The given problem is:

$25^x - 2 \cdot 5^x - 3 = 0$

We need to solve for $x$.

Let's denote:

$y = 5^x$

Using this substitution, the original equation becomes:

$25^x - 2 \cdot 5^x - 3 = 0$

Notice that $25^x = (5^2)^x = (5^x)^2 = y^2$. Thus, we have:

$y^2 - 2y - 3 = 0$

Now, we need to solve this quadratic equation for $y$:

$y^2 - 2y - 3 = 0$

To solve this quadratic, we use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = -2$, and $c = -3$. Plugging in these values:

$y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-3)}}{2 \cdot 1}$

$y = \frac{2 \pm \sqrt{4 + 12}}{2}$

$y = \frac{2 \pm \sqrt{16}}{2}$

$y = \frac{2 \pm 4}{2}$

This gives two possible solutions:

1. $y = \frac{2 + 4}{2} = 3$
2. $y = \frac{2 - 4}{2} = -1$

Since $y = 5^x$, and $y$ must be positive, we discard $y = -1$. Thus:

$y = 3$

Now recall that $y = 5^x$, so:

$5^x = 3$

To solve for $x$, take the logarithm base 5 of both sides:

$x = \log_5 3$

Thus, the answer is:

$\boxed{D) \, \log_5 3}$

Would you like more details on any step, or do you have any questions?

Here are some related questions for further practice:

1. How would you solve an exponential equation involving different bases, such as $2^{3x} = 7^{x+1}$?
2. How do you solve a logarithmic equation, for instance, $\log_2 (x + 1) = 3$?
3. What are the properties of logarithms that are useful for solving equations like $\log_a b = x$?
4. How can substitution help in solving complex exponential equations?
5. How do you convert logarithms of different bases using the change of base formula?

Tip: For exponential and logarithmic equations, substitutions can often simplify the equation to a more familiar form, like a quadratic. This approach can make complex-looking problems easier to handle.