Suppose the growth of a population is proportional to the number present. If the population of a colony dubles in 50 months, in how many months will the population become triple ?

Answer

To solve the problem of how long it will take for a population to triple given that it doubles in 50 months, we can follow these steps:

### Step 1: Set up the differential equation
The growth of the population \( P \) is proportional to the number present, which can be expressed as:
\[ \frac{dP}{dt} = \lambda P \]
where \( \lambda \) is a constant.

### Step 2: Separate variables and integrate
We can separate the variables and integrate both sides:
\[ \int \frac{1}{P} dP = \int \lambda dt \]
This gives us:
\[ \ln P = \lambda t + C \]
where \( C \) is the constant of integration.

### Step 3: Solve for the constant \( C \)
Let the initial population at time \( t = 0 \) be \( P_0 \):
\[ \ln P_0 = \lambda \cdot 0 + C \implies C = \ln P_0 \]

### Step 4: Write the general solution
Substituting \( C \) back into the equation, we have:
\[ \ln P = \lambda t + \ln P_0 \]
This can be rewritten as:
\[ \ln \left(\frac{P}{P_0}\right) = \lambda t \]

### Step 5: Analyze the doubling condition
When the population doubles, \( P = 2P_0 \). Substituting this into the equation:
\[ \ln \left(\frac{2P_0}{P_0}\right) = \lambda \cdot 50 \]
This simplifies to:
\[ \ln 2 = \lambda \cdot 50 \]
From this, we can solve for \( \lambda \):
\[ \lambda = \frac{\ln 2}{50} \]

### Step 6: Analyze the tripling condition
Now we want to find the time \( t \) when the population triples, \( P = 3P_0 \):
\[ \ln \left(\frac{3P_0}{P_0}\right) = \lambda t \]
This simplifies to:
\[ \ln 3 = \lambda t \]
Substituting \( \lambda \) from the previous step:
\[ \ln 3 = \frac{\ln 2}{50} t \]

### Step 7: Solve for \( t \)
Rearranging gives:
\[ t = \frac{50 \ln 3}{\ln 2} \]

### Final Result
Thus, the time it will take for the population to triple is:
\[ t = 50 \cdot \frac{\ln 3}{\ln 2} \text{ months} \]