To remove bacteria living in a water tank, a chemical is to be administered. Let $C _ { 0 }$ be the initial number of bacteria per 1 mL of water in the tank, and let $C$ be the number of bacteria per 1 mL at time $t$ hours after the chemical is administered. The following relationship holds: $$\log \frac { C } { C _ { 0 } } = - k t \quad ( k \text { is a positive constant } )$$ The initial number of bacteria per 1 mL of water is $8 \times 10 ^ { 5 }$, and at time 3 hours after the chemical is administered, the number of bacteria per 1 mL becomes $2 \times 10 ^ { 5 }$. After $a$ hours from administering the chemical, the number of bacteria per 1 mL first becomes $8 \times 10 ^ { 3 }$ or less. Find the value of $a$. (Here, calculate using $\log 2 = 0.3$.) [4 points]
To remove bacteria living in a water tank, a chemical is to be administered. Let $C _ { 0 }$ be the initial number of bacteria per 1 mL of water in the tank, and let $C$ be the number of bacteria per 1 mL at time $t$ hours after the chemical is administered. The following relationship holds:
$$\log \frac { C } { C _ { 0 } } = - k t \quad ( k \text { is a positive constant } )$$
The initial number of bacteria per 1 mL of water is $8 \times 10 ^ { 5 }$, and at time 3 hours after the chemical is administered, the number of bacteria per 1 mL becomes $2 \times 10 ^ { 5 }$. After $a$ hours from administering the chemical, the number of bacteria per 1 mL first becomes $8 \times 10 ^ { 3 }$ or less. Find the value of $a$. (Here, calculate using $\log 2 = 0.3$.) [4 points]