At the beginning of a study, a certain culture of bacteria has a population of 810 . The population grows according to a continuous exponential growth model. After 21 days, there are 972 bacteria. $$ \begin{array}{l}\text { (a) Let } t \text { be the time (in days) since the beginning of the study, and let } y \ \text { be the number of bacteria at time } t \text {. } \ \text { Write a formula relating } y \text { to } t \text {. } \ \text { Use exact expressions to fill in the missing parts of the formula. } \ \text { Do not use approximations. } \ y=\square e \text { milit } \ \text { (b) How many bacteria are there } 22 \text { days after the beginning of the } \ \text { study? } \ \text { Do not round any intermediate computations, and round your } \ \text { answer to the nearest whole number. } \ \text { bacteria }\end{array} $$

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Given that the population of bacteria grows according to a continuous exponential growth model, we can use the formula for exponential growth to relate the number of bacteria $$ y $$ to the time $$ t $$ since the beginning of the study.

The formula for exponential growth is given by:
$$ y = y_0 e^{kt} $$
where:
- $$ y_0 $$ is the initial population of bacteria,
- $$ k $$ is the growth rate,
- $$ t $$ is the time since the beginning of the study.

We are given that the initial population of bacteria is 810 and after 21 days, there are 972 bacteria. We can use this information to find the growth rate $$ k $$ and then use it to find the number of bacteria after 22 days.

Let's first find the growth rate $$ k $$ using the given information:
$$ 972 = 810 e^{k \cdot 21} $$

Now, we can solve for $$ k $$ and then use it to find the number of bacteria after 22 days.
Solve the equation by following steps:
- step0: Solve for $$k$$:
  $$972=810e^{k\times 21}$$
- step1: Reorder the terms:
  $$972=810e^{21k}$$
- step2: Swap the sides:
  $$810e^{21k}=972$$
- step3: Divide both sides:
  $$\frac{810e^{21k}}{810}=\frac{972}{810}$$
- step4: Divide the numbers:
  $$e^{21k}=\frac{6}{5}$$
- step5: Take the logarithm of both sides:
  $$\ln{\left(e^{21k}\right)}=\ln{\left(\frac{6}{5}\right)}$$
- step6: Evaluate the logarithm:
  $$21k=\ln{\left(\frac{6}{5}\right)}$$
- step7: Divide both sides:
  $$\frac{21k}{21}=\frac{\ln{\left(\frac{6}{5}\right)}}{21}$$
- step8: Divide the numbers:
  $$k=\frac{\ln{\left(\frac{6}{5}\right)}}{21}$$
- step9: Simplify:
  $$k=\frac{\ln{\left(6\right)}-\ln{\left(5\right)}}{21}$$
The growth rate $$ k $$ is given by:
$$ k = \frac{\ln(6) - \ln(5)}{21} $$

Now, we can use this growth rate to find the number of bacteria after 22 days. The formula for the number of bacteria after 22 days is:
$$ y = 810 e^{k \cdot 22} $$

Substitute the value of $$ k $$ into the formula and calculate the number of bacteria after 22 days.
Calculate the value by following steps:
- step0: Calculate:
  $$810e^{\frac{\ln{\left(6\right)}-\ln{\left(5\right)}}{21}\times 22}$$
- step1: Use the logarithm product rule:
  $$810e^{\frac{\ln{\left(\frac{6}{5}\right)}}{21}\times 22}$$
- step2: Multiply the numbers:
  $$810e^{\frac{22\ln{\left(\frac{6}{5}\right)}}{21}}$$
- step3: Transform the expression:
  $$810\left(\frac{6}{5}\right)^{\frac{22}{21}}$$
- step4: Simplify:
  $$810\times \frac{6^{\frac{22}{21}}}{5^{\frac{22}{21}}}$$
- step5: Reduce the numbers:
  $$162\times \frac{6^{\frac{22}{21}}}{5^{\frac{1}{21}}}$$
- step6: Multiply:
  $$\frac{162\times 6^{\frac{22}{21}}}{5^{\frac{1}{21}}}$$
- step7: Simplify:
  $$\frac{162\times 6\sqrt[21]{6}}{5^{\frac{1}{21}}}$$
- step8: Simplify:
  $$\frac{162\times 6\sqrt[21]{6}}{\sqrt[21]{5}}$$
- step9: Multiply by the Conjugate:
  $$\frac{162\times 6\sqrt[21]{6}\times \sqrt[21]{5^{20}}}{\sqrt[21]{5}\times \sqrt[21]{5^{20}}}$$
- step10: Multiply the numbers:
  $$\frac{972\sqrt[21]{6\times 5^{20}}}{\sqrt[21]{5}\times \sqrt[21]{5^{20}}}$$
- step11: Multiply the numbers:
  $$\frac{972\sqrt[21]{6\times 5^{20}}}{5}$$
The number of bacteria after 22 days is approximately 980.
(a) The formula relating $$ y $$ to $$ t $$ is: $$ y = 810 e^{kt} $$ (b) After 22 days, there are approximately 980 bacteria.

At the beginning of a study, a certain culture of bacteria has a population of 810 . The population grows according to a continuous exponential growth model.
After 21 days, there are 972 bacteria.

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