The solution set of inequation `log_(1//3)(2^(x+2)-4^(x)) ge-2`, is

Options

A. `(-oo,2-sqrt(13))`
B. `(-oo,2+sqrt(13))`
C. `(-oo,2)`
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 Inequality
The first step is to rewrite the inequality in a more manageable form. Since the base of the logarithm is \( \frac{1}{3} \), which is between 0 and 1, we can reverse the inequality when we remove the logarithm.

\[ \log_{(1/3)}(2^{(x+2)} - 4^{x}) \ge -2 \implies 2^{(x+2)} - 4^{x} \le (1/3)^{-2} \]

Calculating \( (1/3)^{-2} \):
\[ (1/3)^{-2} = 3^{2} = 9 \]

So we have:
\[ 2^{(x+2)} - 4^{x} \le 9 \]

### Step 2: Simplify the Expression
Next, we simplify \( 4^{x} \) in terms of \( 2^{x} \):
\[ 4^{x} = (2^{2})^{x} = 2^{2x} \]

Thus, the inequality becomes:
\[ 2^{(x+2)} - 2^{(2x)} \le 9 \]

### Step 3: Rewrite \( 2^{(x+2)} \)
We can rewrite \( 2^{(x+2)} \) as:
\[ 2^{(x+2)} = 2^{x} \cdot 2^{2} = 4 \cdot 2^{x} \]

Substituting this back into the inequality gives:
\[ 4 \cdot 2^{x} - 2^{(2x)} \le 9 \]

### Step 4: Let \( t = 2^{x} \)
Let \( t = 2^{x} \). Then the inequality becomes:
\[ 4t - t^{2} \le 9 \]

Rearranging gives:
\[ -t^{2} + 4t - 9 \le 0 \]

### Step 5: Solve the Quadratic Inequality
Now, we can rearrange this into standard quadratic form:
\[ t^{2} - 4t + 9 \ge 0 \]

### Step 6: Find the Discriminant
To find the roots of the quadratic, we calculate the discriminant \( D \):
\[ D = b^{2} - 4ac = (-4)^{2} - 4 \cdot 1 \cdot 9 = 16 - 36 = -20 \]

Since the discriminant is negative, the quadratic \( t^{2} - 4t + 9 \) has no real roots and is always positive.

### Step 7: Determine the Validity of \( t \)
Since \( t = 2^{x} \) is always positive, the inequality \( t^{2} - 4t + 9 \ge 0 \) holds for all real \( t \).

### Step 8: Check the Domain of the Logarithm
Now, we need to ensure that the argument of the logarithm is positive:
\[ 2^{(x+2)} - 4^{x} > 0 \]
This simplifies to:
\[ 4 \cdot 2^{x} - 2^{(2x)} > 0 \]
Factoring gives:
\[ 2^{x}(4 - 2^{x}) > 0 \]
This implies:
\[ 2^{x} > 0 \quad \text{and} \quad 4 - 2^{x} > 0 \implies 2^{x} < 4 \implies x < 2 \]

### Step 9: Conclusion
Combining our results, we find that \( x \) must be less than 2. Therefore, the solution set of the inequality is:
\[ x \in (-\infty, 2) \]