To solve the inequality \( 4^x - 3 \cdot 2^x + 2 > 0 \), we can follow these steps:
### Step 1: Rewrite the expression
We know that \( 4^x \) can be rewritten as \( (2^x)^2 \). Thus, we can express the inequality as:
\[
(2^x)^2 - 3 \cdot 2^x + 2 > 0
\]
Let \( t = 2^x \). The inequality now becomes:
\[
t^2 - 3t + 2 > 0
\]
### Step 2: Factor the quadratic expression
Next, we need to factor the quadratic \( t^2 - 3t + 2 \):
\[
t^2 - 3t + 2 = (t - 1)(t - 2)
\]
Thus, the inequality can be rewritten as:
\[
(t - 1)(t - 2) > 0
\]
### Step 3: Determine the critical points
The critical points from the factors are \( t = 1 \) and \( t = 2 \). We will analyze the sign of the product \( (t - 1)(t - 2) \) in the intervals determined by these points: \( (-\infty, 1) \), \( (1, 2) \), and \( (2, \infty) \).
### Step 4: Test the intervals
1. **Interval \( (-\infty, 1) \)**:
- Choose \( t = 0 \):
\[
(0 - 1)(0 - 2) = (-1)(-2) = 2 > 0
\]
So, the product is positive.
2. **Interval \( (1, 2) \)**:
- Choose \( t = 1.5 \):
\[
(1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 < 0
\]
So, the product is negative.
3. **Interval \( (2, \infty) \)**:
- Choose \( t = 3 \):
\[
(3 - 1)(3 - 2) = (2)(1) = 2 > 0
\]
So, the product is positive.
### Step 5: Combine the results
The inequality \( (t - 1)(t - 2) > 0 \) holds for the intervals:
\[
t < 1 \quad \text{or} \quad t > 2
\]
### Step 6: Substitute back for \( t \)
Recall that \( t = 2^x \). We now translate the intervals back to \( x \):
1. For \( t < 1 \):
\[
2^x < 1 \implies x < 0
\]
2. For \( t > 2 \):
\[
2^x > 2 \implies x > 1
\]
### Final Solution
Combining these results, we find:
\[
x < 0 \quad \text{or} \quad x > 1
\]
Thus, the solution in interval notation is:
\[
(-\infty, 0) \cup (1, \infty)
\]
