If for $$n \geq 1$$, $$P_n = \int_1^e (\log x^n) dx$$, then $$P_{10} - 90P_8$$ is equal to:

We are given that for $$ n \geq 1 $$, $$ P_n = \int_1^e (\log x^n) dx $$. The expression $$ \log x^n $$ can be ambiguous, but in this context, given the options and the recurrence that works, it is interpreted as $$ (\log x)^n $$. Thus, $$ P_n = \int_1^e (\log x)^n dx $$.

To solve for $$ P_{10} - 90P_8 $$, we need a reduction formula for $$ P_n $$. Using integration by parts, set $$ u = (\log x)^n $$ and $$ dv = dx $$. Then $$ du = n (\log x)^{n-1} \cdot \frac{1}{x} dx $$ and $$ v = x $$. Applying integration by parts:

$$$ \int (\log x)^n dx = x (\log x)^n - \int x \cdot n (\log x)^{n-1} \cdot \frac{1}{x} dx = x (\log x)^n - n \int (\log x)^{n-1} dx $$$

Now, evaluate the definite integral from 1 to e:

$$$ P_n = \left[ x (\log x)^n \right]_1^e - n \int_1^e (\log x)^{n-1} dx $$$

At $$ x = e $$: $$ e (\log e)^n = e \cdot 1^n = e $$.

At $$ x = 1 $$: $$ 1 \cdot (\log 1)^n = 1 \cdot 0^n $$. For $$ n \geq 1 $$, $$ 0^n = 0 $$, so this term is 0.

Thus,

$$$ P_n = e - n \int_1^e (\log x)^{n-1} dx = e - n P_{n-1} $$$

So we have the recurrence relation:

$$$ P_n = e - n P_{n-1} $$$

We need to compute $$ P_{10} - 90P_8 $$. Using the recurrence for $$ P_{10} $$:

$$$ P_{10} = e - 10 P_9 $$$

Now apply the recurrence for $$ P_9 $$:

$$$ P_9 = e - 9 P_8 $$$

Substitute this into the expression for $$ P_{10} $$:

$$$ P_{10} = e - 10 (e - 9 P_8) = e - 10e + 90 P_8 = -9e + 90 P_8 $$$

Now form the expression $$ P_{10} - 90 P_8 $$:

$$$ P_{10} - 90 P_8 = (-9e + 90 P_8) - 90 P_8 = -9e $$$

Thus, $$ P_{10} - 90P_8 = -9e $$.

Comparing with the options:

A. $$-9$$

B. $$10e$$

C. $$-9e$$

D. $$10$$

The result $$-9e$$ matches option C.

Hence, the correct answer is Option C.