Number of solutions of equation `sqrt(7^(2x^(2-5x-6))) = (sqrt(2))^(3log_2 49)`

Options

A. 2
B. 1
C. 4
D. 5

Answer

To solve the equation \( \sqrt{7^{2x^2 - 5x - 6}} = (\sqrt{2})^{3\log_2 49} \), we will follow these steps:

### Step 1: Simplify the Right-Hand Side (RHS)
The RHS can be simplified as follows:
\[ (\sqrt{2})^{3\log_2 49} = (2^{1/2})^{3\log_2 49} = 2^{(1/2) \cdot 3\log_2 49} = 2^{\frac{3}{2} \log_2 49} \]
Using the property of logarithms, we can rewrite this as:
\[ 2^{\log_2(49^{3/2})} = 49^{3/2} \]
Since \( 49 = 7^2 \), we have:
\[ 49^{3/2} = (7^2)^{3/2} = 7^3 \]
Thus, the RHS simplifies to \( 7^3 \).

### Step 2: Set the Left-Hand Side (LHS) Equal to the RHS
Now we can rewrite the equation:
\[ \sqrt{7^{2x^2 - 5x - 6}} = 7^3 \]
Squaring both sides gives:
\[ 7^{2x^2 - 5x - 6} = 7^6 \]

### Step 3: Equate the Exponents
Since the bases are the same, we can equate the exponents:
\[ 2x^2 - 5x - 6 = 6 \]
Rearranging this gives:
\[ 2x^2 - 5x - 12 = 0 \]

### Step 4: Factor the Quadratic Equation
To factor the quadratic equation \( 2x^2 - 5x - 12 = 0 \):
\[ 2x^2 - 8x + 3x - 12 = 0 \]
Grouping the terms:
\[ (2x^2 - 8x) + (3x - 12) = 0 \]
Factoring out common terms:
\[ 2x(x - 4) + 3(x - 4) = 0 \]
Factoring further:
\[ (x - 4)(2x + 3) = 0 \]

### Step 5: Solve for x
Setting each factor to zero gives us:
1. \( x - 4 = 0 \) → \( x = 4 \)
2. \( 2x + 3 = 0 \) → \( x = -\frac{3}{2} \)

### Step 6: Check for Validity of Solutions
Since the original equation involves a square root, we need to ensure that the values of \( x \) yield non-negative results in the exponent of the left-hand side:
- For \( x = 4 \):
\[ 2(4)^2 - 5(4) - 6 = 32 - 20 - 6 = 6 \quad (\text{valid, since } 6 \geq 0) \]
- For \( x = -\frac{3}{2} \):
\[ 2\left(-\frac{3}{2}\right)^2 - 5\left(-\frac{3}{2}\right) - 6 = 2 \cdot \frac{9}{4} + \frac{15}{2} - 6 = \frac{18}{4} + \frac{30}{4} - \frac{24}{4} = \frac{24}{4} = 6 \quad (\text{valid, since } 6 \geq 0) \]

### Conclusion
Both values \( x = 4 \) and \( x = -\frac{3}{2} \) are valid solutions.

Thus, the number of solutions to the equation is **2**.

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