Transcript Slide 1

WORKSHEET KEY
1)
8
11)
2, ext: -1/3
2)
-5
12)
-8, -2
3)
3.1989
4)
No Solution
5)
9.0952
6)
0.2916
7)
7.0711
8)
6
9)
9
10)
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5.6 – Solving Log Properties
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PAGE 386
1)
{4}
35)
{9}
3)
{1/9}
37)
{5}
5)
{-3, 1/2}
39)
{6}
7)
{-2, -1/2}
41)
{3}
9)
{1.465}
43)
11)
{2.71}
13)
{-1.825}
15)
{-0.128}
17)
{0.805}
19)
{-0.895}
21)
{1.579}
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47)
−𝟓 + 𝟑𝟕
−𝟓 − 𝟑𝟕
, 𝒆𝒙𝒕:
𝟐
𝟐
{5}
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5.6 – Solving Log Properties
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SOLVING LOGARITHMS
DAY 3
Section 5.5a
Pre–Calculus, Revised ©2015
[email protected]
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REVIEW
Solve x2 – 3x + 2 = 0
x  3x  2  0
2
 x  2   x  1 
0
1, 2 
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5.5 – Properties of Logarithms
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EXAMPLE 1
Solve e2x – 3ex + 2 = 0
e
2x
 3e  2  0
u e
x
x
u  3u  2  0
 u  2   u  1  0
2
e
x
 2   e  1  0
x
e 2  0
e 1 0
x
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x
5.5 – Properties of Logarithms
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EXAMPLE 1
Solve e2x – 3ex + 2 = 0
e 2  0
x
e  2
x
ln e  ln 2
e 1 0
x
e 1
x
ln e  ln 1
x ln e  ln 2
x  ln 2
x  0 .6 9 3
x ln e  ln 1
x  ln 1
x  0
x
x
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
0 .6 9 3, 0 
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EXAMPLE 2
Solve e2x – 3ex – 40 = 0
 ln 8 ext : ln  5
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5.5 – Properties of Logarithms
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YOUR TURN
Solve e2x – 7ex + 12 = 0
  1 .0 9 9 ,  1 .3 8 6 
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EXAMPLE 3
Solve 5 – ln(x – 3) = ln(x – 3)
ln  x  3   5  ln  x  3 
ln  x  3   ln  x  3   5
2 ln  x  3   5
ln  x  3  
e
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ln  x  3 
5
2
 e
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5/2
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EXAMPLE 3
Solve 5 – ln(x – 3) = ln(x – 3)
e
ln  x  3 
 e
x3e
x e
5/2
5/2
5/2
3
 1 5 .1 8 3
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YOUR TURN
Solve 8 – ln(2x + 1) = ln(2x + 1)
 2 6 .7 9 9
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EXAMPLE 4
A deposit is made for $100,000 into an account that pays 6% interest.
Find the balance after 10 years if the interest is compounded
continuously.
A = ??
Account
rt
A  Pe
P = $100,000
Principle
e = Use
Natural
e in Base
Calc
0.06

10


A  100000e
r = 0.06
Interest Rate
t = 10
Time
$182, 211.88
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EXAMPLE 5
You have deposited $500 in an account that pays 6.75% interest,
compounded continuously. How long will it take your money to
double?
rt
A  Pe
Doubled
Amount
A  500 e
0.0675 t
1000  500 e
500 e
0.0675 t
500 e
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0.0675 t
 1000
0.0675 t
500

