To determine how long it will take for the bacteria to be 99% gone with a continuous rate of eradication of 0.8% per hour, we can use the exponential decay formula:
\[ N(t) = N_0 e^{-kt} \]
Where:
- \( N(t) \) is the amount of bacteria remaining at time \( t \),
- \( N_0 \) is the initial amount of bacteria,
- \( k \) is the decay constant (rate of eradication), and
- \( t \) is time in hours.
In this case, \( k = 0.008 \) (since 0.8% as a decimal is 0.008), and we want to find \( t \) when \( N(t) \) is 1% of \( N_0 \) (meaning 99% has been eradicated).
Setting up the equation:
\[ N(t) = N_0 \cdot (1 - 0.99) = 0.01N_0 \]
Substituting into the decay formula:
\[ 0.01N_0 = N_0 e^{-0.008t} \]
Dividing both sides by \( N_0 \) (assuming \( N_0 \) is not zero):
\[ 0.01 = e^{-0.008t} \]
To solve for \( t \), we take the natural logarithm of both sides:
\[ \ln(0.01) = -0.008t \]
Calculating \( \ln(0.01) \):
\[ \ln(0.01) \approx -4.60517 \]
Now substituting back into the equation gives:
\[ -4.60517 = -0.008t \]
Now, solving for \( t \):
\[ t = \frac{-4.60517}{-0.008} \approx 575.64625 \]
Rounding to two decimal places:
\[ t \approx 575.65 \text{ hours} \]
Thus, it will take approximately 575.65 hours for the bacteria to be 99% gone.