Chapter 4.6, Problem 1E

Population Growth These exercises use the population growth model.

1. Bacteria Culture A certain culture of the bacterium Streptococcus A initially has 10 bacteria and is observed to double every 1.5 hours.

1. (a) Find an exponential model n(t) = n₀2^t/a for the number of bacteria in the culture after t hours.
2. (b) Estimate the number of bacteria after 35 hours.
3. (c) After how many hours will the bacteria count reach 10,000?

To determine

To find: An exponential model $n\left(t\right)=n_0{2}^{\frac{t}{a}}$ for the number of bacteria in the culture after $t$ hours.

Answer to Problem 1E

The exponential model for the number of bacteria in the culture after $t$ hours is $n\left(t\right)=10{\left(2\right)}^{\frac{2t}{3}}$.

Explanation of Solution

Given:

The initial bacteria population is $10$.

The bacteria population is double in $1.5\text{hours}$.

Formula used: Exponential growth model (doubling time)

$n\left(t\right)=n_0{2}^{\frac{t}{a}}$ (I)

Where,

• $n_0$ is the initial size of the population.
• $a$ is the doubling time.
• $n\left(t\right)$ is size of the population at time $t$.

Calculation:

The given initial bacteria population is $10$. and the bacteria population is double in $1.5\text{hours}$.

Substitute $10$ for $n_0$ and $1.5$ for $a$ in equation (I) to find the model for the number of bacteria in the culture after $t$ hours,

$\begin{array}{c}n\left(t\right)=10\left(2^{\frac{t}{1.5}}\right)\\ =10\left(2^{\frac{2t}{3}}\right)\end{array}$

Thus, the exponential model for the number of bacteria in the culture after $t$ hours is $n\left(t\right)=10{\left(2\right)}^{\frac{2t}{3}}$.

To determine

To find: The number of bacteria in the culture after $35\text{hours}$.

Answer to Problem 1E

The number of bacteria in the culture after $35\text{hours}$ is ${10}^8\left(1.06\right)$.

Explanation of Solution

Given:

From part (a), the exponential model for the number of bacteria in the culture after $t$ hours is given below,

$n\left(t\right)=10{\left(2\right)}^{\frac{2t}{3}}$ (II)

Calculation:

Substitute $35$ for $t$ in equation (II) to find the number of bacteria after $35\text{hours}$.

$\begin{array}{c}n\left(35\right)=10{\left(2\right)}^{\frac{2\left(35\right)}{3}}\\ =10\left(2^{\frac{70}{3}}\right)\\ =10\left(10568983.798\right)\end{array}$

Simplify the above expression,

$\begin{array}{c}n(35)=10\left({10}^7\right)\left(1.06\right)\\ ={10}^8\left(1.06\right)\end{array}$

Thus, the number of bacteria in the culture after $35\text{hours}$ is ${10}^8\left(1.06\right)$.

To determine

To find: The time in which bacteria count reach to $10000$.

Answer to Problem 1E

The bacteria count reach to $10000$ in $14.9\text{hours}$.

Explanation of Solution

Given:

From part (a), the exponential model for the number of bacteria in the culture after $t$ hours is given below,

$n\left(t\right)=10{\left(2\right)}^{\frac{2t}{3}}$

Calculation:

Substitute $10000$ for $n\left(t\right)$ in equation (II) to find the time in which bacteria population reach to $10000$,

$\begin{array}{c}10000=10{\left(2\right)}^{\frac{2t}{3}}\\ 1000=2^{\frac{2t}{3}}\\ {10}^3=2^{\frac{2t}{3}}\end{array}$

Simplify the above expression taking $\log$ to each side,

$\begin{array}{c}\log {10}^3=\log 2^{\frac{2t}{3}}\\ 3\log 10=\frac{2t}{3}\log 2\left(\because \log a^b=b\log a\right)\\ 9=2t\log 2\left(\because \log 10=1\right)\\ t=\frac{9}{2\log 2}\end{array}$

Further simplify the above expression,

$\begin{array}{c}t=\frac{9}{2\left(0.301030\right)}\left(\because \log 2=0.301030\right)\\ =\frac{9}{0.602060}\\ =14.94868\\ \approx 14.9\end{array}$

Thus, the bacteria count reach to $10000$ in $14.9\text{hours}$.