If the equation `x^2-3kx+2e^(2logk)-1=0` has real roots such that the product of roots is `7` then the value of `k` is

Explanation

To solve the equation \( x^2 - 3kx + 2e^{2\log k} - 1 = 0 \) given that the product of the roots is 7, we can follow these steps:

### Step 1: Rewrite the equation
The equation can be rewritten as:
\[ x^2 - 3kx + (2e^{2\log k} - 1) = 0 \]

### Step 2: Simplify \( e^{2\log k} \)
Using the property of logarithms, we know that:
\[ e^{\log a} = a \]
Thus, we can simplify \( e^{2\log k} \):
\[ e^{2\log k} = (e^{\log k})^2 = k^2 \]
So, the equation becomes:
\[ x^2 - 3kx + (2k^2 - 1) = 0 \]

### Step 3: Use the product of roots
For a quadratic equation of the form \( ax^2 + bx + c = 0 \), the product of the roots is given by:
\[ \text{Product of roots} = \frac{c}{a} \]
In our case, \( a = 1 \), \( b = -3k \), and \( c = 2k^2 - 1 \). Therefore, the product of the roots is:
\[ \frac{2k^2 - 1}{1} = 2k^2 - 1 \]
We know from the problem statement that this product equals 7:
\[ 2k^2 - 1 = 7 \]

### Step 4: Solve for \( k \)
Now, we can solve for \( k \):
\[ 2k^2 - 1 = 7 \]
Adding 1 to both sides:
\[ 2k^2 = 8 \]
Dividing both sides by 2:
\[ k^2 = 4 \]
Taking the square root of both sides:
\[ k = 2 \quad \text{or} \quad k = -2 \]

### Step 5: Determine valid values of \( k \)
Since \( k \) is the input of a logarithm, it must be positive. Therefore, we reject \( k = -2 \) and accept:
\[ k = 2 \]

### Final Answer
The value of \( k \) is:
\[ \boxed{2} \]