tree that died in the forest 83 years ago is examined for carbon 14. We assume that the ratio amount of carbon 14 in the tree when it died is compatible with that in the atmosphere at the time but now, 83 years later, we discover that one percent of the carbon 14 has decayed to nitrogen 14. We can model this by an exponential decay equation:

    P (t) = P0e −kt

    where P (t) represents the mass of carbon 14 at time t and P0 represents the mass of carbon 14 at time t = 0. (Here we assume that k is positive, so that the exponential function is decaying, not growing.) In this case, with t = 83, we have that P (t) = 0.99P0. So our equation is

    0.99P0 = P0e −83k.Smith (SHSU) Elementary Functions 2013 6 / 21

    Radioactive half-life

    We discovered that

    0.99P0 = P0e −83k.

    We may divide both sides of this equation by P0 to obtain

    0.99 = e−83k

    and then take the natural log of both sides

    ln(0.99) = −83k.

    Therefore k = − ln 0.9983 ≈ 0.000121 So the amount of carbon 14 t years after the death of the specimen can be approximated by

    P (t) = P0e −0.000121t.

    Smith (SHSU) Elementary Functions 2013 7 / 21

    Applications of logarithms

    Sample Problem. Suppose the amount of carbon 14 in a specimen can be measured by the equation above. From the equation, what is the half-life of carbon 14?

    Solution. In this case, we solve for t in the equation

    1 2P0 = P0e

    −0.000121t.

    Dividing both sides by P0 we have 1 2 = e

    −0.000121t.

    Taking the natural log of both sides we have

    ln(12) = −0.000121t, or (using the power rule of logs)

    − ln(2) = −0.000121t,

    t = ln(2)0.000121 ≈ 5728.5years.

    (The correct half-life for carbon 14 (according to Wikipedia is 5730± 40 years, so this answer is compatible with our understanding of carbon 14.)

    Smith (SHSU) Elementary Functions 2013 8 / 21

    http://en.wikipedia.org/wiki/Carbon-14

  • Applications of logarithms

    In the next presentation we will continue with applications of logarithms to doubling times of investments and similar problems where we are, once again, solving for an exponent.

    (END)

    Smith (SHSU) Elementary Functions 2013 9 / 21

    Elementary Functions Part 3, Exponential Functions & Logarithms

    Lecture 3.6b, Applications of Logarithms: Doubling Time

    Dr. Ken W. Smith

    Sam Houston State University

    2013

    Smith (SHSU) Elementary Functions 2013 10 / 21

    Doubling time for an investment

    Just as radioactive isotopes decay exponentially, investment of money typically grows in proportion to the money invested and so financial investments are an example of exponential growth. In financial investments, we are often interested in the “doubling time” of an investment. It this case we solve for time where the future value of the investment is 2P0 where P0 is the amount of our initial investment. Here are some sample problems.

    Smith (SHSU) Elementary Functions 2013 11 / 21

    Doubling time for an investment

    Worked problems on doubling time. Find the doubling time for the following investments.

    1 An annual interest rate of 9%, compounded monthly. Solution. Here the monthly interest rate is r = 0.09/12 = 0.0075 and time n is measured in months, the length of the interest period. We solve for the exponent n in the equation

    2P0 = P0(1.0075) n,

    or, after dividing by P0,

    2 = (1.0075)n.

    Taking the natural log of both sides we have that

    ln 2 = ln((1.0075)n).

    Using properties of logs, we may rewrite this as

    ln 2 = n ln(1.0075)

    n = ln 2ln 1.0075 ≈ 92.766. Here time is measured in months so the doubling time is almost 93 months or 7 years and 9 months . Smith (SHSU) Elementary Functions 2013 12 / 21

  • Doubling time for an investment

    2 An annual interest rate of 24%, compounded monthly. Solution. We solve for n in the equation

    2P0 = P0(1.02) n.

    (Here r = 0.24/12 = 0.02 and time n is measured in months, the length of the interest period.) Dividing by P0 and taking the natural log of both sides we have that

    ln 2 = ln((1.02)n).

    Using properties of logs, we may rewrite this as

    ln 2 = n ln(1.02).

    and so

    n = ln 2ln 1.02 ≈ 35.0028. Since time here is measured in months then the doubling time is 35 months or 2 years and 11 months .

    Smith (SHSU) Elementary Functions 2013 13 / 21

    Doubling time for an investment

    3 An annual interest rate of 9%, compounded continuously. Solution. We solve for t in the equation 2 = e0.09t where t is time measured in years. Taking the natural log of both sides we have that ln 2 = 0.09t and so t = ln 20.09 =

    100 ln 2 9 which is about 7.702 years .

    Notice that this answer is close to the answer in problem 1.

    Smith (SHSU) Elementary Functions 2013 14 / 21

    Doubling time for an investment

    4 An annual interest rate of 24%, compounded continuously. Solution. We solve for t in the equation 2 = e0.24t. Taking the natural log of both sides we have that ln 2 = 0.06t and so t = ln 20.24 =

    100 ln 2 24 which is about 2.88 years . Notice that this answer

    is close to the answer in problem 2.

    Smith (SHSU) Elementary Functions 2013 15 / 21

    The Rule of 72

    The Rule of 72 In general, the doubling time for an investment solves for t in the exponential equation

    2 = (1 + r)t.

    If we apply the natural log to this equation, we find that

    t = ln 2ln(1+r) .

    For interest rates close to 8%, this formula can be accurately approximated by the formula

    f(t) = 72

    100r .

    In other words, the doubling time is about what one would get by taking the interest rate, in percents, an dividing it into 72. Since it is fairly easy to divide numbers into 72, this rough rule of thumbs shows up in business discussions as “the rule of 72.”

    For example, when does a 4% interest rate double? Since 724 = 18, a 4% rate should double in about 18 years.

    Smith (SHSU) Elementary Functions 2013 16 / 21

  • Applications of logarithms

    Use the Rule of 72 to approximate the following:

    1 The doubling time of a 3% investment,

    2 The doubling time of an 8% investment,

    3 The doubling time of a 9% investment,

    4 The doubling time of a 24% investment,

    Solutions.

    1 Using the Rule of 72 we estimate that a 3% investment should double