Solve for x: `4^(x+1.5) + 9^(x+0.5) = 10.6^x`.

Answer

To solve the equation \( 4^{(x + 1.5)} + 9^{(x + 0.5)} = 10 \cdot 6^x \), we will follow these steps:

### Step 1: Rewrite the equation
We can express \( 4 \) and \( 9 \) in terms of their prime factors:
\[ 4 = 2^2 \quad \text{and} \quad 9 = 3^2 \]
Thus, we can rewrite the equation as:
\[ (2^2)^{(x + 1.5)} + (3^2)^{(x + 0.5)} = 10 \cdot (2 \cdot 3)^x \]
This simplifies to:
\[ 2^{2(x + 1.5)} + 3^{2(x + 0.5)} = 10 \cdot 2^x \cdot 3^x \]

### Step 2: Simplify the powers
Now, we can simplify the left-hand side:
\[ 2^{2x + 3} + 3^{2x + 1} = 10 \cdot 2^x \cdot 3^x \]

### Step 3: Factor out common terms
We can rewrite the equation:
\[ 2^{2x + 3} + 3^{2x + 1} - 10 \cdot 2^x \cdot 3^x = 0 \]

### Step 4: Substitute variables
Let \( a = 2^x \) and \( b = 3^x \). Then, we can rewrite the equation as:
\[ 8a^2 + 3b^2 - 10ab = 0 \]

### Step 5: Rearrange the equation
This is a quadratic equation in terms of \( a \):
\[ 8a^2 - 10ab + 3b^2 = 0 \]

### Step 6: Apply the quadratic formula
Using the quadratic formula \( a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \), where \( A = 8 \), \( B = -10b \), and \( C = 3b^2 \):
\[ a = \frac{10b \pm \sqrt{(-10b)^2 - 4 \cdot 8 \cdot 3b^2}}{2 \cdot 8} \]
\[ = \frac{10b \pm \sqrt{100b^2 - 96b^2}}{16} \]
\[ = \frac{10b \pm \sqrt{4b^2}}{16} \]
\[ = \frac{10b \pm 2b}{16} \]

### Step 7: Solve for \( a \)
This gives us two possible solutions for \( a \):
1. \( a = \frac{12b}{16} = \frac{3b}{4} \)
2. \( a = \frac{8b}{16} = \frac{b}{2} \)

### Step 8: Substitute back for \( x \)
Recall that \( a = 2^x \) and \( b = 3^x \):
1. From \( 2^x = \frac{3 \cdot 3^x}{4} \):
\[ 4 \cdot 2^x = 3 \cdot 3^x \implies \frac{4}{3} = \left(\frac{3}{2}\right)^x \]
Taking logarithms:
\[ x = \frac{\log(4/3)}{\log(3/2)} \]

2. From \( 2^x = \frac{3^x}{2} \):
\[ 2 \cdot 2^x = 3^x \implies 2 = \left(\frac{3}{2}\right)^x \]
Taking logarithms:
\[ x = \frac{\log(2)}{\log(3/2)} \]

### Final Solution
Thus, the solutions for \( x \) are:
\[ x = \frac{\log(4/3)}{\log(3/2)} \quad \text{and} \quad x = \frac{\log(2)}{\log(3/2)} \]