The number of solutions of `(log5+log(x^(2)+1))/(log(x-2))=2` is

Options

A. `1`
B. `2`
C. `3`
D. none

Answer

To solve the equation \(\frac{\log 5 + \log(x^2 + 1)}{\log(x - 2)} = 2\), we will follow these steps:

### Step 1: Rewrite the equation using logarithmic properties
Using the property of logarithms that states \(\log a + \log b = \log(ab)\), we can rewrite the left-hand side:

\[ \frac{\log(5(x^2 + 1))}{\log(x - 2)} = 2 \]

### Step 2: Cross-multiply to eliminate the fraction
Cross-multiplying gives us:

\[ \log(5(x^2 + 1)) = 2 \log(x - 2) \]

### Step 3: Use the power property of logarithms
Using the property \(\log a^b = b \log a\), we can rewrite the right-hand side:

\[ \log(5(x^2 + 1)) = \log((x - 2)^2) \]

### Step 4: Set the arguments of the logarithms equal to each other
Since the logarithmic function is one-to-one, we can set the arguments equal:

\[ 5(x^2 + 1) = (x - 2)^2 \]

### Step 5: Expand both sides
Expanding the right-hand side:

\[ 5x^2 + 5 = x^2 - 4x + 4 \]

### Step 6: Rearrange the equation
Rearranging gives us:

\[ 5x^2 - x^2 + 4x + 5 - 4 = 0 \]

This simplifies to:

\[ 4x^2 + 4x + 1 = 0 \]

### Step 7: Factor the quadratic equation
The quadratic can be factored as:

\[ (2x + 1)^2 = 0 \]

### Step 8: Solve for \(x\)
Setting the factor equal to zero gives:

\[ 2x + 1 = 0 \implies x = -\frac{1}{2} \]

### Step 9: Check the validity of the solution
Since we have logarithmic functions in our original equation, we need to check the domain restrictions. The logarithm \(\log(x - 2)\) requires that \(x - 2 > 0\), or \(x > 2\). The solution \(x = -\frac{1}{2}\) does not satisfy this condition.

### Conclusion
Since there are no valid solutions that satisfy the original equation, the number of solutions is:

\[ \text{Number of solutions} = 0 \]

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