Bacteria increases at the rate proportional to the number of bacteria present. If the original number N doubles in 3 hours, find in how many hours the number of bacteria will by 4N ?

Answer

To solve the problem, we will use the concept of exponential growth, which is described by the differential equation:

\[ \frac{dN}{dt} = kN \]

where \( N \) is the number of bacteria, \( t \) is time, and \( k \) is the constant of proportionality.

### Step 1: Set up the equation

The solution to the differential equation is given by:

\[ N(t) = N_0 e^{kt} \]

where \( N_0 \) is the initial number of bacteria.

### Step 2: Use the given information

We know that the original number \( N_0 \) doubles in 3 hours. This means:

\[ N(3) = 2N_0 \]

Substituting into the equation, we have:

\[ 2N_0 = N_0 e^{3k} \]

### Step 3: Simplify the equation

Dividing both sides by \( N_0 \) (assuming \( N_0 \neq 0 \)) gives:

\[ 2 = e^{3k} \]

### Step 4: Solve for \( k \)

Taking the natural logarithm of both sides:

\[ \ln(2) = 3k \]

Thus, we can solve for \( k \):

\[ k = \frac{\ln(2)}{3} \]

### Step 5: Find the time when \( N = 4N_0 \)

We want to find the time \( t \) when the number of bacteria is \( 4N_0 \):

\[ 4N_0 = N_0 e^{kt} \]

Dividing both sides by \( N_0 \):

\[ 4 = e^{kt} \]

### Step 6: Substitute \( k \)

Substituting \( k = \frac{\ln(2)}{3} \) into the equation:

\[ 4 = e^{\left(\frac{\ln(2)}{3}\right)t} \]

### Step 7: Take the natural logarithm

Taking the natural logarithm of both sides:

\[ \ln(4) = \frac{\ln(2)}{3} t \]

### Step 8: Solve for \( t \)

We know that \( \ln(4) = \ln(2^2) = 2\ln(2) \). Thus, we have:

\[ 2\ln(2) = \frac{\ln(2)}{3} t \]

Dividing both sides by \( \ln(2) \) (assuming \( \ln(2) \neq 0 \)) gives:

\[ 2 = \frac{t}{3} \]

Multiplying both sides by 3:

\[ t = 6 \text{ hours} \]

### Final Answer

The number of bacteria will be \( 4N \) in **6 hours**.

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