91Ó°ÊÓ

We are given the following values: - The initial amount is the initial size of the colony, which we'll call P. - The time it takes for the colony to double in size is 140 minutes. - We want to find the time it takes for the colony to become five times its current size, that means A(t) will be 5P, where P is the initial size.

We know that the colony doubles in size every 140 minutes. We can plug this information into the exponential growth formula: \[ 2P = P(1 + r)^{140} \] We need to solve this equation for the growth rate, r. First, divide both sides by P: \[ 2 = (1 + r)^{140} \] Now, take the 140th root of both sides: \[ \sqrt[140]{2} = 1 + r \] Subtract 1 from both sides to solve for r: \[ r = \sqrt[140]{2} - 1 \]

Now that we have the growth rate, we can use it to find the time it takes for the colony to grow to five times its initial size. We'll plug in this information into the exponential growth formula: \[ 5P = P(1 + (\sqrt[140]{2} - 1))^t \] First, divide both sides by P: \[ 5 = (1 + (\sqrt[140]{2} - 1))^t \] Now, take the natural logarithm (ln) of both sides: \[ \ln(5) = \ln\big((1 + (\sqrt[140]{2} - 1))^t\big) \] Using logarithm properties, bring t down: \[ \ln(5) = t \ln\big(1 + (\sqrt[140]{2} - 1)\big) \] Finally, solve for t by dividing both sides by \( \ln\big(1 + (\sqrt[140]{2} - 1)\big) \): \[ t = \frac{\ln(5)}{\ln\big(1 + (\sqrt[140]{2} - 1)\big)} \] Using a calculator, we find: \[ t \approx 349.969 \] Hence, it will take approximately 350 minutes for the colony of bacteria to become five times its current size.