algebra precalculus - Transforming this exponential function?

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 $\frac{1}{2}$?

Answer

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.