1000
5.6 – Solving Log Properties
500
14
EXAMPLE 5
You have deposited $500 in an account that pays 6.75% interest,
compounded continuously. How long will it take your money to
rt
double?
A  Pe
e
 2
ln 2  0 .0 6 7 5t
0.0675 t
t 
ln 2
0 .0 6 7 5
 10.269 years
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YOUR TURN
How long will it take $30,000 to accumulate to $110,000 in a trust that
earns a 10% annual return compounded continuously?
t  12.993 years
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EXPONENTIAL GROWTH/DECAY
P  P0 e
kt
P = Ending Amount
P0 = Initial Amount
e = The Natural Base
k = Growth or Decay Rate
t = Time
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EXAMPLE 6
A certain bacterium has an exponential growth rate of 25% per day. If
we start with 0.5 grams and provide unlimited resources how much
bacteria can we grow in 14 days?
P = ??
Ending Amount
kt
P  P0 e
P0 = 0.5
Initial Amount
e = The Natural Base
 0.25 14 
P  0.5 e
k = 0.25
Growth or Decay
Rate
t = 14
days
t = Time
P  16.558 gram s
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EXAMPLE 7
What is the total amount of bacteria when the initial amount of bacteria
is 300, k = 0.068, and the time studied is 52 hours?
P  10, 298.798 bacteria
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YOUR TURN
At the start of an experiment, there are 100 bacteria. If the bacteria
follow an exponential growth pattern with rate k = 0.02, what will be
the population after 5 hours?
P  1 1 0 .5 1 7 b a c te ria
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EXAMPLE 8
During its exponential growth phase, a certain bacterium can grow from
5,000 cells to 12,000 cells in 10 hours. What is the growth rate?
P  P0 e
kt
12, 000  5, 000 e
e
 k 10 

 k 10 
12, 000
5, 000
ln
12
 10 k
P = 12,000
Ending Amount
P0== Initial
5,000 Amount
e = The Natural Base
k = ??
Growth or Decay
Rate
Time
tt == 10
5
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EXAMPLE 8
During its exponential growth phase, a certain bacterium can grow from
5,000 cells to 12,000 cells in 10 hours. What is the growth rate?
ln
12
 10 k
5
 12 
ln 

 5 
k
10
k  0.0875
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EXAMPLE 9
During its exponential growth phase, a certain bacterium can grow from
5,000 cells to 15,000 cells in 12 hours. What is the growth rate?
k  0.0916
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YOUR TURN
The population of a certain city in 2000 was 99,500. What is its initial
population in 1975 when its growth rate is at 0.0170. Round to the
nearest whole number.
P0  65, 050
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EXAMPLE 10
If certain isotope has a half-life of 4.2 days. How long will it take for a
150 milligram sample to decay so that only 10 milligrams are left?
P  P0 e
75  150 e
1
e
kt
 k  4.2 
 k  4.2 
2
ln
1
 4.2 k
P = 75
Ending Amount
P0== Initial
150 Amount
e = The Natural Base
k = ??
Growth or Decay
Rate
Time
tt == 4.2
2
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EXAMPLE 10
If certain isotope has a half-life of 4.2 days. How long will it take for a
150 milligram sample to decay so that only 10 milligrams are left?
ln
1
 4.2 k
2
1
ln  
2
k
4.2
k   0.1650
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EXAMPLE 10
If certain isotope has a half-life of 4.2 days. How long will it take for a
150 milligram sample to decay so that only 10 milligrams are left?
10  150 e
10
e
  .1650 t 
  .1650 t 
150
1
ln
  .1650 t
15
 1 
ln 

1
5


 t
  .1 6 5 0 
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P  P0 e
kt
P = 10
P0 = 150
e = The Natural Base
k = –.1650
t = ??
t  16.412 days
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EXAMPLE 11
The half-life of carbon-14 is 5,730 years. The skeleton of a mastodon has
42% of its original Carbon-14. When did the mastodon die?
kt
P = ½ (half life)
P

P
e
1
 k 5730 
0
 1e
P0 = 1 (full life)
2
1
 k 5730 
e = The Natural Base
e
2
k = ??
1
ln  5730 k
t = 5,730
2
1
ln  
 2  5730k

