Solve for x :
`4^(x-1)xx(0.5)^(3-2x)=((1)/(8))^(-x)`

Answer

To solve the equation \( 4^{(x-1)} \cdot (0.5)^{(3-2x)} = \left(\frac{1}{8}\right)^{-x} \), we will follow these steps:

### Step 1: Rewrite the bases
First, we will express all terms with a common base of 2.

- \( 4 = 2^2 \), so \( 4^{(x-1)} = (2^2)^{(x-1)} = 2^{2(x-1)} = 2^{2x - 2} \).
- \( 0.5 = \frac{1}{2} = 2^{-1} \), so \( (0.5)^{(3-2x)} = (2^{-1})^{(3-2x)} = 2^{-(3-2x)} = 2^{2x - 3} \).
- \( 8 = 2^3 \), so \( \left(\frac{1}{8}\right)^{-x} = (2^{-3})^{-x} = 2^{3x} \).

Now, substituting these into the equation gives us:
\[ 2^{2x - 2} \cdot 2^{2x - 3} = 2^{3x} \]

### Step 2: Combine the left side
Using the property of exponents that states \( a^m \cdot a^n = a^{m+n} \), we can combine the left side:
\[ 2^{(2x - 2) + (2x - 3)} = 2^{3x} \]
This simplifies to:
\[ 2^{4x - 5} = 2^{3x} \]

### Step 3: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:
\[ 4x - 5 = 3x \]

### Step 4: Solve for x
Now, we will solve for \( x \):
\[ 4x - 3x = 5 \]
\[ x = 5 \]

### Final Answer
Thus, the solution for \( x \) is:
\[ \boxed{5} \]