If `10^(x-1) + 10^(-x-1) = (1)/(3)`, then x equals to

Options

A. `+-"log"_(10)3`
B. `2"log"_(3)10`
C. `"log"_(3)3`
D. `"log"_(2) 10`

Answer

To solve the equation \(10^{(x-1)} + 10^{(-x-1)} = \frac{1}{3}\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ 10^{(x-1)} + 10^{(-x-1)} = \frac{1}{3} \]
We can rewrite \(10^{(x-1)}\) as \(\frac{10^x}{10}\) and \(10^{(-x-1)}\) as \(\frac{1}{10 \cdot 10^x}\). Thus, we can factor out \(\frac{1}{10}\):
\[ \frac{10^x + \frac{1}{10^x}}{10} = \frac{1}{3} \]

### Step 2: Multiply both sides by 10
To eliminate the fraction, we multiply both sides by 10:
\[ 10^x + \frac{1}{10^x} = \frac{10}{3} \]

### Step 3: Let \(y = 10^x\)
Now, we can let \(y = 10^x\). The equation becomes:
\[ y + \frac{1}{y} = \frac{10}{3} \]

### Step 4: Multiply through by \(y\)
To eliminate the fraction, we multiply through by \(y\):
\[ y^2 + 1 = \frac{10}{3}y \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ y^2 - \frac{10}{3}y + 1 = 0 \]

### Step 6: Clear the fraction by multiplying by 3
To eliminate the fraction, multiply the entire equation by 3:
\[ 3y^2 - 10y + 3 = 0 \]

### Step 7: Use the quadratic formula
We can now use the quadratic formula \(y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 3\), \(b = -10\), and \(c = 3\):
\[ y = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 3 \cdot 3}}{2 \cdot 3} \]
\[ y = \frac{10 \pm \sqrt{100 - 36}}{6} \]
\[ y = \frac{10 \pm \sqrt{64}}{6} \]
\[ y = \frac{10 \pm 8}{6} \]

### Step 8: Calculate the two possible values for \(y\)
Calculating the two possible values:
1. \(y = \frac{18}{6} = 3\)
2. \(y = \frac{2}{6} = \frac{1}{3}\)

### Step 9: Solve for \(x\)
Now we substitute back for \(y = 10^x\):
1. If \(y = 3\), then \(10^x = 3 \implies x = \log_{10}(3)\)
2. If \(y = \frac{1}{3}\), then \(10^x = \frac{1}{3} \implies x = \log_{10}\left(\frac{1}{3}\right) = -\log_{10}(3)\)

### Final Answer
Thus, the values of \(x\) are:
\[ x = \log_{10}(3) \quad \text{or} \quad x = -\log_{10}(3) \]