If for `n ge 1, P_(n) = int_(1)^( e) (log x)^(n) dx`, then `P_(10) - 90P_(8)` is equal to

Options

A. `-9`
B. `10e`
C. `-9e`
D. `10`

Answer

To solve the problem, we need to evaluate \( P_{10} - 90P_{8} \) where \( P_n = \int_{1}^{e} (\log x)^{n} \, dx \).

### Step-by-Step Solution:

1. **Define \( P_n \)**:
\[ P_n = \int_{1}^{e} (\log x)^{n} \, dx \]

2. **Use Integration by Parts**:
We can use integration by parts to evaluate \( P_n \). Let:
- \( u = (\log x)^{n} \) → \( du = n (\log x)^{n-1} \cdot \frac{1}{x} \, dx \)
- \( dv = dx \) → \( v = x \)

Then, the integration by parts formula \( \int u \, dv = uv - \int v \, du \) gives us:
\[ P_n = \left[ x (\log x)^{n} \right]_{1}^{e} - \int_{1}^{e} x \cdot n (\log x)^{n-1} \cdot \frac{1}{x} \, dx \]
Simplifying, we have:
\[ P_n = \left[ e (\log e)^{n} - 1 \cdot (\log 1)^{n} \right] - n P_{n-1} \]
Since \( \log e = 1 \) and \( \log 1 = 0 \):
\[ P_n = e - n P_{n-1} \]

3. **Recursive Relation**:
We have established a recursive relation:
\[ P_n = e - n P_{n-1} \]

4. **Calculate \( P_{10} \) and \( P_{8} \)**:
We can use the recursive relation to find \( P_{10} \) and \( P_{8} \):
- For \( n = 10 \):
\[ P_{10} = e - 10 P_{9} \]
- For \( n = 9 \):
\[ P_{9} = e - 9 P_{8} \]
- For \( n = 8 \):
\[ P_{8} = e - 8 P_{7} \]

5. **Substituting Back**:
We can substitute back to express \( P_{10} \) in terms of \( P_{8} \):
\[ P_{9} = e - 9 P_{8} \implies P_{10} = e - 10(e - 9 P_{8}) = e - 10e + 90 P_{8} = -9e + 90 P_{8} \]

6. **Final Calculation**:
Now we can compute \( P_{10} - 90P_{8} \):
\[ P_{10} - 90P_{8} = (-9e + 90 P_{8}) - 90 P_{8} = -9e \]

### Conclusion:
Thus, the value of \( P_{10} - 90P_{8} \) is:
\[ \boxed{-9e} \]