Uninhibited Exponential Growth Model
A colony of bacteria grows according to the law of uninhibited growth. The size of the colony, measured in grams, at time, $t$, measured in days, $B(t)=103e^{0.057t}$
a. What is the initial size of the bacteria colony?
b. What is the growth rate for this bacteria colony? $◻\mathrm{\%}$
c. What is the size of the bacteria colony after 7 days? $◻g$
d. How long will it take the colony size to reach 176 grams? $◻$ days
d. What is the doubling time for this colony? $◻$ days

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Step 1 ：Substitute $t=0$ into the equation to find the initial size of the bacteria colony: $B(0)=103e^{0.057\cdot 0}=103e^0=103$

Step 2 ：\boxed{103}

Step 3 ：The growth rate for this bacteria colony is given by the coefficient of $t$ in the exponent of $e$, which is 0.057. To convert this to a percentage, we multiply by 100, giving a growth rate of 5.7%.

Step 4 ：\boxed{5.7\%}

Step 5 ：Substitute $t=7$ into the equation to find the size of the bacteria colony after 7 days: $B(7)=103e^{0.057\cdot 7}=103e^{0.399}=103\cdot 1.49=153.27$

Step 6 ：\boxed{153.27}

Step 7 ：Set $B(t)=176$ and solve for $t$ to find out how long it will take the colony size to reach 176 grams: $176=103e^{0.057t}$. Dividing both sides by 103 gives: $1.7087=e^{0.057t}$. Taking the natural logarithm of both sides gives: $\ln (1.7087)=0.057t$. Dividing both sides by 0.057 gives: $t=\frac{\ln (1.7087)}{0.057}=8.5$

Step 8 ：\boxed{8.5}

Step 9 ：Set $B(t)=2\cdot 103$ and solve for $t$ to find the doubling time for this colony: $206=103e^{0.057t}$. Dividing both sides by 103 and taking the natural logarithm gives: $t=\frac{\ln (2)}{0.057}=12.2$

Step 10 ：\boxed{12.2}