Warm Up 10/20

Solve.

1. 43x–1 = 8x+1

2. 32x–1 = 20

3. log7(5x + 3) = 3

4. log(3x + 1) – log 4 = 2

5. log4(x – 1) + log

4(3x – 1) = 2

x ≈ 1.86

x = 68

x = 133

x = 3

x = 5 3

Be seated before the bell rings

DESK

homework Warm-up (in

your notes)

Agenda:

Warmup

Notes 7.6 notes

Notebook Table of content

Page 1

1

25) Exponential Functions, Growth, and

Decay

26) Inverses & Logarithmic

Functions

27) Properties of Logarithms

28) Exponential, Logarithmic, and

inequalities functions

29) Natural

base e

7.6 Natural Base

e

HW ;

p. 534;11-22,30

7.6 Natural Base e

Compound Interest: When the interest due at the end of

the payment is added to the principal

Formula:

Principal

Rate

Number of times it gets

compounded per yr

year

Amount

Interest can compounded :

Annually : 1 per yr

Semiannually: Twice per yr

Quarterly: 4 times per yr

Monthly: 12 times per yr

Daily: 365 times per yr

Example 1: Find the amount that results from each investment

$100 is invested at 4 % compounded quarterly after a period of 2 years.

2:

Find the amount that results from each investment

$100 is invested at 4 % compounded monthly after a

period of 2 years.

(12)(2).04

100 1 $108.3112

A

Example 2:

investment of $1 is compounded n times for 1 year. At a

rate of 100%

( )(1)

11

n

An

What do you think the value of this will be as n get very

large?

As n gets very large, interest is

continuously compounded.

A= (1 + )n = 2.7182818….

This number is called e.

Like , the constant e is an

irrational number.

n 1

The natural logarithmic

function f(x) = ln x is the

inverse of the natural

exponential function f(x) =

ex.

4. Simplify.

A. ln e0.15t B. e3ln(x +1)

ln e0.15t = 0.15t e3ln(x +1) = (x + 1)3

ln e2x + ln ex = 2x + x = 3x

C. ln e2x + ln ex

Simplify.

a. ln e3.2 b. e2lnx

c. ln ex +4y

ln e3.2 = 3.2 e2lnx = x2

ln ex + 4y = x + 4y

Check It Out! Example 2

Principal

Rate

year

Find the amount A that results from investing a principal P of

$2000 at an annual rate r of 8% compounded continuously for a

time t of 1 year.

(.08)(1)2000A e $2166.57

A) $300 is invested at 11% compound

continuously for a period of 2 ¼ years.

B) $300 is invested at 11% compound quarterly

for a period of 2 ¼ years.

(.11)(2.25)300A e $384.26

(4)(2.25).11

300 14

A

How would you invest your $300 ?

B) Start with $300. Add 2.50 each month for a

period of 2 ¼ years.

300 27 3A $381

$382.96

Growth & Decay rtA Pe

Time

(sec,hours,days)

Growth/decay Rate

K < 0 (decay)

kt

oA A e

Initial

amount

Ending

Amount

K > 0 Growth

A colony of bacteria grows according to the law of

uninhibited growth according to the function

where N is measured in grams and t is measure in days. 0.0590 ,tN t e

(a) Determine the initial amount of bacteria.

(b) What is the growth rate of bacteria?

(c) What is the populations after 5 days?

(d) How long will it take for the population to reach 140 grams?

90 grams

5% growth rate

0.05( )5( ) 90N t e 115.56 grams

0.05( )140 90 te 8.83 days

0.05( )180 90 te 13.86 days

6. Zombies grows according to the law of uninhibited

growth to the function

Where N is the number of Zombies

& t is # in days.

0.37( ) 5 tN t e

(a) At the beginning of the outbreak,

how many people were affected?

Zombies at Vista

(b) What is the growth rate of the

Zombies?

(c) What is the population in 10 days?

(d) How long will it take for the population to

reach 600 Zombies?

(e) What is the double time for the

population?

We have exponential decay when:

Time

(sec,hour

s,days)

Decay Rate: K < 0 (decay)

kt

oA A e

Initial

amount

Ending

Amount

In 1940 a group of boys walking in the woods near

the village of Lascaux in France suddenly realized

their dog was missing. They soon found him in a

hole too deep for him to climb out. One of the boys

was lowered into the hole to rescue the dog and

stumbled upon one of the greatest archaeological

discoveries of all time. What he discovered was a

cave whose walls were covered with drawings of

wild horses, cattle and a fierce-looking beast

resembling a modern bull. In addition, the cave

contained the charcoal remains of a small fire, and

from these remains scientists were able to

determine that the cave was occupied 15,000

years ago.

Charcoal left from the

logs contains Carbon-14

Radiocarbon Dating

All living things contain

Carbon 14. When they die,

the C-14 begins to decay.

We can determine how

long something has been

dead by the amount of C-14

left.

7. How old is the cave?: the amount of C-14 remaining in

the samples of the Lascaux charcoal was 15% of the

amount such trees would contain when living. The half-

life of C-14 is approximately 5600 years.

kt

oeAtA

56001

2

k

o oA A e

1

ln 56002

k

(a) Find the decay rate

56001ln ln

2

ke

1ln

2

5600k

.000123776 k

So assuming the paintings were made at the time of the fire in the cave,

they are approximately 15,000 years old.

7B. The amount of C-14 remaining in the samples of the Lascaux

charcoal was 15% of the amount such trees would contain when

living. The half-life of C-14 is approximately 5600 years.

kt

oeAtA

000123776.0k

oA15.0 kteln15.0ln

327,15000123776.0

15.0ln

t

(b) How old is the cave?

8. After the release of radioactive material into the atmosphere

from a nuclear power plant, the hay was contaminated by iodine

131 ( half-life, 8 days). If it is all right to feed the hay to cows

when 10% of the iodine 131 remains, how long did the farmers

need to wait to use this hay?

kt

oeAtA

8

0

1

2

k

oA A e

81

2

ke

81ln ln

2

ke

1ln 8

2k

1ln

2

8k

.0866433976k

0.10

k t

oA A e

.10

k te

ln.10 ln

k te

ln.10 ktln .10

tk

26.575 days

8B . Some of the farmers wanted to wait until only 5% of the original

iodine 131 was present. How long did they need to wait?

After the release of radioactive material into the atmosphere from a

nuclear power plant, the hay was contaminated by iodine 131 ( half-life,

8 days). If it is all right to feed the hay to cows when 10% of the iodine

131 remains, how long did the farmers need to wait to use this hay?

.0866433976k

0.05

k t

oA A e

.05

k te

ln.05 ln

k te

ln.05 kt

ln .05t

k

34.58 dayst

4. The Starship Enterprise has finished its 43 year mission to map the

unknown universe. Captain Kirk has ordered the Enterprise to return to

Earth. A course has been mapped out and it is calculated that they will

need 17.4 grams of Startonium to get home. Startonium is a radioactive

fuel with a half life of 17 years. When the Starship Enterprise left on its

mission (43 years ago) 85 grams of Startonium was loaded into a special

holding tank for the trip home. Captain Kirk is delighted that they have

plenty of fuel in store because the crew is anxious to get home. Startonium

is difficult to find and they will not need to make a time-consuming detour

to buy more fuel. Mr. Spock reminds Captain Kirk that because Startonium

is a radioactive fuel it has been decaying ever since they left Earth.

Will the Starship Enterprise be able to reach Earth using the Startonium

that is left?

If not, how much will they need to buy in order to reach Earth? If they can

reach Earth with what is left, how much will remain when they reach

home?