To solve the equation \( 5^{2x} \cdot 25^x = 100 \), follow these steps:

1. **Express \( 25^x \) in terms of \( 5 \):**

\[ 25 = 5^2 \]
\[ 25^x = (5^2)^x = 5^{2x} \]

So the equation becomes:
\[ 5^{2x} \cdot 5^{2x} = 100 \]

2. **Combine the exponents:**

\[ 5^{2x} \cdot 5^{2x} = 5^{2x + 2x} = 5^{4x} \]

So now the equation is:
\[ 5^{4x} = 100 \]

3. **Take the logarithm of both sides:**

\[ \log(5^{4x}) = \log(100) \]

Using the property of logarithms \( \log(a^b) = b\log(a) \), we get:
\[ 4x \log(5) = \log(100) \]

4. **Solve for \( x \):**

\[ x = \frac{\log(100)}{4 \log(5)} \]

5. **Simplify the logarithmic expression:**

We know that \( \log(100) = \log(10^2) = 2 \log(10) = 2 \) (since \( \log(10) = 1 \)).
So,
\[ x = \frac{2}{4 \log(5)} = \frac{1}{2 \log(5)} \]

Therefore, the solution to the equation \( 5^{2x} \cdot 25^x = 100 \) is:
\[ x = \frac{1}{2 \log(5)} \]

If you have any questions or would like more details, please let me know!

Here are some related questions for further exploration:
1. How do you solve equations involving multiple exponential terms?
2. What is the significance of the base in exponential equations?
3. How can logarithms be used to solve exponential equations more efficiently?
4. What are the common properties of exponents used in solving equations?
5. How do you handle exponential equations with different bases?
6. How can you verify the solution to an exponential equation?
7. What are some real-world applications of solving exponential equations?
8. How does changing the exponent affect the solution to an equation?

**Tip:** When working with exponential equations, always try to express all terms with the same base to simplify the calculations and solve the equation more easily.

Learn how to solve the equation 5^(2x)*25^(x)=100 step-by-step. Understand the relationship between exponential equations, logarithms, and different bases. Suitable for advanced high school and college students.

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