The population grows in at the rate of 8% per year. Find the time taken for the population to become double . (Given : log 2 = 0.6912)

Answer

To solve the problem of finding the time taken for a population to double given a growth rate of 8% per year, we can follow these steps:

### Step 1: Set up the differential equation
The population \( P \) grows at a rate of 8% per year, which can be expressed as:
\[ \frac{dP}{dt} = 0.08P \]
This means the rate of change of the population with respect to time is proportional to the current population.

### Step 2: Rearrange the equation
We can rearrange the equation to separate the variables:
\[ \frac{dP}{P} = 0.08 \, dt \]

### Step 3: Integrate both sides
Next, we integrate both sides:
\[ \int \frac{dP}{P} = \int 0.08 \, dt \]
This gives us:
\[ \ln P = 0.08t + C \]
where \( C \) is the constant of integration.

### Step 4: Solve for the constant of integration
To find \( C \), we consider the initial condition. Let \( P_0 \) be the initial population at \( t = 0 \):
\[ \ln P_0 = 0.08(0) + C \implies C = \ln P_0 \]
So, we can rewrite the equation as:
\[ \ln P = 0.08t + \ln P_0 \]

### Step 5: Exponentiate to eliminate the natural logarithm
Exponentiating both sides gives:
\[ P = e^{0.08t} \cdot P_0 \]

### Step 6: Set up the equation for doubling the population
We want to find the time \( t \) when the population doubles, i.e., when \( P = 2P_0 \):
\[ 2P_0 = e^{0.08t} \cdot P_0 \]
Dividing both sides by \( P_0 \) (assuming \( P_0 \neq 0 \)):
\[ 2 = e^{0.08t} \]

### Step 7: Take the natural logarithm of both sides
Taking the natural logarithm gives:
\[ \ln 2 = 0.08t \]

### Step 8: Solve for \( t \)
Now we can solve for \( t \):
\[ t = \frac{\ln 2}{0.08} \]

### Step 9: Substitute the value of \( \ln 2 \)
Given \( \ln 2 = 0.6912 \):
\[ t = \frac{0.6912}{0.08} \]

### Step 10: Calculate \( t \)
Calculating the value:
\[ t = 8.64 \text{ years} \]

### Final Answer
The time taken for the population to become double is \( 8.64 \) years.

---