Solve` (3/4)^(6x+10-x^(2)) lt 27/64`.

Explanation

To solve the inequality \( \left(\frac{3}{4}\right)^{6x + 10 - x^2} < \frac{27}{64} \), we will follow these steps:

### Step 1: Rewrite the inequality
We start with the given inequality:
\[ \left(\frac{3}{4}\right)^{6x + 10 - x^2} < \frac{27}{64} \]

### Step 2: Express \( \frac{27}{64} \) in terms of base \( \frac{3}{4} \)
Notice that:
\[ \frac{27}{64} = \frac{3^3}{4^3} = \left(\frac{3}{4}\right)^3 \]
Thus, we can rewrite the inequality as:
\[ \left(\frac{3}{4}\right)^{6x + 10 - x^2} < \left(\frac{3}{4}\right)^3 \]

### Step 3: Compare the exponents
Since the base \( \frac{3}{4} \) is less than 1, we reverse the inequality when comparing the exponents:
\[ 6x + 10 - x^2 > 3 \]

### Step 4: Rearrange the inequality
Rearranging gives:
\[ 6x + 10 - x^2 - 3 > 0 \]
This simplifies to:
\[ -x^2 + 6x + 7 > 0 \]

### Step 5: Multiply by -1
Multiplying the entire inequality by -1 (which reverses the inequality sign) gives:
\[ x^2 - 6x - 7 < 0 \]

### Step 6: Factor the quadratic
Next, we factor the quadratic:
\[ x^2 - 6x - 7 = (x - 7)(x + 1) = 0 \]
The roots of the equation are \( x = 7 \) and \( x = -1 \).

### Step 7: Determine the intervals
We will analyze the sign of the product \( (x - 7)(x + 1) \) in the intervals determined by the roots:
- Interval 1: \( (-\infty, -1) \)
- Interval 2: \( (-1, 7) \)
- Interval 3: \( (7, \infty) \)

### Step 8: Test the intervals
1. For \( x < -1 \) (e.g., \( x = -2 \)):
\[ (-2 - 7)(-2 + 1) = (-9)(-1) = 9 > 0 \]
2. For \( -1 < x < 7 \) (e.g., \( x = 0 \)):
\[ (0 - 7)(0 + 1) = (-7)(1) = -7 < 0 \]
3. For \( x > 7 \) (e.g., \( x = 8 \)):
\[ (8 - 7)(8 + 1) = (1)(9) = 9 > 0 \]

### Step 9: Conclusion
The inequality \( (x - 7)(x + 1) < 0 \) holds for the interval:
\[ x \in (-1, 7) \]

### Final Answer
Thus, the solution to the inequality is:
\[ x \in (-1, 7) \]

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