Two substances `A (t_(1//2) = 5 min)` and `B(t_(1//2) = 15 min)` follow first order kinetics and are taken in such a way that initially `[A] = 4[B]`. The time after which the concentration of both the substance will be equal is `5x min`. Find the value of `x`.

Explanation

To solve the problem step by step, we need to find the time after which the concentrations of substances A and B become equal, given their half-lives and initial concentrations.

### Step 1: Understand the given data
- Substance A has a half-life (t₁/₂) of 5 minutes.
- Substance B has a half-life (t₁/₂) of 15 minutes.
- The initial concentration of A is 4 times that of B, i.e., [A]₀ = 4[B]₀.

### Step 2: Write the first-order kinetics equations
For a first-order reaction, the concentration at time t can be expressed as:
\[ [A] = [A]₀ e^{-k_A t} \]
\[ [B] = [B]₀ e^{-k_B t} \]

### Step 3: Relate the rate constants to half-lives
The rate constant (k) for a first-order reaction is related to the half-life by the formula:
\[ k = \frac{0.693}{t_{1/2}} \]

Thus, we can calculate:
- For substance A:
\[ k_A = \frac{0.693}{5} \, \text{min}^{-1} \]
- For substance B:
\[ k_B = \frac{0.693}{15} \, \text{min}^{-1} \]

### Step 4: Set up the equation for equal concentrations
We need to find the time t when [A] = [B]. Therefore, we set up the equation:
\[ [A]₀ e^{-k_A t} = [B]₀ e^{-k_B t} \]

Substituting [A]₀ = 4[B]₀, we get:
\[ 4[B]₀ e^{-k_A t} = [B]₀ e^{-k_B t} \]

Dividing both sides by [B]₀ (assuming [B]₀ ≠ 0):
\[ 4 e^{-k_A t} = e^{-k_B t} \]

### Step 5: Simplify the equation
Rearranging gives:
\[ \frac{4}{1} = e^{-(k_B - k_A)t} \]
Taking the natural logarithm of both sides:
\[ \ln(4) = -(k_B - k_A)t \]

### Step 6: Substitute the values of k_A and k_B
Substituting the expressions for k_A and k_B:
\[ \ln(4) = -\left(\frac{0.693}{15} - \frac{0.693}{5}\right)t \]

### Step 7: Simplify the expression
Calculating the difference:
\[ \frac{0.693}{5} - \frac{0.693}{15} = 0.693 \left(\frac{1}{5} - \frac{1}{15}\right) \]
Finding a common denominator (15):
\[ = 0.693 \left(\frac{3 - 1}{15}\right) = 0.693 \left(\frac{2}{15}\right) \]

Thus, we have:
\[ \ln(4) = -\left(-\frac{0.693 \cdot 2}{15}\right)t \]
This simplifies to:
\[ \ln(4) = \frac{0.693 \cdot 2}{15}t \]

### Step 8: Solve for t
Now, we know that:
\[ \ln(4) = 2 \ln(2) \]
So:
\[ 2 \ln(2) = \frac{0.693 \cdot 2}{15}t \]
Dividing both sides by 2:
\[ \ln(2) = \frac{0.693}{15}t \]

Now, solving for t:
\[ t = \frac{15 \ln(2)}{0.693} \]

### Step 9: Calculate the value of x
We know that the time after which the concentrations of both substances will be equal is given as \( 5x \) minutes. Thus:
\[ 5x = \frac{15 \ln(2)}{0.693} \]

To find x:
\[ x = \frac{15 \ln(2)}{5 \cdot 0.693} = \frac{3 \ln(2)}{0.693} \]

Using \( \ln(2) \approx 0.693 \):
\[ x \approx \frac{3 \cdot 0.693}{0.693} = 3 \]

### Final Answer
The value of \( x \) is **3**.