6. Suppose $y(t)=40 e^{2 t}+8$ represents the number of bacteria present at time $t$ minutes. At what time will the population reach 100 bacteria? (Note: Answers are expressed in terms of the natural logarithm.)

Step 1 ：We are given the equation \(y(t)=40 e^{2 t}+8\) and we need to find the time \(t\) when the population \(y(t)\) reaches 100. This means we need to solve the equation \(40 e^{2 t}+8 = 100\) for \(t\).

Step 2 ：Subtract 8 from both sides of the equation to isolate the exponential term: \(40 e^{2 t} = 100 - 8 = 92\).

Step 3 ：Divide by 40 to get \(e^{2t}\) alone on one side: \(e^{2t} = \frac{92}{40} = 2.3\).

Step 4 ：Take the natural logarithm of both sides to solve for \(t\): \(2t = \ln(2.3)\).

Step 5 ：Finally, divide by 2 to solve for \(t\): \(t = \frac{\ln(2.3)}{2}\).

Step 6 ：\(\boxed{t = \frac{\ln(2.3)}{2}}\) is the time at which the population will reach 100 bacteria.