The number of bacteria in a culture decreases according to a continuous exponential decay model. The initial population in a study is 7600 bacteria, and there are 1748 bacteria left after 12 minutes.
(a) Let $t$ be the time (in minutes) since the beginning of the study, and let $y$ be the number of bacteria at time $t$.
Write a formula relating $y$ to $t$.
Use exact expressions to fill in the missing parts of the formula.
Do not use approximations.
\[
y=\square e^{\square t}
\]
Answer
The final formula relating y to t is: \(\boxed{y = 7600 * e^{\frac{\ln(\frac{1748}{7600})}{12} * t}}\)
Step 1 ：Substitute the given values into the formula: \(1748 = 7600 * e^{12k}\)
Step 2 ：Divide both sides by 7600: \(\frac{1748}{7600} = e^{12k}\)
Step 3 ：Take the natural logarithm of both sides: \(\ln(\frac{1748}{7600}) = 12k\)
Step 4 ：Divide both sides by 12 to solve for k: \(k = \frac{\ln(\frac{1748}{7600})}{12}\)
Step 5 ：Substitute the value of k into the formula: \(y = 7600 * e^{\frac{\ln(\frac{1748}{7600})}{12} * t}\)
Step 6 ：The final formula relating y to t is: \(\boxed{y = 7600 * e^{\frac{\ln(\frac{1748}{7600})}{12} * t}}\)
