Page 188 - Calculus Demystified

Transcendental Functions
CHAPTER 6
Taking the natural logarithm of both sides yields 175
ln 2 = ln(e K·4 ) = 4K.
We conclude that K =[ln 2]/4. As a result,
([ln 2]/4)t
B(t) = 5000 · e .
We simplify this equation by noting that
e ([ln 2]/4)t = (e ln 2 t/4 = 2 t/4 .
)
In conclusion,
B(t) = 5000 · 2 t/4 .
The number of bacteria at noon (time t = 3) is then given by
B(3) = 5000 · 2 3/4 ≈ 8409.

It is important to realize that population growth problems cannot be described
using just arithmetic. Exponential growth is nonlinear, and advanced analytical
ideas (such as calculus) must be used to understand it.

EXAMPLE 6.31
Suppose that a certain petri dish contains 6000 bacteria at 9:00 p.m. and
10000 bacteria at 11:00 p.m. How many of the bacteria were there at
7:00 p.m.?
SOLUTION
We know that
B(t) = P · e Kt .

The algebra is always simpler if we take one of the times in the initial data to
correspond to t = 0. So let us say that 9:00 p.m. is t = 0. Then 11:00 p.m. is
t = 2 and 7:00 p.m. is t =−2. The initial data then tell us that
K·0
6000 = P · e (∗)
10000 = P · e K·2 . (∗∗)

From equation (∗) we may immediately conclude that P = 6000. Substituting
this into (∗∗) gives
K 2
10000 = 6000 · (e ) .
We conclude that
√
5
e K = √ .
3