PRECALCULUS I. EXPONENTIAL & LOG MODELS. Dr. Claude S. Moore Danville Community College. FIVE COMMON TYPES OF MATHEMATICAL MODELS. 1. Exponential Growth 2. Exponential Decay 3. Gaussian Model 4. Logistics Growth 5. Logarithmic Model. 1. EXPONENTIAL GROWTH.
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EXPONENTIAL & LOG MODELS
Dr. Claude S. MooreDanville Community College
1. Exponential Growth
2. Exponential Decay
3. Gaussian Model
4. Logistics Growth
5. Logarithmic Model
Find the annual rate (%) for a $10,000
investment to double in 5 years with
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%.
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 .
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 .
4. 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 .
5. The time, t, elapsed since death and the body temperature, T, at room temperature of 70 degrees is given by
If the body temperature at 9:00 a.m. was 85.7 degrees, estimate time of death.
5. 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 hrs
So time of death was 1.5 hrs before 9 a.m Thus the time of death was 7:30 a.m.
Study and Learn before time runs OUT.