The solution set of inequation `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 inequation \( \log_{1/3}(2^{x+2} - 4^x) \ge -2 \), we will follow these steps:

### Step 1: Rewrite the Inequation
We start with the given inequation:
\[ \log_{1/3}(2^{x+2} - 4^x) \ge -2 \]
Using the property of logarithms, we can rewrite this as:
\[ 2^{x+2} - 4^x \ge (1/3)^{-2} \]
Calculating \( (1/3)^{-2} \):
\[ (1/3)^{-2} = 3^2 = 9 \]
Thus, we can rewrite the inequation as:
\[ 2^{x+2} - 4^x \ge 9 \]

### Step 2: Simplify the Expression
Next, we simplify \( 4^x \):
\[ 4^x = (2^2)^x = (2^x)^2 \]
So, our inequation becomes:
\[ 2^{x+2} - (2^x)^2 \ge 9 \]
Now, we can express \( 2^{x+2} \) as:
\[ 2^{x+2} = 4 \cdot 2^x \]
Substituting this back into the inequation gives us:
\[ 4 \cdot 2^x - (2^x)^2 \ge 9 \]

### Step 3: Let \( t = 2^x \)
Let \( t = 2^x \). The inequation now becomes:
\[ 4t - t^2 \ge 9 \]
Rearranging this gives:
\[ -t^2 + 4t - 9 \ge 0 \]
Multiplying through by -1 (and reversing the inequality) results in:
\[ t^2 - 4t + 9 \le 0 \]

### Step 4: Analyze the Quadratic
Next, we need to find the discriminant \( D \) of the quadratic \( t^2 - 4t + 9 \):
\[ D = b^2 - 4ac = (-4)^2 - 4 \cdot 1 \cdot 9 = 16 - 36 = -20 \]
Since the discriminant is negative, the quadratic \( t^2 - 4t + 9 \) does not intersect the x-axis and is always positive (as the coefficient of \( t^2 \) is positive).

### Step 5: Conclusion about \( t \)
Since \( t^2 - 4t + 9 > 0 \) for all \( t \), there are no real values of \( t \) that satisfy the inequality \( t^2 - 4t + 9 \le 0 \). Thus, the solution set for \( t \) (and consequently for \( x \)) is all real numbers.

### Step 6: Check the Logarithm Condition
However, we need to ensure that the expression inside the logarithm is positive:
\[ 2^{x+2} - 4^x > 0 \]
This simplifies to:
\[ 2^{x+2} > 4^x \]
Substituting \( 4^x \) gives:
\[ 4 \cdot 2^x > (2^x)^2 \]
Factoring out \( 2^x \) gives:
\[ 2^x(4 - 2^x) > 0 \]
Since \( 2^x > 0 \) for all \( x \), we need:
\[ 4 - 2^x > 0 \implies 2^x < 4 \implies 2^x < 2^2 \]
Taking logarithms gives:
\[ x < 2 \]

### Final Solution
Thus, the solution set for the original inequation is:
\[ x \in (-\infty, 2) \]