`log_(10)^(2) x + log_(10) x^(2) = log_(10)^(2) 2 - 1`

Options

A. `1/5`
B. `1/6`
C. `1/7`
D. None

Answer

To solve the equation \( \log_{10}^2 x + \log_{10} x^2 = \log_{10}^2 2 - 1 \), we can follow these steps:

### Step 1: Rewrite the logarithmic expressions
We know that \( \log_{10} x^2 = 2 \log_{10} x \). Let \( A = \log_{10} x \). Thus, we can rewrite the equation as:
\[ A^2 + 2A = \log_{10}^2 2 - 1 \]

### Step 2: Rearrange the equation
Now, we can rearrange the equation to bring all terms to one side:
\[ A^2 + 2A - \log_{10}^2 2 + 1 = 0 \]

### Step 3: Identify coefficients for the quadratic formula
This is a quadratic equation in the form \( Ax^2 + Bx + C = 0 \), where:
- \( A = 1 \)
- \( B = 2 \)
- \( C = -\log_{10}^2 2 + 1 \)

### Step 4: Apply the quadratic formula
Using the quadratic formula \( A = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \):
\[ A = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-\log_{10}^2 2 + 1)}}{2 \cdot 1} \]
\[ A = \frac{-2 \pm \sqrt{4 + 4(\log_{10}^2 2 - 1)}}{2} \]
\[ A = \frac{-2 \pm \sqrt{4\log_{10}^2 2}}{2} \]
\[ A = -1 \pm \sqrt{\log_{10}^2 2} \]

### Step 5: Simplify the expression for A
Since \( \sqrt{\log_{10}^2 2} = |\log_{10} 2| \) and since \( \log_{10} 2 \) is positive, we have:
\[ A = -1 + \log_{10} 2 \quad \text{or} \quad A = -1 - \log_{10} 2 \]

### Step 6: Convert back to x
Recall that \( A = \log_{10} x \). Therefore:
1. For \( A = -1 + \log_{10} 2 \):
\[ \log_{10} x = -1 + \log_{10} 2 \implies \log_{10} x = \log_{10} \frac{2}{10} \implies x = \frac{2}{10} = \frac{1}{5} \]

2. For \( A = -1 - \log_{10} 2 \):
\[ \log_{10} x = -1 - \log_{10} 2 \implies \log_{10} x = \log_{10} \frac{1}{10 \cdot 2} \implies x = \frac{1}{20} \]

### Final Solutions
Thus, the values of \( x \) are:
\[ x = \frac{1}{5} \quad \text{and} \quad x = \frac{1}{20} \]

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