High School Algebra II Unlocked (2016)

Lesson 5.5. Logarithms in the Real World

Any exponential relationship is, when viewed in the other direction, a logarithmic relationship. So, any real-life situation involving exponential growth (such as growth of certain populations) or decay (such as certain rates of erosion) can be represented by a logarithmic function.

Scientists have studied and described the exponential decay of various substances in terms of their half-life, which is the amount of time it takes half of an original quantity of the substance to decay. The formula N = represents this relationship, where A represents the original quantity of the substance, N represents the quantity of the substance left after an amount of time, t, and t_1/2 represents the half-life of that particular substance.

The ratio is the total
amount of elapsed
time divided by the
amount of time it
takes the substance
to reduce by half,
so this ratio is really
just the number of
times the substance
reduces by half in that
time period. So, the
original amount, A,
is halved (multiplied
by 1/2) repeatedly,
this number of times,
to end up with N, the
remaining amount.

Iodine-131 has a half-life of about 8 days. Write an equation that gives the amount of time, in days, elapsed since an iodine-131 sample weighed 10 milligrams, given its current weight in milligrams. Using technology, graph both the exponential function and the logarithmic function that represent this relationship. Solve for the time it takes the 10-milligram sample to reduce to 1.25 milligrams.

The relationship between the original quantity of an iodine-131 sample, A, and its quantity, N, after t seconds is also given by the formula N = A⋅e^{−10−6 ∙ t}. Rewrite this formula to solve for t as a function of N, then show that the resulting function produces the same answer for the amount of time it takes a 10-milligram sample to reduce to 1.25 milligrams.

Notice that time is
defined in seconds in
this formula, unlike the
first equation we will
write, which uses days.

We’ll use the half-life formula, N = , for the first equation, with time t expressed in days. Iodine-131 has a half-life of about 8 days, so t_1/2 = 8. The original amount of the sample is 10 milligrams, so A = 10. Our exponential equation is therefore N = 10 ⋅ (1/2)^t/8. Let’s solve this equation for t in terms of N.

When using specific
variables to represent
certain quantities, we
solve for the inverse
function by solving
for the other variable.
We do not switch
variables the way we
do with x and y.

The graphs of N = 10 ⋅ (1/2)^t/8 and t = −8log₂(N/10) are shown below.

This logarithmic
function can also be
written as
t = −8(log₂N − log₂10)
using the logarithmic
quotient property and
then simplified as
t = −8log₂N + 8log₂10.

The exponential function graph may seem more intuitively logical, because time increases along the horizontal axis from left to right, but the logarithmic function graph also accurately represents the relationship. You can see from the graph that, for a quantity,N, of 10 milligrams, the time elapsed is 0 days, and for a quantity, N, of 5 milligrams, the time elapsed is 8 days. This fits with the fact that iodine-131 has a half-life of 8 days; half of the original sample size is left after 8 days.

Notice that, as with
any pair of inverse
functions, these two
graphs are reflections
of one another across
the diagonal line
y = x. Any point (t, N)
in the graph on the left
has a corresponding
point (N, t) in the
graph on the right.

To solve for the time it takes the 10-milligram sample to decay to a remaining mass of 1.25 milligrams, substitute 1.25 for N in t = −8log₂(N/10).

t = −8log₂(1.25/10)

We could expand
either graph to
see what t-value
corresponds to an
N-value of 1.25. Zoom
in on a graphing
calculator for a pretty
accurate answer.

There will be 1.25 milligrams left of the sample after 24 days.

The decimal 0.125 is
equal to 1/8, or 8⁻¹,
which is the same
as 2⁻³. So, log₂0.125
= log₂2⁻³ = −3.

The second formula is N = A⋅e^{−10−6 ∙ t}, where t is time elapsed, in seconds. Again, A = 10, so the equation is N = 10 ⋅ e^{−10−6 ∙ t}. Let’s solve this equation for t.

To solve for the time it takes the sample to decay to just 1.25 milligrams, substitute 1.25 for N in this new formula and solve for t.

10⁻⁶ is 0.000001,
so the formula can
also be written as
N = 10e^−0.000001t.

The exponent
10⁻⁶ ⋅ t, or 0.000001t,
is the opposite,
of −10⁻⁶ ⋅ t, or
−0.000001t. The −6
in the exponent of
10 only influences
the magnitude of
the coefficient of t,
so it is not affected
when you rewrite

10⁻⁶ ⋅ 10⁶ =
10^{−6 + 6} = 10⁰ = 1, so
(10⁻⁶ ⋅ t) ⋅ 10⁶ = t.

