Once an antibiotic is introduced to treat a bacterial infection, the number of bacteria decreases exponentially. For example, beginning with 1 million bacteria, the amount present $$ t $$ days from the time is introduced is given by the function $$ A(t)=1,000,000(2)^{-t / 5} $$. Rounding to the nearest thousand, determine how many bacteria are present after $$ \begin{array}{ll} ext { (a) } 2 ext { days } & ext { (b) } 1 ext { week } \ ext { Part } 1 ext { of } 3 & ext { (c) } 2 ext { weeks } \ ext { (a) After } 2 ext { days, approximately } \ ext { Part } 2 ext { of } 3 \ ext { (b) After } 1 ext { week, approximately } \ ext { Part } 3 ext { of } 3\end{array} $$ (c) After 2 weeks, approximately bacteria remain. bain. remain. (b)

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Once an antibiotic is introduced to treat a bacterial infection, the number of bacteria decreases exponentially. For example, beginning with 1 million bacteria, the amount present $$ t $$ days from the time is introduced is given by the function $$ A(t)=1,000,000(2)^{-t / 5} $$. Rounding to the nearest thousand, determine how many bacteria are present after $$ \begin{array}{ll}	ext { (a) } 2 	ext { days } & 	ext { (b) } 1 	ext { week } \ 	ext { Part } 1 	ext { of } 3 & 	ext { (c) } 2 	ext { weeks } \ 	ext { (a) After } 2 	ext { days, approximately } \ 	ext { Part } 2 	ext { of } 3 \ 	ext { (b) After } 1 	ext { week, approximately } \ 	ext { Part } 3 	ext { of } 3\end{array} $$ (c) After 2 weeks, approximately bacteria remain. bain. remain. (b)

UpStudy AI · Accepted Answer

We are given
$$
A(t)=1{,}000{,}000\,(2)^{-t/5}.
$$

**(a) After 2 days:**  
Substitute $$ t=2 $$:
$$
A(2)=1{,}000{,}000\,(2)^{-2/5}.
$$
Compute the exponent:
$$
2^{-2/5}=\exp\Bigl(-\frac{2}{5}\ln 2\Bigr)=\exp(-0.4\ln 2).
$$
Since $$\ln 2\approx0.693$$, we have
$$
-0.4\ln 2\approx -0.277.
$$
Thus,
$$
2^{-2/5}\approx\exp(-0.277)\approx0.758.
$$
Then,
$$
A(2)\approx1{,}000{,}000\times0.758\approx758{,}000.
$$
Rounding to the nearest thousand gives approximately $$\boxed{758{,}000}$$ bacteria.

---

**(b) After 1 week (7 days):**  
Substitute $$ t=7 $$:
$$
A(7)=1{,}000{,}000\,(2)^{-7/5}.
$$
Compute the exponent:
$$
2^{-7/5}=\exp\Bigl(-\frac{7}{5}\ln 2\Bigr)=\exp(-1.4\ln 2).
$$
With $$\ln 2\approx0.693$$:
$$
-1.4\ln 2\approx -0.970,
$$
so
$$
2^{-7/5}\approx\exp(-0.970)\approx0.379.
$$
Thus,
$$
A(7)\approx1{,}000{,}000\times0.379\approx379{,}000.
$$
Rounded to the nearest thousand, there are approximately $$\boxed{379{,}000}$$ bacteria.

---

**(c) After 2 weeks (14 days):**  
Substitute $$ t=14 $$:
$$
A(14)=1{,}000{,}000\,(2)^{-14/5}=1{,}000{,}000\,(2)^{-2.8}.
$$
Compute the exponent:
$$
2^{-2.8}=\exp(-2.8\ln 2).
$$
With $$\ln 2\approx0.693$$:
$$
-2.8\ln 2\approx -1.940,
$$
so
$$
2^{-2.8}\approx\exp(-1.940)\approx0.144.
$$
Thus,
$$
A(14)\approx1{,}000{,}000\times0.144\approx144{,}000.
$$
Rounded to the nearest thousand, there are approximately $$\boxed{144{,}000}$$ bacteria.
After 2 days, there are approximately 758,000 bacteria. After 1 week, there are approximately 379,000 bacteria. After 2 weeks, there are approximately 144,000 bacteria.

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