Answer
a) $B(m)=4000(2)^{\frac{m}{10}}$
b) $80$ minutes
c) $ 67,108,864,000$
Work Step by Step
a) Let $B(m)$ be the number of Salmonella bacteria on the counter after $m$ minutes. Then, the initial bacteria population is $B_0= 4000$, the doubling constant is $b= 2$ and the exponent is $a=\frac{m}{10}$, since the bacteria double every 10 minutes. Hence, our model is.
$$ B(m) = B_0(b)^a= 4000(2)^{\frac{m}{10}}$$ b) First solve for $m$ and set $B(m)= 1000000$ to find $m$. \begin{equation}
\begin{aligned}
B(m) &= 4000(2)^{\frac{m}{10}}\\
\ln B(m) & =\ln\left(4000(2)^{\frac{m}{10}} \right)\\
\ln B(m) & =\ln 2^{\frac{m}{10}}+\ln\left(4000 \right)\\
\ln B(m)- \ln\left(4000 \right) & =\frac{m}{10}\ln (2)\\
\therefore m&= \frac{10\cdot \ln B(m)-10\ln(4000)}{\ln (2)}\\
m& = \frac{10\cdot \ln (1000000)-10\ln(4000)}{\ln (2)}\\
&\approx 80.
\end{aligned}
\end{equation} The bacteria population will be $1$ million after about $80$ minutes.
c) There are $4$ hours between $2$ pm to $6$ pm. This is equivalent to $240$ minutes. Find $B(240)$. \begin{equation}
\begin{aligned}
B(240)&=4000(2)^{\frac{240}{10}}\\
&=4000(2)^{24}\\
&= 67,108,864,000
\end{aligned}
\end{equation} The bacteria population will become $67,108,864,000$ if still unnoticed until $6$ pm.
