# Exp/Log Models ```PRECALCULUS I
EXPONENTIAL &amp;
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
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%.
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 .
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 .
BACTERIA GROWTH
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 .
TIME OF DEATH
5. The time, t, elapsed since death and the
body temperature, T, at room temperature
of 70 degrees is given by
85.7  70
t  2.5 ln
98.6  70
If the body temperature at 9:00 a.m. was
85.7 degrees, estimate time of death.
TIME OF DEATH concluded
5. If the body temperature at 9:00 a.m.
was 85.7 degrees, estimate time of death.
85.7  70
t  2.5 ln
98.6  70
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.
```