5730
5730
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k 
ln 0.5
5730
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EXAMPLE 11
The half-life of carbon-14 is 5,730 years. The skeleton of a mastodon has
42% of its original Carbon-14. When did the mastodon die?
P  P0 e
  ln 0.5 5730  t 
kt
P = 0.42 (total left)
P0 = 1
0.42  1e
e = The Natural Base
  ln 0.5 5730  t 
k = (ln 0.5)/5730
0.42  e
ln 0.5
t ln e  5 7 3 0  t = ??
 5 7 3 0  ln 0.42 
5730
 4970.778  ln 0.5t
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t  7171.3171 years
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YOUR TURN
The half-life of carbon-14 is 5,730 years. If it is determined that an old
bone contains 85% of its original carbon-14, how old is the bone?
t  1, 343.4859 years
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NEWTON’S LAW OF COOLING
T F  T R   T0  T R  e
 kt
TF = Final Temperature
TR = Temperature of the Environment
T0 = Initial Temperature of the Object
e = The Natural Base
k = Growth or Decay Rate
t = Time
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EXAMPLE 12
A container of ice cream arrives home from the supermarket at a
temperature of 65°F. It is placed in the freezer which has a
temperature of 20°F. Determine the final temperature at which it will
be still considered “freezing,” if the rate of change is 0.107°F per
minute for 12.35 minutes.
TF == ??
Final Temperature
 kt
TR = Environment
20°
Temp
T F  T R   T0  T R  e
T0 = Initial
65° Temperature
  0.107 12.35 
T F  20   65  20  e
e = The Natural Base
T F  2 0  1 2 
k = Growth
0.107 or Decay Rate
TF  3 2
t = Time
12.35
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YOUR TURN
The cooling model for tea served in a 6 oz. cup uses Newton’s Law of
Cooling equation. The original temperature was 200°F and current
environment temperature of the tea is at 68°F. Determine the
temperature if the decay rate is at 0.41 per minute and waiting time
is 6 minutes.
T F  79.277 F
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EXAMPLE 13
When an object is removed from a furnace and placed in an environment
with a constant decay rate of 0.3114 and the room temperature of 80°F,
its core temperature is 1500°F. If the final temperature is at 378°F,
about how long is it out of the furnace (in hours)?
 kt
T F  T R   T0  T R  e
TF == 378°
Final Temperature
  0.3114 t 
80°
TR = Environment
Temp
378  80  1500  80  e
 0.3114 t
T0 = Initial
1500° Temperature
378  80  1420 e
 0.3114 t
e = The Natural Base
298  1420 e
 0.3114 t
k = Growth
0.3114 or Decay Rate
1420  e
298

t
=
??
Time
1420
1420
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5.6 – Solving Log Properties
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EXAMPLE 13
When an object is removed from a furnace and placed in an environment
with a constant decay rate of 0.3114 and the room temperature of 80°F,
its core temperature is 1500°F. If the final temperature is at 378°F,
about how long is it out of the furnace (in hours)?
 kt
T F  T R   T0  T R  e
298

1420  e
1420
298
1420
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 0.3114 t
1420
e
 0.3114 t
t  5 .0 1 2 h o u rs
5.6 – Solving Log Properties
 298 
ln 
   0.3114 t
 1420 
 298 
ln 

 1 4 2 0   0.3114 t

 0.3114
 0.3114
35
EXAMPLE 14
Pete was driving on a hot day when the car starts overheating and stops
running. It overheats to 280°F and can be driven again at 230°F.
Suppose it takes 60 minutes until Pete can drive if is 80°F outside,
what is the decay factor? Round to three decimal places.
 kt
T F  T R   T0  T R  e
k  0 .0 0 4 7
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YOUR TURN
Devin baked a yam at 350°, and when Devin removed it from the oven,
he let the yam cool, which has a room temperature of 68°F. After 10
minutes, the yam has cooled to 240°F. What is the decay factor?
T F  T R   T0  T R  e
 kt
k  0 .0 4 9
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ASSIGNMENT
Worksheet
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