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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.

25^x - 2.5^x - 3 = 0, denkleminin kökü aşağıdakilerden hangisidir?

Learn how to solve the exponential equation 25^x - 2.5^x - 3 = 0 using substitution and the quadratic formula. The problem involves core concepts such as exponential and logarithmic equations. Detailed step-by-step solution provided for students in Grades 10-12.

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