The solution set of the equation
`"log"_(1//3)(2^(x+2)-4^(x)) ge -2`, is

Options

A. `(-oo, 2)`
B. `(-oo, 2 + sqrt13)`
C. `(2,oo)`
D. none of these

Answer

To solve the inequality \( \log_{1/3}(2^{x+2} - 4^x) \ge -2 \), we will follow these steps:

### Step 1: Rewrite the logarithmic inequality
We start by rewriting the logarithmic inequality:
\[ \log_{1/3}(2^{x+2} - 4^x) \ge -2 \]
This can be rewritten in exponential form:
\[ 2^{x+2} - 4^x \le (1/3)^{-2} \]
Calculating \( (1/3)^{-2} \):
\[ (1/3)^{-2} = 9 \]
Thus, we have:
\[ 2^{x+2} - 4^x \le 9 \]

### Step 2: Simplify the expression
Next, we simplify \( 4^x \):
\[ 4^x = (2^2)^x = (2^x)^2 \]
So we can rewrite the inequality as:
\[ 2^{x+2} - (2^x)^2 \le 9 \]
Let \( y = 2^x \). Then \( 2^{x+2} = 4y \), and our inequality becomes:
\[ 4y - y^2 \le 9 \]

### Step 3: Rearrange the inequality
Rearranging gives:
\[ -y^2 + 4y - 9 \le 0 \]
Multiplying through by -1 (which reverses the inequality):
\[ y^2 - 4y + 9 \ge 0 \]

### Step 4: Find the roots of the quadratic
To find the roots of the quadratic equation \( y^2 - 4y + 9 = 0 \), we use the quadratic formula:
\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -4, c = 9 \):
\[ y = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{4 \pm \sqrt{16 - 36}}{2} = \frac{4 \pm \sqrt{-20}}{2} \]
Since the discriminant is negative (\(-20\)), there are no real roots.

### Step 5: Analyze the quadratic
Since the quadratic \( y^2 - 4y + 9 \) opens upwards (as the coefficient of \( y^2 \) is positive) and has no real roots, it is always positive:
\[ y^2 - 4y + 9 > 0 \quad \forall y \in \mathbb{R} \]

### Step 6: Check the domain of the logarithm
Next, we need to ensure that the argument of the logarithm is positive:
\[ 2^{x+2} - 4^x > 0 \]
This simplifies to:
\[ 4y - y^2 > 0 \]
Factoring gives:
\[ y(4 - y) > 0 \]
This inequality holds when \( y > 0 \) and \( y < 4 \). Thus:
\[ 0 < 2^x < 4 \implies 0 < x < 2 \]

### Final Solution
Combining the conditions, we find:
\[ x < 2 \]
Therefore, the solution set is:
\[ (-\infty, 2) \]