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:
- Brackets.
- Parenthesis.
- Exponents.
- Multiplication.
- Division.
- Addition.
- Subtraction.
Definitions:
We are given the following Exponential Growth Function (Doubling Time Model), [tex]\displaystyle\mathsf{P_{(t)}\:=\:P_0\cdot2^{(t/d)}}[/tex] where:
- [tex]\displaystyle\sf{P_t\:\:\rightarrow}[/tex] The population of bacteria after “t ” number of hours.
- [tex]\displaystyle\sf{P_0 \:\:\rightarrow}[/tex] The initial population of bacteria.
- [tex]\displaystyle{t \:\:\rightarrow}[/tex] Time unit (in hours).
- [tex]\displaystyle{\textit d \:\:\rightarrow}[/tex] 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.
- [tex]\displaystyle\sf{P_0\:=}[/tex] 9,300.
- t = 10 hours.
- d = 3.
Step 2: Find value.
1. Substitute the values into the given exponential function.
[tex]\displaystyle\mathsf{P_{(t)} = P_0\cdot2^{(t/d)}}[/tex]
[tex]\displaystyle\mathsf{\longrightarrow P_{(10)} = 9300\cdot2^{(10/3)}}[/tex]
2. Evaluate using the BPEMDAS order of operations.
[tex]\displaystyle\mathsf{P_{(10)} = 9300\cdot2^{(10/3)}\quad \Longrightarrow BPEMDAS:\:(Parenthesis\:\:and\:\:Division).}[/tex]
[tex]\displaystyle\sf P_{(10)} = 9300\cdot2^{(3.333333)}\quad\Longrightarrow BPEMDAS:\:(Exponent).}[/tex]
[tex]\displaystyle\sf P_{(10)} = 9300\cdot(10.079368399)\quad \Longrightarrow BPEMDAS:(Multiplication).}[/tex]
[tex]\boxed{\displaystyle\mathsf{P_{(10)} \approx 93,738.13\:\:\:or\:\:93,738}}[/tex]
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:
- [tex]\displaystyle\mathsf{P_{(t)} = 93,738 }[/tex].
- [tex]\displaystyle\mathsf{P_0 = 9,300}[/tex].
- d = 3.
2. Substitute the values into the given exponential function.
[tex]\displaystyle\mathsf{P_{(t)} = P_0\cdot2^{(t/d)}}[/tex]
[tex]\displaystyle\mathsf{\longrightarrow 93,378 = 9,300\cdot2^{(t/3)}}[/tex]
3. Divide both sides by 9,300:
[tex]\displaystyle\mathsf{\longrightarrow \frac{93,378}{9,300} = \frac{9,300\cdot2^{(t/3)}}{9,300}}[/tex]
[tex]\displaystyle\mathsf{\longrightarrow 10.07936840 = 2^{(t/3)}}[/tex]
4. Transform the right-hand side of the equation into logarithmic form.
[tex]\boxed{\displaystyle\mathsf{\underbrace{ x = a^y}_{Exponential\:Form} \longrightarrow \underbrace{y = log_a x}_{Logarithmic\:Form}}}[/tex]
[tex]\displaystyle\mathsf{\longrightarrow 10.07936840 = \bigg[\:\frac{t}{3}\:\bigg]log(2)}[/tex]
5. Take the log of both sides of the equation (without rounding off any digits).
[tex]\displaystyle\mathsf{log(10.07936840) = \bigg[\:\frac{t}{3}\:\bigg]log(2)}[/tex]
[tex]\displaystyle\mathsf{\longrightarrow 1.003433319 = \bigg[\:\frac{t}{3}\:\bigg]\cdot(0.301029996)}[/tex]
6. Divide both sides by (0.301029996).
[tex]\displaystyle\mathsf{\frac{1.003433319}{0.301029996} = \frac{\bigg[\:\frac{t}{3}\:\bigg]\cdot(0.301029996) }{0.301029996}}[/tex]
[tex]\displaystyle\mathsf{\longrightarrow 3.3333333 = \frac{t}{3}}[/tex]
7. Multiply both sides of the equation by 3 to isolate "t."
[tex]\displaystyle\mathsf{(3)\cdot(3.3333333) = \bigg[\:\frac{t}{3}\:\bigg]\cdot(3)}[/tex]
[tex]\boxed{\displaystyle\mathsf{t\approx10}}[/tex]
Hence, it will take about 10 hours for the population of bacteria to increase to 93,378.
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