Approximately how many hours have passed when there are 4500 bacteria left alive?

Given: [tex]4500 = 75,000 e^{-0.1733 t}[/tex]

A. 16.4 hours
B. 16.6 hours
C. 16.0 hours
D. 16.2 hours



Answer :

To determine how many hours have passed when the number of live bacteria is reduced from 75,000 to 4500, we can use the following exponential decay equation:
[tex]\[ 4500 = 75,000 \cdot e^{-0.1733t} \][/tex]

We'll solve for [tex]\( t \)[/tex], which represents the time in hours.

1. Start with the given equation:
[tex]\[ 4500 = 75,000 \cdot e^{-0.1733t} \][/tex]

2. Divide both sides of the equation by 75,000 to isolate the exponential term:
[tex]\[ \frac{4500}{75000} = e^{-0.1733t} \][/tex]

3. Simplify the fraction on the left side:
[tex]\[ \frac{4500}{75000} = 0.06 \][/tex]

So the equation then becomes:
[tex]\[ 0.06 = e^{-0.1733t} \][/tex]

4. Take the natural logarithm (ln) of both sides to solve for the exponent:
[tex]\[ \ln(0.06) = \ln(e^{-0.1733t}) \][/tex]

Using the property of logarithms [tex]\(\ln(e^x) = x\)[/tex]:
[tex]\[ \ln(0.06) = -0.1733t \][/tex]

5. Calculate the natural logarithm of 0.06:
[tex]\[ \ln(0.06) \approx -2.81341 \][/tex]

So, the equation now is:
[tex]\[ -2.81341 = -0.1733t \][/tex]

6. Solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-2.81341}{-0.1733} \][/tex]

7. Perform the division:
[tex]\[ t \approx 16.23434 \][/tex]

Therefore, the time [tex]\( t \)[/tex] that has passed is approximately 16.2 hours.

So the correct choice is:
[tex]\[ \boxed{\text{D. 16.2 hours}} \][/tex]