Solve: ` 15^x-25* 3^x-9*5^x+225 >= 0`.

Answer

To solve the inequality \( 15^x - 25 \cdot 3^x - 9 \cdot 5^x + 225 \geq 0 \), we will follow these steps:

### Step 1: Rewrite the terms
We start by rewriting \( 15^x \) in terms of its prime factors:
\[ 15^x = (3 \cdot 5)^x = 3^x \cdot 5^x \]
Now, we substitute this back into the inequality:
\[ 3^x \cdot 5^x - 25 \cdot 3^x - 9 \cdot 5^x + 225 \geq 0 \]

### Step 2: Factor out common terms
Next, we can factor out \( 3^x \) and \( 5^x \) from the first two terms:
\[ 3^x (5^x - 25) - 9 \cdot 5^x + 225 \geq 0 \]
Now, we can rewrite the expression:
\[ 3^x (5^x - 25) - 9(5^x - 25) \geq 0 \]
This can be factored as:
\[ (3^x - 9)(5^x - 25) \geq 0 \]

### Step 3: Set up the critical points
Now we will find the critical points by setting each factor to zero:
1. \( 3^x - 9 = 0 \) implies \( 3^x = 9 \) or \( x = 2 \).
2. \( 5^x - 25 = 0 \) implies \( 5^x = 25 \) or \( x = 2 \).

### Step 4: Analyze the intervals
The critical point \( x = 2 \) divides the number line into intervals. We will test the sign of the product \( (3^x - 9)(5^x - 25) \) in each interval:
1. **Interval 1**: \( (-\infty, 2) \)
- Choose \( x = 0 \):
\[ (3^0 - 9)(5^0 - 25) = (-8)(-24) > 0 \]
2. **Interval 2**: \( (2, \infty) \)
- Choose \( x = 3 \):
\[ (3^3 - 9)(5^3 - 25) = (27 - 9)(125 - 25) = (18)(100) > 0 \]
3. **At the point**: \( x = 2 \)
\[ (3^2 - 9)(5^2 - 25) = (9 - 9)(25 - 25) = 0 \]

### Step 5: Combine the results
The product \( (3^x - 9)(5^x - 25) \geq 0 \) holds true in the intervals:
- \( (-\infty, 2] \)
- \( [2, \infty) \)

Thus, the solution set is:
\[ x \in (-\infty, 2] \cup [2, \infty) = \mathbb{R} \]

### Final Answer
The solution to the inequality \( 15^x - 25 \cdot 3^x - 9 \cdot 5^x + 225 \geq 0 \) is:
\[ x \in \mathbb{R} \]