Inverse, Exponential, and Logarithmic Functions

Problem : If P = 300e ^2t , at what time will P=600?

Solution for Problem 1 >>

600		= 300e 2t
2		= e 2t

use natural log to solve for t

ln(2)		= 2t
t		= .347

Close

Problem : A strain of bacteria multiply in such a fashion that they double in number every 4 hours. Find an expression that describes this kind of growth.

Solution for Problem 2 >>

A function that grows or decays by a fixed percentage over time exhibits growth of the form

B(t) = B 0 e kt

After 4 hours, B(t) = 2B ₀ , so 2B ₀ = B ₀ e ^k(4) , 2 = e ^4k Use natural log to solve for k

ln(2)		= 4k
k		= .173
B(t)		= B 0 e .173t

Close

Problem : Find a function that meets the following:

= 0.693y

Solution for Problem 3 >>

Any function for which

= ky

must be of the form y = e ^kt so y = e ^0.693t satisfies the condition above. To verify this, see that

e 0.693t = 0.693 e 0.693t

or 0.693y

Close

Problem : The half-life of a substance is the time that it takes for the mass of that substance to decay to 50% of its original value. If a certain substance has a half-life of 30 minutes, what is an equation that describes its decay?

Solution for Problem 4 >>

This is another case of exponential decay, because a certain percentage of the substance decays at regular time intervals. So, the general form is

C(t) = C 0 e kt

At t = 30 , 1/2 of the original remains, so

C 0		= C 0 e k(30)
		= e 30k
ln		= 30k
k		- 0.0231
C(t)		= C 0 e -0.0231t

Close

Problem : After 40 minutes, only 34% of a radioactive compound is remaining. What is the expression describing its decay?

Solution for Problem 5 >>

C(t)		= C 0 e kt
.34C 0		= C 0 e k(40)
.34		= e 40k
ln.34		= 40k
k		- 0.027
C(t)		= C 0 e -0.027t

Close