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, $B(t)=103 e^{0.057 t}$ a. What is the initial size of the bacteria colony? b. What is the growth rate for this bacteria colony? $\square \%$ c. What is the size of the bacteria colony after 7 days? $\square g$ d. How long will it take the colony size to reach 176 grams? $\square$ days d. What is the doubling time for this colony? $\square$ days Note: You can earn partial credit on this problem. Preview My Answers Submit Answers

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

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: \(B(7)=103 e^{0.057 \cdot 7}=103 e^{0.399}=103 \cdot 1.49=153.27\)

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: \(176=103 e^{0.057 t}\). Dividing both sides by 103 gives: \(1.7087=e^{0.057 t}\). Taking the natural logarithm of both sides gives: \(\ln(1.7087)=0.057 t\). 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: \(206=103 e^{0.057 t}\). Dividing both sides by 103 and taking the natural logarithm gives: \(t=\frac{\ln(2)}{0.057}=12.2\)

Step 10 ：\boxed{12.2}