The solution set of the inequality `|9^x-3^(x+1) + 15| < 2.9^x - 3^x` is

Options

A. `(-oo,1)`
B. `(1,oo)`
C. `(-oo,1]`
D. `[2,oo)`

Answer

To solve the inequality \( |9^x - 3^{x+1} + 15| < 2 \cdot 9^x - 3^x \), we can follow these steps:

### Step 1: Rewrite the terms
First, we can express \( 9^x \) in terms of \( 3^x \):
\[ 9^x = (3^2)^x = (3^x)^2 \]
Let \( t = 3^x \). Then, we can rewrite the inequality as:
\[ |t^2 - 3t + 15| < 2t^2 - t \]

### Step 2: Simplify the inequality
Now, we can simplify the inequality:
\[ |t^2 - 3t + 15| < 2t^2 - t \]
This can be split into two cases based on the definition of absolute value.

### Step 3: Case 1: \( t^2 - 3t + 15 < 2t^2 - t \)
In this case, we can simplify:
\[ t^2 - 3t + 15 < 2t^2 - t \]
Rearranging gives:
\[ 0 < t^2 + 2t - 15 \]
Factoring the quadratic:
\[ 0 < (t - 3)(t + 5) \]
The critical points are \( t = 3 \) and \( t = -5 \). The solution to this inequality is:
\[ t < -5 \quad \text{or} \quad t > 3 \]

### Step 4: Case 2: \( -(t^2 - 3t + 15) < 2t^2 - t \)
Now, we consider the second case:
\[ -t^2 + 3t - 15 < 2t^2 - t \]
Rearranging gives:
\[ 0 < 3t^2 - 4t + 15 \]
This quadratic is always positive because its discriminant \( (-4)^2 - 4 \cdot 3 \cdot 15 < 0 \).

### Step 5: Combine the results
From Case 1, we have \( t < -5 \) or \( t > 3 \). Since \( t = 3^x \) must be positive, we discard \( t < -5 \). Thus, we only consider:
\[ t > 3 \]
This translates back to:
\[ 3^x > 3 \implies x > 1 \]

### Final Solution
The solution set of the inequality is:
\[ x > 1 \quad \text{or} \quad (1, \infty) \]