Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function...

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Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v , then u = v For any logarithmic function f(x) = log b x: If log b u = log b v , then u = v

Transcript of Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function...

Page 1: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Date: 3.5 Equation Solving and Modeling (3.5)

One-to-One Properties

For any exponential function f(x) = bx:

If bu = bv , then u = v

For any logarithmic function f(x) = logbx:

If logbu = logbv , then u = v

Page 2: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Solve: 31

20 52

x

320 1 5

220 20

x

31 1

42

x

23 1

2

1

2

x

Use one-to-one property

23

x

6x

Page 3: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Solve: 2log 2x 2 2lolog g10x Use one-to-one property

2 210x 2 100x

-OR-rewrite

exponentially

2 100x

10x

Page 4: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Solve: rewrite exponentially log

2(x 4) 3

23 x 4 4 +4

12 x

Check:

log2(12 4) 3

log28 3

3 3

Page 5: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

log2 x(x - 7) = 3 use product rule & rewrite

23= x(x - 7) rewrite exponentially

8 = x2 - 7x simplify

0 = x2 - 7x - 8 set equal to zero

0 = (x-8)(x+1) factor

Solve: log2 x + log2(x - 7) = 3

Page 6: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

x - 8 = 0 or x + 1 =0 set each factor = 0

x = 8 or x = -1 solve

Checklog2 x + log2 (x - 7) = 3 log2 8 + log2 (8 - 7) = 3 3 + log2 1 = 3 3 + 0 = 3

3 = 3

Checklog2 x + log2 (x - 7) = 3 log2 -1 + log2 (-1 - 7) = 3 The number -1 does not check.Negative numbers do not have logarithms.

?

continued:

0 = (x-8)(x+1)

Page 7: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

+7 +7 add 7 to both sides

3x+2= 34 ln 3 x+2 = ln 34 take ln of both sides

(x+2) ln 3 = ln 34 rewrite exponent

ln 3 ln 3 divide both sides by ln 3

x+2 = 3.21 simplify

-2 -2 subtract 2 to both sides

x = 1.21

Solve: 3x+2-7 = 27

Page 8: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Solve: 0.6

400150

1 95 xe

0.61 95150

400 xe

0.61 958

3xe

1 1

0.6

395

5 xe0.65 95

5 3

1

9 95

xe

19

Page 9: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Solve: 0.61

57xe

0.6ln ln1

57xe

1ln 0.6

57x

0.6

1l

0.6

n 0.657 x

6.74 x

Page 10: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

. 7e2x 63 7 7

e2x 9

lne2x ln9

2x ln9 2 2

x 1.10

Solve:

Page 11: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Orders of MagnitudeThe common logarithm of a positive quantity is its’ order of magnitude.

(the difference in their powers of ten)

DAY 2

Determine the order of magnitude:A kilometer and a meter.

A dollar and a penny

A horse weighing 400 kg and a mouse weighing 40 g.

New York City with 7 million people and Earmuff Junction with a population of 7.

A kilometer is 3 orders of magnitude longer than a meter.

A dollar is 2 orders of magnitude greater than a penny.

The horse is 4 orders of magnitude heavier than the mouse.

New York is 6 orders of magnitude bigger than Earmuff Junction.

Page 12: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Earthquake IntensitiesThe Richter scale magnitude R of an earthquake is

a is the amplitude in micrometers (μm) of the vertical ground motion at the receiving station, T is the period of the associated seismic wave in seconds, and B accounts for the weakening of seismic wave with increasing distance from the epicenter of the earthquake.

loga

R BT

Measure of AcidityThe acidity of a water-based solution is measured by the concentration of hydrogen ion [H+] in the solution (in moles per liter). The measure of acidity used is pH, the opposite of the common log of the hydrogen-ion concentration: pH log[H ]

Page 13: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Newton’s Law of Cooling

The temperature, T, of a heated object at time t is given by

T(t) = Tm + (T0 - Tm)e-kt

Where Tm is the constant temperature of the surrounding medium, T0 is the initial temperature of the heated object, and k is a constant that is associated with the cooling object.

Page 14: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Example

A cake removed from the oven has a temperature of 2100F. It is left to cool in a room that has a temperature of 700F. After 30 minutes, the temperature of the cake is 1400F.

a. Use Newton’s Law of Cooling to find a model for the temperature of the cake, T after t minutes.

b. What is the temperature of the cake after 40 minutes?

c. When will the temperature of the cake be 900F?

Page 15: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

T = 70 + (210 - 70)e-kt T0=700 Tm =2100

T = 70 + 140ekt

After 30 minutes the cake is 1400F140 = 70 + 140e-k30 T=1400 t=30

-70 -70 70 = 140e-30k

140 1401/2 = e-30k

ln1/2 = ln e-30k take the ln of both sides

ln(1/2) = -30k

a. T = Tm + (T0 - Tm)e-kt Newton’s Law of Cooling

Page 16: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

ln(1/2) = -30k -30 -300.0231 = kThe temperature of the cake is modeled by:T = 70 + 140e-0.0231t

b. Find the temperature after 40 minutes T = 70 + 140e-0.0231(40)

T = 1260Fc. Find when the temperature of the cake will

be 900F.90 = 70 + 140e-0.0231t

Page 17: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

90 = 70 + 140e-0.0231t

-70 -70 20 = 140e-0.0231t

140 140 1/7 = e-0.0231t

ln1/7 = ln e-0.0231t take the ln of both sides

ln(1/7) = -0.0231t-0.0231 -0.0231 84 = tThe temperature of the cake will be 900F

after 84 minutes.

Page 18: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Complete Student CheckpointAn object is heated to 100C. It is left to cool in a room that has a temperature of 30C. After 5 minutes, the temperature of the object is 80C.a. Use Newton’s Law of Cooling to find a model for the temperature of the object, T, after t minutes.

m 0 mT ( T ) ktT T e 580 30 100( )30 ke

57080 30 ke 30 30

57050 ke 70 70

55

7ke

5ln l5

7n ke

5ln 5

7k

5 5

0.0673 k

0.06737030 tT e

Page 19: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Complete Student CheckpointAn object is heated to 100C. It is left to cool in a room that has a temperature of 30C. After 5 minutes, the temperature of the object is 80C.b. What is the temperature of the object after 20 minutes?

c. When will the temperature of the object be 35C?

T 30 70e 0.0673(20) 48

35 30 70e 0.0673t

30 30 5 70e 0.0673t

70 70

1

14e 0.0673t

ln

1

14ln e 0.0673t

ln

1

14 0.0673t

0.0673 0.0673

39 t

The temperature of the object will be 35C after39 minutes.

Page 20: Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function f(x) = b x : If b u = b v, then u = v For any logarithmic.

Equation Solving and Modeling