Unveiling Logarithmic Functions in Real Life and Their Models

$x$	$y$
$1$	$3$
$2$	$\approx - 0.47$
$3$	$\approx - 2.49$
$4$	$\approx - 3.93$
$5$	$\approx - 5.05$

x

y

1

3

2

\approx - 0.47

3

\approx - 2.49

4

\approx - 3.93

5

\approx - 5.05

$x$	$y$
$1$	$1 k$
$2$	$2 k$
$3$	$4 k$
$4$	$8 k$
$5$	$16 k$

x

y

1

1 k

2

2 k

3

4 k

4

8 k

5

16 k

$x$	$40 e^{- 0.154 t}$	$y$
$0$	$40 e^{- 0.154 (0)}$	$40$
$2$	$40 e^{- 0.154 (2)}$	$\approx 29.4$
$4$	$40 e^{- 0.154 (4)}$	$\approx 21.6$
$6$	$40 e^{- 0.154 (6)}$	$\approx 15.9$
$8$	$40 e^{- 0.154 (8)}$	$\approx 11.7$
$10$	$40 e^{- 0.154 (10)}$	$\approx 8.6$

x

40 e^{- 0.154 t}

y

0

40 e^{- 0.154 (0)}

40

2

40 e^{- 0.154 (2)}

\approx 29.4

4

40 e^{- 0.154 (4)}

\approx 21.6

6

40 e^{- 0.154 (6)}

\approx 15.9

8

40 e^{- 0.154 (8)}

\approx 11.7

10

40 e^{- 0.154 (10)}

\approx 8.6

$x$	$x ln 2$	$ln y = x ln 2$
$- 2$	$- 2 ln 2$	$\approx - 1.39$
$- 1$	$- 1 ln 2$	$\approx - 0.69$
$0$	$0 ln 2$	$0$
$1$	$1 ln 2$	$\approx 0.69$
$2$	$2 ln 2$	$\approx 1.39$

x

x ln 2

ln y = x ln 2

- 2

- 2 ln 2

\approx - 1.39

- 1

- 1 ln 2

\approx - 0.69

0

0 ln 2

0

1

1 ln 2

\approx 0.69

2

2 ln 2

\approx 1.39

$x$	$y$	$ln y$
$20$	$12$	$ln 12 \approx 2.48$
$30$	$15$	$ln 15 \approx 2.71$
$40$	$25$	$ln 25 \approx 3.22$
$50$	$40$	$ln 40 \approx 3.69$
$60$	$100$	$ln 100 \approx 4.61$

x

y

ln y

20

12

ln 12 \approx 2.48

30

15

ln 15 \approx 2.71

40

25

ln 25 \approx 3.22

50

40

ln 40 \approx 3.69

60

100

ln 100 \approx 4.61

$x$	$y$
$20$	$12$
$30$	$15$
$40$	$25$
$50$	$40$
$60$	$100$

x

y

20

12

30

15

40

25

50

40

60

100

The table corresponds to an exponential function with the following form.

y = a b^{x - 1}

Write the exponential function that corresponds to the table.

$x$	$y$
$1$	$36$
$2$	$18$
$3$	$9$
$4$	$4.5$
$5$	$2.25$
$6$	$1.125$

Looking at the table, we see that the initial value is 36. We know that the table of values corresponds to an exponential function. Let's now identify the constant multiplier.

The constant multiplier is 12. We can use this and the initial value 36 to write our exponential function. y= 36( 1/2)^(x-1)

The amount of

Carbon - 14

in micrograms in a preserved vegetable that died

t

years ago can be modeled by an exponential function.

C (t) = 1.29 (\frac{1}{2})^{\frac{t}{5 5 0 0}}

Find the amount of

Carbon - 14

present in a vegetable that died

2000

years ago. Round the answer to the nearest integer.

If the amount of

Carbon - 14

found in a preserved vegetable is

1.25

micrograms, how many years ago did it die? Round the answer to the nearest integer.

We want to find the amount of Carbon-14 present in a vegetable that died 2000 years ago. To do so, we will substitute t=2000 in the equation for the exponential function. Then, we will evaluate the right-hand side. Let's do it!

We found that a preserved vegetable that died 2000 years ago would contain about 1 microgram of Carbon-14.

A preserved vegetable contains 1.25 micrograms of Carbon-14. We want to find how many years ago it died. To do so, we will substitute 1.25 for C(t) in the given formula and solve for t.

