If `3^(x)+2^(2x) ge 5^(x)`, then the solution set for x, is

Options

A. `(-oo,2]`
B. `[2,oo)`
C. `[0,2]`
D. {2}

Answer

To solve the inequality \( 3^x + 2^{2x} \geq 5^x \), we will follow these steps:

### Step 1: Rewrite the inequality
We start with the given inequality:
\[ 3^x + 2^{2x} \geq 5^x \]
We can rewrite \( 2^{2x} \) as \( (2^x)^2 \):
\[ 3^x + (2^x)^2 \geq 5^x \]

### Step 2: Substitute \( y = 2^x \)
Let \( y = 2^x \). Then \( 3^x = \left(\frac{3}{2}\right)^x y \) and \( 5^x = \left(\frac{5}{2}\right)^x y \). The inequality becomes:
\[ \left(\frac{3}{2}\right)^x y + y^2 \geq \left(\frac{5}{2}\right)^x y \]

### Step 3: Rearranging the inequality
Rearranging gives us:
\[ y^2 + \left(\frac{3}{2}\right)^x y - \left(\frac{5}{2}\right)^x y \geq 0 \]
Factoring out \( y \):
\[ y \left(y + \left(\frac{3}{2}\right)^x - \left(\frac{5}{2}\right)^x\right) \geq 0 \]

### Step 4: Analyzing the factors
1. The term \( y \) (which is \( 2^x \)) is always positive for all real \( x \).
2. We need to analyze the second factor:
\[ y + \left(\frac{3}{2}\right)^x - \left(\frac{5}{2}\right)^x \geq 0 \]

### Step 5: Finding critical points
To find when the second factor is zero, we can set:
\[ \left(\frac{3}{2}\right)^x = \left(\frac{5}{2}\right)^x \]
Taking logarithms gives:
\[ x \log\left(\frac{3}{2}\right) = x \log\left(\frac{5}{2}\right) \]
This equality holds when \( x = 0 \).

### Step 6: Testing intervals
Now we will test intervals around the critical point \( x = 2 \):
- For \( x < 2 \): Choose \( x = 1 \):
\[ 3^1 + 2^{2 \cdot 1} = 3 + 4 = 7 \quad \text{and} \quad 5^1 = 5 \quad \Rightarrow \quad 7 \geq 5 \quad \text{(True)} \]
- For \( x = 2 \):
\[ 3^2 + 2^{2 \cdot 2} = 9 + 16 = 25 \quad \text{and} \quad 5^2 = 25 \quad \Rightarrow \quad 25 \geq 25 \quad \text{(True)} \]
- For \( x > 2 \): Choose \( x = 3 \):
\[ 3^3 + 2^{2 \cdot 3} = 27 + 64 = 91 \quad \text{and} \quad 5^3 = 125 \quad \Rightarrow \quad 91 < 125 \quad \text{(False)} \]

### Conclusion
The inequality holds for \( x \leq 2 \). Therefore, the solution set is:
\[ x \in (-\infty, 2] \]