Question

A culture of bacteria has an initial population of 9300 bacteria and doubles every 3

Answer 1

Answer:

The approximate population of bacteria in the culture after 10 hours is 93,738.  

Step-by-step explanation:

General Concepts:

• Exponential Functions.
• Exponential Growth.
• Doubling Time Model.
• Logarithmic Form.

BPEMDAS Order of Operations:

1. Brackets.
2. Parenthesis.
3. Exponents.
4. Multiplication.
5. Division.
6. Addition.
7. Subtraction.

Definitions:

We are given the following Exponential Growth Function (Doubling Time Model), $\displaystyle\mathsf{P_{(t)}\:=\:P_0\cdot2^{(t/d)}}$ where:

• $\displaystyle\sf{P_t\:\:\rightarrow}$ The population of bacteria after “t ” number of hours.
• $\displaystyle\sf{P_0 \:\:\rightarrow}$ The initial population of bacteria.
• $\displaystyle{t \:\:\rightarrow}$  Time unit (in hours).
• $\displaystyle{\textit d \:\:\rightarrow}$  Doubling time, which represents the amount of time it takes for the population of bacteria to grow exponentially to become twice its initial quantity.  

Solution:

Step 1: Identify the given values.

• $\displaystyle\sf{P_0\:=}$ 9,300.
• t = 10 hours.
• d = 3.  

Step 2: Find value.

1. Substitute the values into the given exponential function.

  $\displaystyle\mathsf{P_{(t)} = P_0\cdot2^{(t/d)}}$

  $\displaystyle\mathsf{\longrightarrow P_{(10)} = 9300\cdot2^{(10/3)}}$

2. Evaluate using the BPEMDAS order of operations.

  $\displaystyle\mathsf{P_{(10)} = 9300\cdot2^{(10/3)}\quad \Longrightarrow BPEMDAS:\:(Parenthesis\:\:and\:\:Division).}$

  $\displaystyle\sf P_{(10)} = 9300\cdot2^{(3.333333)}\quad\Longrightarrow BPEMDAS:\:(Exponent).}$

  $\displaystyle\sf P_{(10)} = 9300\cdot(10.079368399)\quad \Longrightarrow BPEMDAS:(Multiplication).}$

 $\boxed{\displaystyle\mathsf{P_{(10)} \approx 93,738.13\:\:\:or\:\:93,738}}$

Hence, the population of bacteria in the culture after 10 hours is approximately 93,738.  

Double-check:

We can solve for the amount of time (t ) it takes for the population of bacteria to increase to 93,738.

1. Identify given:

• $\displaystyle\mathsf{P_{(t)} = 93,738 }$.
• $\displaystyle\mathsf{P_0 = 9,300}$.
• d = 3.

2. Substitute the values into the given exponential function.

  $\displaystyle\mathsf{P_{(t)} = P_0\cdot2^{(t/d)}}$

  $\displaystyle\mathsf{\longrightarrow 93,378 = 9,300\cdot2^{(t/3)}}$

3. Divide both sides by 9,300:

  $\displaystyle\mathsf{\longrightarrow \frac{93,378}{9,300} = \frac{9,300\cdot2^{(t/3)}}{9,300}}$

  $\displaystyle\mathsf{\longrightarrow 10.07936840 = 2^{(t/3)}}$

4. Transform the right-hand side of the equation into logarithmic form.

  $\boxed{\displaystyle\mathsf{\underbrace{ x = a^y}_{Exponential\:Form} \longrightarrow \underbrace{y = log_a x}_{Logarithmic\:Form}}}$    

  $\displaystyle\mathsf{\longrightarrow 10.07936840 = \bigg[\:\frac{t}{3}\:\bigg]log(2)}$  

5. Take the log of both sides of the equation (without rounding off any digits).  

  $\displaystyle\mathsf{log(10.07936840) = \bigg[\:\frac{t}{3}\:\bigg]log(2)}$

  $\displaystyle\mathsf{\longrightarrow 1.003433319 = \bigg[\:\frac{t}{3}\:\bigg]\cdot(0.301029996)}$

6. Divide both sides by (0.301029996).

  $\displaystyle\mathsf{\frac{1.003433319}{0.301029996} = \frac{\bigg[\:\frac{t}{3}\:\bigg]\cdot(0.301029996) }{0.301029996}}$

  $\displaystyle\mathsf{\longrightarrow 3.3333333 = \frac{t}{3}}$

7. Multiply both sides of the equation by 3 to isolate "t."

  $\displaystyle\mathsf{(3)\cdot(3.3333333) = \bigg[\:\frac{t}{3}\:\bigg]\cdot(3)}$

  $\boxed{\displaystyle\mathsf{t\approx10}}$

Hence, it will take about 10 hours for the population of bacteria to increase to 93,378.    

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