The solution set of the equation `log_(1//5)(2x+5)+log_(5)(16-x^(2))le1` is

Options

A. `(-(5)/(2),1)`
B. `[-1,4
]`
C. `[1,4]`
D. `[-(5)/(2),4]`

Answer

To solve the inequality \( \log_{1/5}(2x+5) + \log_{5}(16-x^2) \leq 1 \), we will follow these steps:

### Step 1: Rewrite the logarithm
We can rewrite the logarithm with base \(1/5\) in terms of base \(5\):
\[ \log_{1/5}(2x+5) = -\log_{5}(2x+5) \]
Thus, the inequality becomes:
\[ -\log_{5}(2x+5) + \log_{5}(16-x^2) \leq 1 \]

### Step 2: Combine the logarithms
Using the property of logarithms \( \log_{a}(m) - \log_{a}(n) = \log_{a}(\frac{m}{n}) \), we can combine the logarithms:
\[ \log_{5}(16-x^2) - \log_{5}(2x+5) \leq 1 \]
This simplifies to:
\[ \log_{5}\left(\frac{16-x^2}{2x+5}\right) \leq 1 \]

### Step 3: Exponentiate to eliminate the logarithm
Exponentiating both sides with base \(5\) gives:
\[ \frac{16-x^2}{2x+5} \leq 5^1 \]
This simplifies to:
\[ \frac{16-x^2}{2x+5} \leq 5 \]

### Step 4: Clear the fraction
Multiply both sides by \(2x + 5\) (noting that \(2x + 5 > 0\) for valid \(x\)):
\[ 16 - x^2 \leq 5(2x + 5) \]
Expanding the right side:
\[ 16 - x^2 \leq 10x + 25 \]

### Step 5: Rearrange the inequality
Rearranging gives:
\[ -x^2 - 10x + 16 - 25 \leq 0 \]
This simplifies to:
\[ -x^2 - 10x - 9 \leq 0 \]
Multiplying through by \(-1\) (and reversing the inequality):
\[ x^2 + 10x + 9 \geq 0 \]

### Step 6: Factor the quadratic
Factoring the quadratic:
\[ (x + 1)(x + 9) \geq 0 \]

### Step 7: Find the critical points
The critical points are \(x = -1\) and \(x = -9\). We will test intervals around these points to determine where the product is non-negative.

### Step 8: Test intervals
1. For \(x < -9\) (e.g., \(x = -10\)): \((x + 1)(x + 9) = (-)(-) > 0\)
2. For \(-9 < x < -1\) (e.g., \(x = -5\)): \((x + 1)(x + 9) = (-)(+) < 0\)
3. For \(x > -1\) (e.g., \(x = 0\)): \((x + 1)(x + 9) = (+)(+) > 0\)

### Step 9: Combine the results
The solution set is where the product is non-negative:
\[ x \in (-\infty, -9] \cup [-1, \infty) \]

### Step 10: Check the domain of the original logarithms
We also need to ensure that the arguments of the logarithms are positive:
1. \(2x + 5 > 0 \Rightarrow x > -\frac{5}{2}\)
2. \(16 - x^2 > 0 \Rightarrow -4 < x < 4\)

### Final Solution
Considering the domain restrictions, the final solution set is:
\[ x \in [-1, 4) \cup (-\infty, -9] \]