Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function...
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Transcript of Date: 3.5 Equation Solving and Modeling (3.5) One-to-One Properties For any exponential function...
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
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
Solve: 2log 2x 2 2lolog g10x Use one-to-one property
2 210x 2 100x
-OR-rewrite
exponentially
2 100x
10x
Solve: rewrite exponentially log
2(x 4) 3
23 x 4 4 +4
12 x
Check:
log2(12 4) 3
log28 3
3 3
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
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)
+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
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
Solve: 0.61
57xe
0.6ln ln1
57xe
1ln 0.6
57x
0.6
1l
0.6
n 0.657 x
6.74 x
. 7e2x 63 7 7
e2x 9
lne2x ln9
2x ln9 2 2
x 1.10
Solve:
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.
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 ]
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.
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?
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
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
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.
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
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.
Equation Solving and Modeling