If the roots of the equation `x^(2)-4x-log_3 a=0` , are real , the least value of a is :

Options

A. 81
B. `1//81`
C. `1//64`
D. None of these

Answer

To find the least value of \( a \) such that the roots of the equation \( x^2 - 4x - \log_3 a = 0 \) are real, we need to ensure that the discriminant of the quadratic equation is non-negative.

### Step-by-Step Solution:

1. **Identify the coefficients**:
The given quadratic equation is \( x^2 - 4x - \log_3 a = 0 \). Here, we can identify:
- \( a = 1 \) (coefficient of \( x^2 \))
- \( b = -4 \) (coefficient of \( x \))
- \( c = -\log_3 a \) (constant term)

2. **Calculate the discriminant**:
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values we identified:
\[ D = (-4)^2 - 4 \cdot 1 \cdot (-\log_3 a) \]
\[ D = 16 + 4\log_3 a \]

3. **Set the discriminant greater than or equal to zero**:
For the roots to be real, we need:
\[ 16 + 4\log_3 a \geq 0 \]

4. **Rearrange the inequality**:
Subtract 16 from both sides:
\[ 4\log_3 a \geq -16 \]
Divide both sides by 4:
\[ \log_3 a \geq -4 \]

5. **Convert the logarithmic inequality to exponential form**:
The inequality \( \log_3 a \geq -4 \) can be rewritten in exponential form:
\[ a \geq 3^{-4} \]
Calculating \( 3^{-4} \):
\[ 3^{-4} = \frac{1}{3^4} = \frac{1}{81} \]

6. **Conclusion**:
Therefore, the least value of \( a \) such that the roots of the equation are real is:
\[ a \geq \frac{1}{81} \]
Thus, the least value of \( a \) is \( \frac{1}{81} \).

### Final Answer:
The least value of \( a \) is \( \frac{1}{81} \).