The number of values of `x` satisfying `1+log_(5)(x^(2)+1)gelog_(5)(x^(2)+4x+1)` is

Options

A. only one
B. two
C. three
D. infinitely many

Answer

To solve the inequality \( 1 + \log_{5}(x^{2} + 1) \geq \log_{5}(x^{2} + 4x + 1) \), we will follow these steps:

### Step 1: Rearranging the Inequality
We start by moving the logarithmic terms to one side of the inequality:
\[ 1 \geq \log_{5}(x^{2} + 4x + 1) - \log_{5}(x^{2} + 1) \]
Using the property of logarithms, \( \log_{a}(m) - \log_{a}(n) = \log_{a}(\frac{m}{n}) \), we can rewrite the inequality as:
\[ 1 \geq \log_{5}\left(\frac{x^{2} + 4x + 1}{x^{2} + 1}\right) \]

### Step 2: Exponentiating Both Sides
To eliminate the logarithm, we exponentiate both sides with base 5:
\[ 5^1 \geq \frac{x^{2} + 4x + 1}{x^{2} + 1} \]
This simplifies to:
\[ 5 \geq \frac{x^{2} + 4x + 1}{x^{2} + 1} \]

### Step 3: Cross-Multiplying
Next, we cross-multiply to eliminate the fraction:
\[ 5(x^{2} + 1) \geq x^{2} + 4x + 1 \]
Expanding both sides gives:
\[ 5x^{2} + 5 \geq x^{2} + 4x + 1 \]

### Step 4: Rearranging the Terms
Now, we rearrange the terms to bring everything to one side:
\[ 5x^{2} - x^{2} - 4x + 5 - 1 \geq 0 \]
This simplifies to:
\[ 4x^{2} - 4x + 4 \geq 0 \]

### Step 5: Factoring the Quadratic
We can factor out a 4:
\[ 4(x^{2} - x + 1) \geq 0 \]
Now, we need to analyze the quadratic \( x^{2} - x + 1 \).

### Step 6: Finding the Discriminant
To determine the nature of the roots of the quadratic, we calculate the discriminant \( D \):
\[ D = b^{2} - 4ac = (-1)^{2} - 4(1)(1) = 1 - 4 = -3 \]
Since the discriminant is negative, the quadratic has no real roots and is always positive.

### Step 7: Conclusion
Since \( 4(x^{2} - x + 1) \geq 0 \) holds for all real \( x \), the original inequality is satisfied for all real values of \( x \).

Thus, the number of values of \( x \) satisfying the inequality is **infinite**.