Solve the exponential equation: (5^x)2 - 6(5^x) + 5 = 0

The given equation is $5^{2x} - 6 \times 5^x + 5 = 0$.
We can rewrite $5^{2x}$ as $(5^x)^2$.
So the equation becomes:

$$(5^x)^2 - 6(5^x) + 5 = 0$$

Let $y = 5^x$.
Substitute $y$ into the equation:

$$y^2 - 6y + 5 = 0$$

This is a quadratic equation that can be factored. We need two numbers that multiply to 5 and add up to -6. These numbers are -1 and -5.

$$(y - 1)(y - 5) = 0$$

This gives two possible values for $y$:

$$y - 1 = 0 \implies y = 1$$ $$y - 5 = 0 \implies y = 5$$

Case 1: $y = 1$

$$5^x = 1$$

Since any non-zero number raised to the power of 0 is 1, we can write 1 as $5^0$.

$$5^x = 5^0$$

Therefore,

$$x = 0$$

Case 2: $y = 5$

$$5^x = 5$$

Since 5 can be written as $5^1$.

$$5^x = 5^1$$

Therefore,

$$x = 1$$

The values of $x$ that satisfy the equation are $\boxed{x = 0 \text{ or } x = 1}$.