Now we will use the definition of a logarithm to rewrite the exponential equation as a logarithmic equation. Definition b^c= a ⇔ c=log_b a [1em] Equation ( 1/2)^(t5500)= 125/129 ⇔ t/5500=log_(12) 125/129 We will now solve the logarithmic equation for t. To do this, we will use the Change of Base Formula so that we can use a calculator to find the numeric value of the logarithm on the right-hand side. Recall that log a is equivalent to log_(10) a.

The vegetable died about 250 years ago.

During the summer, the number of surfboards sold in a Hawaiian surf shop grows exponentially. The owner knows that the number of boards sold during this season can be modeled by an exponential function.

B (t) = a (b)^{t - 1}

Here,

B (t)

represents the number of boards sold and

t

is the number of days that have passed since the season started. On the first day of the season,

2

boards where sold. In the tenth day,

5

boards were sold. Find the values of

a

and

b

and write the exponential function. If necessary, round to five decimal places.

We want to find the values of a and b and write the equation of the exponential function that models the given situation. Let's start by finding a. We know that 2 boards were sold on the first day of the season, so we can substitute t=1 and B(t)=2 in the given equation and solve for a.

We found that a=2. Let's put it into our equation. B(t)=2(b)^(t-1) We also know that 5 boards were sold on the tenth day of the season. This means that we can substitute B(t)=5 and t=10 into our partial equation and solve for b.

Now that we have the value of b, we can write the complete exponential function. B(t)=2(1.10717)^(t-1)

B (t) = 2 (1.10717)^{t - 1}

Here,

B (t)

represents the number of boards sold and

t

is the number of days that have passed since the season started. Select the graph that corresponds to the exponential function.

To determine which of the graphs corresponds with the given exponential function, we will draw the graph of the function and compare it to the diagrams in the options. Let's start by make a table of values. Since the domain of an exponential function is the set of all real numbers, we can assign any value to the variable t.

t	2(1.10717)^(t-1)	B(t)
- 6	2(1.10717)^(- 6-1)	≈ 1
- 4	2(1.10717)^(- 4-1)	≈ 1.2
- 2	2(1.10717)^(- 2-1)	≈ 1.5
0	2(1.10717)^(0-1)	≈ 1.8
2	2(1.10717)^(2-1)	≈ 2.2
4	2(1.10717)^(4-1)	≈ 2.7
6	2(1.10717)^(6-1)	≈ 3.3

We can now plot the points obtained in the table and connect them with a smooth curve.

Our graph matches the one given in option A.

The Spanish government is opening factories in small towns to encourage people to move there. From the moment a massive factory was opened, the population of a small town started to increase year after year. The population

P

of this town, in thousands of people, can be modeled by an exponential function.

P (t) = 10 (1.02)^{t}

Here,

t

represents the number of years that have passed since the factory was opened. How many people will there be in the town

50

years after the factory's opening? Round the answer to the nearest thousand.

We want to find how many people there will be in the town 50 years after the factory was opened. To do so, we will substitute t=50 in the given equation for the exponential function. Then, we will evaluate the right-hand side. Let's do it!

The population will be 26.915880... thousand people. Therefore, rounded to the nearest thousand, the population will be about 27 000 people.

A store sells skateboards. The table shows the numbers of skateboards sold $s$ during the $n th$ month that the store has been open.

$n$	$s$
$1$	$12$
$2$	$16$
$3$	$25$
$4$	$36$
$5$	$50$

Which of the following is a scatter plot on the $(n, ln s)$ plane of the data shown in the table?

Let's start by calculating the natural logarithm of the s-values in the table.

n	s	ln s
1	12	ln 12≈ 2.5
2	16	ln 16≈ 2.8
3	25	ln 25≈ 3.2
4	36	ln 36≈ 3.6
5	50	ln 50≈ 3.9

Now, let's plot the points (n,ln s) on a coordinate plane were the horizontal axis represents the variable n and the vertical axis represents the natural logarithm of s.

This graph corresponds to choice C.

Applications of Exponential and Logarithmic Functions

Catch-Up and Review

Graphing in the $(x, ln y)$ Coordinate Plane

Tables, Logarithmic Models, and Exponential Models

Hint

Solution

Carbon $14$ in Vegetables

Hint

Solution

Uranium $238$

Hint

Solution

Modeling a Village's Population

Hint

Solution

Relation Between Exponential and Linear Models

Visual Near Point

Answer

Hint

Solution

Using Technology

Applications of Exponential and Logarithmic Functions

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