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

a It is known that there is

1

microgram of Carbon

14

present in a vegetable that died

2000

years ago. To calculate how much Carbon

14

was present in the living vegetable,

2000

will be substituted for

t

and

1

will be substituted for

C (2000) .

Then, the equation can be solved for

I .

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

Substitute

$t = 2000$

C (2000) = I (\frac{1}{2})^{\frac{2 0 0 0}{5 5 0 0}}

Substitute

$C (2000) = 1$

1 = I (\frac{1}{2})^{\frac{2 0 0 0}{5 5 0 0}}

▼

Solve for

I

ReduceFrac

$\frac{a}{b} = \frac{a / 5 0 0}{b / 5 0 0}$

1 = I (\frac{1}{2})^{\frac{4}{1 1}}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

1 = I (\frac{1}{2 ^{\frac{4}{1 1}}})

MultEqn

$LHS \cdot 2^{\frac{4}{1 1}} = RHS \cdot 2^{\frac{4}{1 1}}$

2^{\frac{4}{1 1}} = I (1)

IdPropMult

Identity Property of Multiplication

2^{\frac{4}{1 1}} = I

RearrangeEqn

Rearrange equation

I = 2^{\frac{4}{1 1}}

UseCalc

Use a calculator

I = 1.286664 \dots

RoundDec

Round to $2$ decimal place(s)

I \approx 1.29

Therefore, about

1.29

micrograms of Carbon

14

were present in the living vegetable.

b Similar to Part A, it is known that there is

1

microgram of Carbon

14

present in a vegetable that died

10000

years ago. This sentence can also be expressed mathematically.

C (10000) = 1

With this information, the amount of Carbon

14

that was present in the living vegetable

I

can be found by substituting

10000

for

t

and

1

for

C (10000) .

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

Substitute

$t = 10000$

C (10000) = I (\frac{1}{2})^{\frac{1 0 0 0 0}{5 5 0 0}}

Substitute

$C (10000) = 1$

1 = I (\frac{1}{2})^{\frac{1 0 0 0 0}{5 5 0 0}}

▼

Solve for

I

ReduceFrac

$\frac{a}{b} = \frac{a / 5 0 0}{b / 5 0 0}$

1 = I (\frac{1}{2})^{\frac{2 0}{1 1}}

PowQuot

$(\frac{a}{b})^{m} = \frac{a ^{m}}{b ^{m}}$

1 = I (\frac{1}{2 ^{\frac{2 0}{1 1}}})

MultEqn

$LHS \cdot 2^{\frac{2 0}{1 1}} = RHS \cdot 2^{\frac{2 0}{1 1}}$

2^{\frac{2 0}{1 1}} = I (1)

IdPropMult

Identity Property of Multiplication

2^{\frac{2 0}{1 1}} = I

RearrangeEqn

Rearrange equation

I = 2^{\frac{2 0}{1 1}}

UseCalc

Use a calculator

I = 3.526365 \dots

RoundDec

Round to $2$ decimal place(s)

I \approx 3.53

Therefore, about

3.53

micrograms of Carbon

14

were present in the living vegetable.

c Half the initial amount of Carbon

14

present in the vegetable is

\frac{I}{2} .

Therefore, the half-life of Carbon

14

present in the vegetable can be found by substituting

\frac{I}{2}

for

C (t),

and isolating the variable

t .

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

Substitute

$C (t) = \frac{I}{2}$

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

DivEqn

$LHS / I = RHS / I$

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

$a = a^{1}$

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

Next, the Property of Equality for Exponential Equations can be used.

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

Finally, the value of

t

can be found by multiplying both sides of the obtained equation by

5500 .

1 = \frac{t}{5 5 0 0} \Leftrightarrow t = 5500

Therefore, the half-life of Carbon

14

implied by the given function is

5500

years. Heichi is feeling re-energized after solving for these solutions.

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

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

{{ article.displayTitle }}

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

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}