0
$\begingroup$

Here's the problem:

A biologist places agar, a gel made from seaweed, in a Petri dish and infects it with bacteria. She uses the measurement of the growth ring to estimate the number of bacteria present. The biologist finds that the bacteria increase in population at an exponential rate of $20%$ every $2$ days.

a) If the culture starts with a population of $5000$ bacteria, what is the transformed exponential function in the form $P = a(c)^{bx}$ that represents the population, $P$, of the bacteria over time, $x$, in days?

My problem is creating the exponential function.

What I did so far is $P = 5000(0.20)^{???x}$

Where $5000$ is the starting population, $0.20$ is the exponential rate, and I'm not sure how to transform the $x$ so that I could get a rate of "every $2$ days."

The textbook had an answer of:

$P = 5000(1.2)^{\frac{1}{2}x}$

I wanna understand how I could get this answer.

Why is the constant $1.2$? I'm assuming they added a $1$ for the first day of growth; but why would they do that?

Also, how is "every $2$ days" modeled by multiplying $x$ by $\frac{1}{2}$?

$\endgroup$
  • 2
    $\begingroup$ You speak about an increase : $1$ being the initiel value and $0.2$ the increase every two days, then, after two days you have $1+0.2=1.2$ $\endgroup$ – Claude Leibovici Apr 7 at 5:49
2
$\begingroup$

We know that $P(0) = 5000$, so if the form is $P = a(c)^{bx}$ then $5000 = a(c)^{b*0} = a$. Then we know that $\frac{P(x+2)}{P(x)} = 1.2$, so $\frac{5000c^{b(x+2)}}{5000c^{bx}} = 1.2$ which yields $c^{2b} = 1.2$. At this point, let $c > 1$ be whatever you want and calculate $b$. In this case, we can let $c=1.2$ since that's the rate that the problem provided. Then $b$ must be $\frac{1}{2}$.

The intuition is as you said, since the rate of growth is $.2$ for every two years, then if that's the rate of growth we choose for $c$, then $b$ must be halved to account for that, as $x$ would be growing "two times as fast". $c$ also must be greater than $1$, as the growth rate adds on top of what was there before.

$\endgroup$
  • $\begingroup$ How does $\frac {P(x+2)}{P(x)} = .2$ yield $c^{2b} = 1.2?$ $\endgroup$ – Jenny B Apr 7 at 6:18
  • $\begingroup$ Whoops, I meant 1.2 there. $\endgroup$ – Vasting Apr 7 at 6:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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