precalculus i exponential & log models dr. claude s. moore danville community college
TRANSCRIPT
![Page 1: PRECALCULUS I EXPONENTIAL & LOG MODELS Dr. Claude S. Moore Danville Community College](https://reader035.vdocuments.us/reader035/viewer/2022072006/56649cf95503460f949ca3de/html5/thumbnails/1.jpg)
PRECALCULUS I
EXPONENTIAL & LOG MODELS
Dr. Claude S. MooreDanville Community
College
![Page 2: PRECALCULUS I EXPONENTIAL & LOG MODELS Dr. Claude S. Moore Danville Community College](https://reader035.vdocuments.us/reader035/viewer/2022072006/56649cf95503460f949ca3de/html5/thumbnails/2.jpg)
FIVE COMMON TYPES OF MATHEMATICAL MODELSFIVE COMMON TYPES OF
MATHEMATICAL MODELS
1. Exponential Growth
2. Exponential Decay
3. Gaussian Model
4. Logistics Growth
5. Logarithmic Model
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1. EXPONENTIAL GROWTH
Find the annual rate (%) for a $10,000
investment to double in 5 years with
continuous compounding.
A = P ert with P = 10000, A = 20000, t = 5
20000 = 10000er(5) or 2 = e5r
ln 2 = ln e5r gives ln 2 = 5r(ln e) = 5r
r = (ln 2)/5 = 0.1386 or r is 13.9%.
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2. EXPONENTIAL DECAY
The half life of carbon 14 is 5730 years.
Find the equation y = a e bx if a = 3 grams.
0.5(3) = 3 eb(5730) or 0.5 = e5730b
ln 0.5 = ln e5730b gives
ln 0.5 = 5730b(ln e) = 5730b
b = (ln 0.5)/5730 = -0.12097
Thus the equation is y = 3 e -0.12097x .
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EXPONENTIAL EQUATION
3. Write the exponential equation of the line that passes through (0,5) and (4,1).
The equation is of the form y = a e bx .
(0,5) yields 5 = a eb(0) or 5 = a e0 or a = 5.
(4,1) yields 1 = 5 eb(4) or 0.2 = eb(4)
ln 0.2 = ln e4b gives ln 0.2 = 4b(ln e) = 4b
b = (ln 0.2)/4 = -0.402359
Thus the equation is y = 5 e -0.402359x .
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BACTERIA GROWTH4. The number of bacteria N is given by the model N = 250 e kt with t in hours.
If N = 280 when t = 10, estimate time for bacteria to double.
The point (10,280) yields 280 = 250 eb(10)
1.12 = e10b or ln 1.12 = ln e10b
ln 1.12 = 10b(ln e) = 10b
b = (ln 1.12)/10 = 0.0113329
Thus the equation is y = 250 e 0.0113329t .
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TIME OF DEATH5. The time, t, elapsed since death and the
body temperature, T, at room temperature of 70 degrees is given by
706.98
707.85ln5.2
t
If the body temperature at 9:00 a.m. was 85.7 degrees, estimate time of death.
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TIME OF DEATH concluded5. If the body temperature at 9:00 a.m.
was 85.7 degrees, estimate time of death.
t = -2.5 ln 0.54895 = 1.499 or t = 1.5 hrsSo time of death was 1.5 hrs before 9 a.m
Thus the time of death was 7:30 a.m.
706.98
707.85ln5.2
t
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Study and Learn
before time runs OUT.