Solve the equation: 9^x - 10 * 3^x + 9 = 0

We know that $9 = 3^2$. So, $9^x$ can be written as $(3^2)^x = 3^{2x} = (3^x)^2$.
Substituting this into the equation:

$$(3^x)^2 - 10 \times 3^x + 9 = 0$$

Let $y = 3^x$. Substituting $y$ into the equation transforms it into a quadratic equation:

$$y^2 - 10y + 9 = 0$$

We can factor this quadratic equation. We need two numbers that multiply to 9 and add up to $-10$. These numbers are $-1$ and $-9$.

$$(y - 1)(y - 9) = 0$$

This gives two possible values for $y$:

$$y - 1 = 0 \implies y = 1$$ $$y - 9 = 0 \implies y = 9$$

Case 1: $y = 1$

$$3^x = 1$$

Since any non-zero number raised to the power of 0 is 1, we have $3^0 = 1$.
Therefore,

$$x = 0$$

Case 2: $y = 9$

$$3^x = 9$$

Since $9 = 3^2$, we can write:

$$3^x = 3^2$$

Therefore,

$$x = 2$$

The solutions for $x$ are 0 and 2.

The final answers are $x=0, x=2$.