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12. In a petri dish, bacteria triples every 3 hours. If there were initially 300 bacteria and there are currently 5000 bacteria, determine the equation that would be used to find how long the bacteria have been in the dish.

Given:
[tex]\[ A_0 = 300 \][/tex]
[tex]\[ A = 5000 \][/tex]
[tex]\[ T = 3 \text{ hours} \][/tex]

Using the formula for exponential growth:
[tex]\[ A = A_0 \cdot (3)^{\frac{t}{T}} \][/tex]

Where:
[tex]\[ A = \text{final amount of bacteria} \][/tex]
[tex]\[ A_0 = \text{initial amount of bacteria} \][/tex]
[tex]\[ T = \text{time period in which the bacteria triples} \][/tex]
[tex]\[ t = \text{total time elapsed} \][/tex]

Plugging in the given values:
[tex]\[ 5000 = 300 \cdot (3)^{\frac{t}{3}} \][/tex]

Solve for [tex]\( t \)[/tex].


Sagot :

Sure, let's determine the equation needed to find out how long the bacteria have been in the dish. Here is a step-by-step explanation:

1. Start with the given data:
- Initial number of bacteria, [tex]\( A_0 = 300 \)[/tex]
- Current number of bacteria, [tex]\( A = 5000 \)[/tex]
- The bacteria triples every [tex]\( T = 3 \)[/tex] hours

2. Write the exponential growth equation:
[tex]\[ A = A_0 \cdot (3^{t/T}) \][/tex]
where [tex]\( t \)[/tex] is the time in hours.

3. Plug in the known values:
[tex]\[ 5000 = 300 \cdot (3^{t/3}) \][/tex]

4. Solve for [tex]\( t \)[/tex]:

- Divide both sides by 300:
[tex]\[ \frac{5000}{300} = 3^{t/3} \][/tex]

- Simplify the left-hand side:
[tex]\[ \frac{5000}{300} = 16.\overline{6} \][/tex]
So,
[tex]\[ 16.\overline{6} = 3^{t/3} \][/tex]

- To isolate [tex]\( t \)[/tex], take the natural logarithm of both sides:
[tex]\[ \ln(16.\overline{6}) = \ln(3^{t/3}) \][/tex]

- Use the properties of logarithms ([tex]\( \ln(a^b) = b\ln(a) \)[/tex]):
[tex]\[ \ln(16.\overline{6}) = \frac{t}{3} \ln(3) \][/tex]

- Solve for [tex]\( t \)[/tex]:
[tex]\[ t = 3 \cdot \frac{\ln(16.\overline{6})}{\ln(3)} \][/tex]

This is the equation that you would use to determine how long the bacteria have been in the dish.
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