The 10-milligram sample decays to 1.25 milligrams in about 2,079,441 seconds. Let’s convert seconds to days, to compare this answer with the one we got using the first formula.

In other words, we must divide 2,079,441 by 60 to get the number of minutes, divide again by 60 to get the number of hours, and divide by 24 to finally find the equivalence in days.

This is the same answer (24 days) that we got using the formula t = −8log₂(N/10).

DRILL

CHAPTER 5 PRACTICE QUESTIONS

Click here to download a PDF of Chapter 5 Practice Questions.

Directions: Complete the following open-ended problems as specified by each question stem. For extra practice after answering each question, try using an alternative method to solve the problem or check your work.

1. Evaluate each of the following without a calculator:

(a) log₅ 125

(b) log₈ 64

(c) log₃

(d) log₉ 9

(e) log

2. For each of the following, find the inverse function:

(a) y =

(b) f(x) = 16 − x², x ≥ 0

(c) y = , 0 ≤ x ≤ 3

(d) g(x) = 3 ⋅ 8⁻x

(e) h(x) = 4 ⋅ 6^2x − 8

3. For each of the following, rewrite using properties of logarithms to express as a sum or difference:

(a) log₂

(b) ln

(c) ln

(d) log_z

4. Graph the function f(x) = log_1/2(x + 7) − 3.

5. For each of the following, solve for x without a calculator:

(a) 9^{x + 2} = 27²

(b) log₂ = x² − 4x

(c) log₂(x² + 12) = log₂ 7x

(d) 2 log₅ x = log₅ 9, x > 0

6. For each of the following, use the properties of logarithms to rewrite the given expressions as a single logarithm:

(a) log₅(x² − 1) − 6 log₅(x + 1)

(b)

(c)

(d) 3log₂ − c log₂b + 2 log₂(de)

7. For each of the following, solve for x. Make sure that your solutions do not make any terms in the original equations undefined.

(a) 4^{x + 3} = 7^x

(b) ln(x + 5) = ln(x − 1) − ln(x + 1)

(c) log₁₀(x² + 5x) − log₁₀ x = 2

(d) ln(4x − 3) = ln25

8. A population of fungi grows exponentially. A population that initially has 10,000 grows to 25,000 after 2 hours.

(a) Use the exponential growth function, A(t) = A_oe^kt to find the value of k rounded to 2 decimal places. What will be the fungus population after 6 hours?

(b) Write a logarithmic function that gives t as a function of A(t), which we can now write as A.

(c) If a scientist estimates that the fungus population in this sample is 63,000, then how many hours have passed since the initial measure of 10,000 individuals in the population?

9. A certain species of marmoset is listed as vulnerable for its conservation status. There are currently about 8000 of this species, which is 46% of the population 6 years ago, and the decrease in population has been exponential. Let t represent the number of years since 6 years ago (so t = 6 represents the current year).

(a) Write an exponential function that gives the population of this marmoset species in a given year, as defined by t. (Use the formula for continuous growth/decay, A = A₀e^kt, where A is the population after t years, given that the initial population, at t = 0, is A₀.)

(b) Rewrite the equation to give t as a function of the population.

(c) If the population drops below 2500, the species will be considered endangered. If the population continues to decrease at the same exponential rate, in how many years will this marmoset species be considered endangered?

SOLUTIONS TO CHAPTER 5 PRACTICE QUESTIONS

1. (a) 3, (b) 2, (c) −2, (d) 1, (e) −1

(a) Recall that log₅125 is the same as saying “5 raised to what power is equal to 125.” 5 raised to the third power equals 125, so log₅125 equals 3.

(b) Recall that log₈64 is the same as saying “8 raised to what power is equal to 64.” 8 squared equals 64, so log₈64 equals 2.

(c) Recall that log₃ is the same as saying “3 raised to what power is equal to 1/9.” 3 raised to the power of −2 equals 1/9, so log₃1/9 equals −2.

(d) Recall that log₉9 is the same as saying “9 raised to what power is equal to 9.” 9 raised to the first power equals 9, so log₉9 equals 1.

(e) Recall that a logarithm without a provided base has a base of 10. log is the same as saying “10 raised to what power is equal to 1/10.” 10 raised to the −1 power is 1/10, so log equals −1.

2. (a) (x² + 5)/3 = y, x ≥ 0; (b) f ⁻¹(x) = ; (c) y = , 0 ≤ x ≤ 3; (d) g⁻¹(x) = −log₈(x/3); (e) h⁻¹(x) = (log₆(x/4) + 2)/2

We can also use the logarithmic quotient property to write this inverse function as g⁻¹(x) = −(log₈x − log₈3), which simplifies as g⁻¹(x) = −log₈x + log₈3.

