College Algebra, Chapter 9, 9.3, Section 9.3, Problem 70

A culture initially has 5000 bacteria and its size increases by $8\%$ every hour. How many bacteria are present at the end of 5 hours? Find a formula for the number of bacteria present after $n$ hours.

If $a_1 = 5000$ and $a_2 = a_1 + 0.08a_1 = 1.08a_1$, then the common ratio is $r = 1.08$. Therefore, the formula for the number of bacteria present after $n$ hours is


$
\begin{equation}
\begin{aligned}

S_n =& a \frac{1 - r^n}{1 - r}
\\
\\
S_n =& 5000 \left[ \frac{1 - (1.08)^n}{1 - (1.08)} \right]
\\
\\
S_n =& \frac{5000}{-0.08} [1 - (1.08)^n]
\\
\\
S_n =& -62500 [1 - (1.08)^n]

\end{aligned}
\end{equation}
$


Then, the total number of bacteria present after 5 hours is


$
\begin{equation}
\begin{aligned}

S_5 =& -62500 [1 - (1.08)^5]
\\
\\
S_5 =& 29333

\end{aligned}
\end{equation}
$