For every p being real number the solution set of the inequality `log_10P+(log_(10)10p)^2+(log_(10)100p)^2lelog_10(10)^9`

Options

A. `pge10^-4`
B. `pgt10^(1//2)`
C. `10^-4leple10^(1//2)`
D. none of the above

Answer

To solve the inequality \( \log_{10} P + (\log_{10} (10P))^2 + (\log_{10} (100P))^2 \leq \log_{10} (10^9) \), we will follow these steps:

### Step 1: Simplify the logarithmic expressions
We start by rewriting the logarithmic terms:
- \( \log_{10} (10P) = \log_{10} 10 + \log_{10} P = 1 + \log_{10} P \)
- \( \log_{10} (100P) = \log_{10} 100 + \log_{10} P = 2 + \log_{10} P \)

Let \( a = \log_{10} P \). Then we can rewrite the inequality as:
\[ a + (1 + a)^2 + (2 + a)^2 \leq 9 \]

### Step 2: Expand the squares
Now, we expand the squares:
- \( (1 + a)^2 = 1 + 2a + a^2 \)
- \( (2 + a)^2 = 4 + 4a + a^2 \)

Substituting these back into the inequality gives:
\[ a + (1 + 2a + a^2) + (4 + 4a + a^2) \leq 9 \]

### Step 3: Combine like terms
Combining all the terms, we get:
\[ a + 1 + 2a + a^2 + 4 + 4a + a^2 \leq 9 \]
This simplifies to:
\[ 2a^2 + 7a + 5 \leq 9 \]

### Step 4: Rearrange the inequality
Rearranging the inequality gives:
\[ 2a^2 + 7a + 5 - 9 \leq 0 \]
or
\[ 2a^2 + 7a - 4 \leq 0 \]

### Step 5: Factor the quadratic
Next, we factor the quadratic equation:
\[ 2a^2 + 7a - 4 = 0 \]
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 2, b = 7, c = -4 \)
- The discriminant \( D = 7^2 - 4 \cdot 2 \cdot (-4) = 49 + 32 = 81 \)

Calculating the roots:
\[ a = \frac{-7 \pm \sqrt{81}}{2 \cdot 2} = \frac{-7 \pm 9}{4} \]
This gives us the roots:
\[ a_1 = \frac{2}{4} = \frac{1}{2}, \quad a_2 = \frac{-16}{4} = -4 \]

### Step 6: Determine the intervals
The quadratic \( 2a^2 + 7a - 4 \) opens upwards (since the coefficient of \( a^2 \) is positive). Thus, it is less than or equal to zero between the roots:
\[ -4 \leq a \leq \frac{1}{2} \]

### Step 7: Convert back to \( P \)
Since \( a = \log_{10} P \), we can write:
\[ -4 \leq \log_{10} P \leq \frac{1}{2} \]
This implies:
\[ 10^{-4} \leq P \leq 10^{\frac{1}{2}} \]
or
\[ 10^{-4} \leq P \leq \sqrt{10} \]

### Final Answer
Thus, the solution set for the inequality is:
\[ P \in [10^{-4}, \sqrt{10}] \]