We can also write this inverse function as h⁻¹(x) = log₆, using the logarithmic root property.

3. See explanations.

4. See graph.

The value of the base, 1/2, is less than 1, so this function is always decreasing.

The function y = log_1/2x has a vertical asymptote of the y-axis for its left end, because an x-value of 0 makes it undefined, and it passes through the points (1, 0), (2, −1), (4, −2), and (8, −3), because log_1/21 = 0, log_1/22 = −1, log_1/24 = −2, and log_1/28 = −3. The graph of f(x) = log_1/2(x + 7) − 3 is the graph of y = log_1/2x shifted 7 units to the left and 3 units down. So, the graph of f(x) has a vertical asymptote of x = −7 and passes through the points with x-coordinates 7 less and y-coordinates 3 less than those of points on y= log_1/2x.

(1 − 7, 0 − 3) → (−6, −3)
(2 − 7, −1 − 3) → (−5, −4)
(4 − 7, −2 − 3) → (−3, −5)
(8 − 7, −3 − 3) → (1, −6)

The graph of f(x) = log_1/2(x + 7) − 3 passes through (−6, −3), (−5, −4), (−3, −5), and (1, −6). The graph of f(x) is shown below, with asymptote indicated by a dashed line.

5. (a) 1, (b) 2, (c) 3, 4 (d) 3

6. See the following logarithms.

7. (a) 3ln4 / (ln7 − ln4); (b) no solution; (c) 95, (d) 7

However, the argument of a logarithm cannot be negative, so each of these values make two of the logarithmic expressions in the original equation undefined. There are no solutions to the given equation.

An x-value of 0 would make log₁₀x undefined in the original equation, so that is an extraneous solution. The only solution to the given equation is x = 95.

8. (a) 157,998; (b) t = 2.17 ln(A/10,000); (c) 4

(a) The question provides the function, the initial value (t = 0) of 10,000 (A_o), and the amount after t = 2 hours. Plug in the given values to arrive at the value of k:

Next, with the found value of k rounded to 2 decimal places, evaluate using the function when t = 6 hours:

(b) We must solve the exponential equation A(t) = 10,000e^0.46t for t.

The function that gives t, time in hours passed, as a function of A, the current fungus population in the sample, is t = 2.17 ln (A/10,000).

(c) If a scientist estimates the population in the sample to be 63,000, then A = 63,000. Substitute this value and solve for t.

If the fungus population in the sample is 63,000, then 4 hours have passed since the initial population count of 10,000.

9. (a) A = 17,391e^−0.129t; (b) t = −7.75ln(A/17,391); (c) 9

(a) If the current population of this marmoset species, A, is 46% of the population 6 years ago, A₀, then A/A₀ is equal to 46%, or 0.46. Solve the equation A = A₀e^kt for A/A₀ then substitute the appropriate values.

By using the information about the population 6 years ago and this year, we have found the exponential decay rate. Substituting back into the formula, we have A = A₀e^−0.129t. To finish creating an exponential function of A in terms of t, we must find A₀, the population 6 years ago. The current population, 8000, is 46% of A₀.

So, the population, A, of the marmoset species t years after 6 years ago is given by the function A = 17,391e^−0.129t.

(b) Solve the equation found in part (a) for t to rewrite the formula.

A function for t in terms of A is t = −7.75ln(A/17,391).

(c) To find when the population reaches 2500, substitute 2500 for A and solve for t.

The marmoset species population will drop below 2500 about 15 years after 6 years ago, which is 9 years from now.

Alternatively, you could redefine t = 0 as the current year, with A₀ = 8000 and t representing years from now, in the exponential function.

This solution also tells you that the marmoset species will be considered endangered about 9 years from now.

REFLECT

Congratulations on completing Chapter 5!
Here’s what we just covered.
Rate your confidence in your ability to

• Find the inverse relation of a given function, including exponential functions, and determine any necessary domain restrictions on the function to make its inverse a function

1 2 3 4 5

• Evaluate or estimate numerical logarithms, where they exist

1 2 3 4 5

• Solve single-variable equations containing logarithmic expressions

1 2 3 4 5

• Use binary, natural, and common logarithms

1 2 3 4 5

• Use logarithmic identities

1 2 3 4 5

• Graph logarithmic functions

1 2 3 4 5

• Write and use logarithmic functions to represent real-life situations and solve problems

1 2 3 4 5

If you rated any of these topics lower than you’d like, consider reviewing the corresponding lesson before moving on, especially if you found yourself unable to correctly answer one of the related end-of-chapter questions.

Access your online student tools for a handy, printable list of Key Points for this chapter. These can be helpful for retaining what you’ve learned as you continue to explore these topics.