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, we need to find the time at which the concentrations of substances A and B become equal, given their initial concentrations and half-lives.

### Step-by-Step Solution:

1. **Define Initial Concentrations**:
- Let the initial concentration of substance B be \( [B]_0 = a \).
- Since \( [A]_0 = 4[B]_0 \), we have \( [A]_0 = 4a \).

2. **Determine the Rate Constants**:
- For first-order kinetics, the half-life \( t_{1/2} \) is related to the rate constant \( k \) by the formula:
\[ k = \frac{0.693}{t_{1/2}} \]
- For substance A:
\[ k_A = \frac{0.693}{5 \text{ min}} = 0.1386 \text{ min}^{-1} \]
- For substance B:
\[ k_B = \frac{0.693}{15 \text{ min}} = 0.0462 \text{ min}^{-1} \]

3. **Write the Concentration Equations**:
- The concentration of A after time \( t \) is given by:
\[ [A] = [A]_0 e^{-k_A t} = 4a e^{-0.1386 t} \]
- The concentration of B after time \( t \) is given by:
\[ [B] = [B]_0 e^{-k_B t} = a e^{-0.0462 t} \]

4. **Set the Concentrations Equal**:
- We want to find the time \( t \) when \( [A] = [B] \):
\[ 4a e^{-0.1386 t} = a e^{-0.0462 t} \]
- Dividing both sides by \( a \) (assuming \( a \neq 0 \)):
\[ 4 e^{-0.1386 t} = e^{-0.0462 t} \]

5. **Rearranging the Equation**:
- Taking the natural logarithm of both sides:
\[ \ln(4) - 0.1386 t = -0.0462 t \]
- Rearranging gives:
\[ \ln(4) = (0.1386 - 0.0462)t \]
- Simplifying:
\[ \ln(4) = 0.0924 t \]

6. **Solving for Time \( t \)**:
- Now, we can solve for \( t \):
\[ t = \frac{\ln(4)}{0.0924} \]
- Calculating \( \ln(4) \):
\[ \ln(4) \approx 1.386 \]
- Therefore:
\[ t \approx \frac{1.386}{0.0924} \approx 15 \text{ min} \]

7. **Relating Time to \( x \)**:
- We are given that \( t = 5x \):
\[ 5x = 15 \implies x = 3 \]

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

---