Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The problem: $\;f(t) = 4 \cdot 2^{\,t/5}$

I know continuous is e^k, but this problem doesn't seem to work for that. Is the initial value, 4, able to be used to find the growth rates?

share|cite|improve this question

The relative continuous growth rate of $f(t)$ is defined as $$\frac{f'(t)}{f(t)}.$$ Your function is $f(t)=4 \cdot 2^{t/5}$, with $f'(t)=4\cdot (1/5)\ln(2)2^{t/5}.$ So its relative growth rate is $(1/5)\ln(2)$. Note how the initial value 4 "cancelled out" in finding the relative continuous growth rate.

However you must be sure if instead you just want the continuous growth rate, i.e. the derivative. Some texts when they write "growth rate" implicitly mean relative growth rate, and this is a common usage for exponential functions, to mean relative growth rate and just call it the growth rate.

share|cite|improve this answer
the growth rates need to be a percent though. – Tyler Zika Dec 12 '12 at 3:58
The growth rate is as I stated. It may be expressed as a percent, but that is irrelevant, since for example 30% means exactly the same as $30 \cdot \frac{1}{100}$ which is $.3$. – coffeemath Dec 12 '12 at 9:03

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.