If `log_(2)[log_(7)(x^(2)-x+37)]=1`, , then what could be the value of x ?

Options

A. 3
B. 5
C. 4
D. None of these

Answer

To solve the equation \( \log_{2}[\log_{7}(x^{2} - x + 37)] = 1 \), we will follow these steps:

### Step 1: Rewrite the logarithmic equation
Starting with the equation:
\[ \log_{2}[\log_{7}(x^{2} - x + 37)] = 1 \]
We can rewrite this in exponential form:
\[ \log_{7}(x^{2} - x + 37) = 2 \]

### Step 2: Convert the logarithm to its exponential form
Now, we will convert the logarithmic equation \( \log_{7}(x^{2} - x + 37) = 2 \) into its exponential form:
\[ x^{2} - x + 37 = 7^{2} \]
Calculating \( 7^{2} \):
\[ x^{2} - x + 37 = 49 \]

### Step 3: Rearrange the equation
Next, we will rearrange the equation to bring all terms to one side:
\[ x^{2} - x + 37 - 49 = 0 \]
This simplifies to:
\[ x^{2} - x - 12 = 0 \]

### Step 4: Factor the quadratic equation
Now, we will factor the quadratic equation:
\[ x^{2} - x - 12 = (x - 4)(x + 3) = 0 \]

### Step 5: Solve for x
Setting each factor equal to zero gives us:
\[ x - 4 = 0 \quad \text{or} \quad x + 3 = 0 \]
Thus, we find:
\[ x = 4 \quad \text{or} \quad x = -3 \]

### Step 6: Verify the solutions
Since we are dealing with logarithms, we need to ensure that the argument of the logarithm is positive. We check both solutions:
1. For \( x = 4 \):
\[ x^{2} - x + 37 = 4^{2} - 4 + 37 = 16 - 4 + 37 = 49 \quad (\text{valid, as } 49 > 0) \]
2. For \( x = -3 \):
\[ x^{2} - x + 37 = (-3)^{2} - (-3) + 37 = 9 + 3 + 37 = 49 \quad (\text{valid, as } 49 > 0) \]

Both solutions are valid, but since logarithms are typically defined for positive values, we conclude that the acceptable solution is:
\[ \boxed{4} \]