If `x` satisfies the inequality `log_(25)x^(2)+(log_(5)x)^(2)lt2`, then `x epsilon`

Options

A. `((1)/(25),5)`
B. `(1,2)`
C. `(4,5)`
D. `(0,1)`

Answer

To solve the inequality \( \log_{25}(x^2) + (\log_{5}(x))^2 < 2 \), we will follow these steps:

### Step 1: Rewrite the logarithm
We know that \( 25 = 5^2 \), so we can rewrite \( \log_{25}(x^2) \) using the change of base formula:
\[ \log_{25}(x^2) = \frac{\log_{5}(x^2)}{\log_{5}(25)} = \frac{\log_{5}(x^2)}{2} = \frac{2\log_{5}(x)}{2} = \log_{5}(x) \]
Thus, we can rewrite the inequality as:
\[ \log_{5}(x) + (\log_{5}(x))^2 < 2 \]

### Step 2: Let \( y = \log_{5}(x) \)
Substituting \( y \) into the inequality gives us:
\[ y + y^2 < 2 \]
Rearranging this, we have:
\[ y^2 + y - 2 < 0 \]

### Step 3: Factor the quadratic inequality
Next, we factor the quadratic expression:
\[ y^2 + y - 2 = (y - 1)(y + 2) \]
Thus, the inequality becomes:
\[ (y - 1)(y + 2) < 0 \]

### Step 4: Determine the intervals
To solve the inequality \( (y - 1)(y + 2) < 0 \), we find the critical points where the expression equals zero:
- \( y - 1 = 0 \) gives \( y = 1 \)
- \( y + 2 = 0 \) gives \( y = -2 \)

Now, we test the intervals determined by these critical points: \( (-\infty, -2) \), \( (-2, 1) \), and \( (1, \infty) \).

1. For \( y < -2 \) (e.g., \( y = -3 \)):
\[ (-3 - 1)(-3 + 2) = (-4)(-1) = 4 > 0 \]
2. For \( -2 < y < 1 \) (e.g., \( y = 0 \)):
\[ (0 - 1)(0 + 2) = (-1)(2) = -2 < 0 \]
3. For \( y > 1 \) (e.g., \( y = 2 \)):
\[ (2 - 1)(2 + 2) = (1)(4) = 4 > 0 \]

Thus, the solution to the inequality is:
\[ -2 < y < 1 \]

### Step 5: Substitute back for \( x \)
Now we revert back to \( x \) using \( y = \log_{5}(x) \):
\[ -2 < \log_{5}(x) < 1 \]

### Step 6: Convert logarithmic inequalities to exponential form
1. From \( \log_{5}(x) > -2 \):
\[ x > 5^{-2} = \frac{1}{25} \]
2. From \( \log_{5}(x) < 1 \):
\[ x < 5^1 = 5 \]

### Final Solution
Combining these results, we have:
\[ \frac{1}{25} < x < 5 \]
Thus, the solution set is:
\[ x \in \left( \frac{1}{25}, 5 \right) \]