The number of values of x satisfying
` 1 +"log"_(5) (x^(2) + 1) ge "log"_(5) (x^(2) + 4x +1)`, is

Options

A. 1
B. 2
C. 3
D. infinitely many

Explanation

To solve the inequality

\[ 1 + \log_5(x^2 + 1) \geq \log_5(x^2 + 4x + 1), \]

we will follow these steps:

### Step 1: Rewrite the inequality
We can rewrite the inequality using properties of logarithms. Since \(1 = \log_5(5)\), we can express the left side as:

\[ \log_5(5) + \log_5(x^2 + 1) \geq \log_5(x^2 + 4x + 1). \]

Using the property \(\log_a(b) + \log_a(c) = \log_a(bc)\), we can combine the left side:

\[ \log_5(5(x^2 + 1)) \geq \log_5(x^2 + 4x + 1). \]

### Step 2: Remove the logarithm
Since the logarithm function is increasing, we can remove the logarithm (assuming both sides are defined and positive):

\[ 5(x^2 + 1) \geq x^2 + 4x + 1. \]

### Step 3: Simplify the inequality
Now, we simplify the inequality:

\[ 5x^2 + 5 \geq x^2 + 4x + 1. \]

Subtract \(x^2 + 4x + 1\) from both sides:

\[ 5x^2 - x^2 - 4x + 5 - 1 \geq 0, \]

which simplifies to:

\[ 4x^2 - 4x + 4 \geq 0. \]

### Step 4: Factor the quadratic
We can factor out a 4 from the left-hand side:

\[ 4(x^2 - x + 1) \geq 0. \]

Dividing both sides by 4 (which does not change the direction of the inequality):

\[ x^2 - x + 1 \geq 0. \]

### Step 5: Analyze the quadratic
Next, we will find the roots of the quadratic equation \(x^2 - x + 1 = 0\) using the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{1 \pm \sqrt{1 - 4}}{2} = \frac{1 \pm \sqrt{-3}}{2}. \]

Since the discriminant is negative (\(-3\)), this quadratic has no real roots, meaning it does not cross the x-axis.

### Step 6: Determine the sign of the quadratic
Since the coefficient of \(x^2\) is positive, the quadratic \(x^2 - x + 1\) is always positive for all real \(x\):

\[ x^2 - x + 1 > 0 \quad \text{for all } x \in \mathbb{R}. \]

### Step 7: Check the domain of the logarithm
Next, we need to ensure that the expressions inside the logarithms are positive:

1. \(x^2 + 1 > 0\) is always true for all real \(x\).
2. \(x^2 + 4x + 1 > 0\) needs to be checked. The roots are:

\[ x = \frac{-4 \pm \sqrt{16 - 4}}{2} = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}. \]

This quadratic is positive outside the interval \((-2 - \sqrt{3}, -2 + \sqrt{3})\).

### Step 8: Final solution
Thus, the solution set for \(x\) is:

\[ (-\infty, -2 - \sqrt{3}) \cup (-2 + \sqrt{3}, \infty). \]

Since there are infinitely many values of \(x\) in these intervals, the answer is:

**Infinitely many values of \(x\)**.