To determine how long it takes for the size of the sample to double, we start with the given exponential growth function:
[tex]\[
y = y_0 e^{0.0623 t}
\][/tex]
Here, [tex]\( y_0 \)[/tex] is the initial number of bacteria, and [tex]\( y \)[/tex] is the number of bacteria after time [tex]\( t \)[/tex] hours. We need to find the time [tex]\( t \)[/tex] at which the number of bacteria has doubled.
This means we want to find the time [tex]\( t \)[/tex] when [tex]\( y = 2 y_0 \)[/tex]. Substituting this condition into our equation gives:
[tex]\[
2 y_0 = y_0 e^{0.0623 t}
\][/tex]
We can simplify this by dividing both sides by [tex]\( y_0 \)[/tex]:
[tex]\[
2 = e^{0.0623 t}
\][/tex]
To solve for [tex]\( t \)[/tex], we need to eliminate the exponential function by taking the natural logarithm of both sides. Recall that the natural logarithm [tex]\( \ln \)[/tex] and the exponential function [tex]\( e \)[/tex] are inverse functions:
[tex]\[
\ln(2) = \ln(e^{0.0623 t})
\][/tex]
Using the property of logarithms that [tex]\(\ln(e^a) = a\)[/tex], we get:
[tex]\[
\ln(2) = 0.0623 t
\][/tex]
Next, we solve for [tex]\( t \)[/tex] by dividing both sides by the rate 0.0623:
[tex]\[
t = \frac{\ln(2)}{0.0623}
\][/tex]
Given that [tex]\(\ln(2) \)[/tex] is approximately 0.6931471805599453, we calculate:
[tex]\[
t = \frac{0.6931471805599453}{0.0623}
\][/tex]
Finally, this yields:
[tex]\[
t \approx 11.1 \text{ hours}
\][/tex]
Therefore, it takes approximately 11.1 hours for the size of the bacteria sample to double.
