Solve for x : ` sqrt (32^(0) +(2)/(3)) = ( 0.6) ^(2-3x)`

Answer

To solve the equation \( \sqrt{32^{0} + \frac{2}{3}} = (0.6)^{2 - 3x} \), we will follow these steps:

### Step 1: Simplify \( 32^{0} \)
Any number raised to the power of 0 is equal to 1. Therefore, we have:
\[ 32^{0} = 1 \]
So, the equation becomes:
\[ \sqrt{1 + \frac{2}{3}} = (0.6)^{2 - 3x} \]

### Step 2: Simplify the left side
Now, we need to simplify \( 1 + \frac{2}{3} \):
\[ 1 + \frac{2}{3} = \frac{3}{3} + \frac{2}{3} = \frac{5}{3} \]
Thus, the equation now is:
\[ \sqrt{\frac{5}{3}} = (0.6)^{2 - 3x} \]

### Step 3: Square both sides
Next, we will square both sides to eliminate the square root:
\[ \frac{5}{3} = (0.6)^{2 - 3x} \cdot (0.6)^{2 - 3x} = (0.6)^{2(2 - 3x)} = (0.6)^{4 - 6x} \]

### Step 4: Rewrite \( 0.6 \)
We can express \( 0.6 \) as a fraction:
\[ 0.6 = \frac{3}{5} \]
So, we rewrite the equation:
\[ \frac{5}{3} = \left(\frac{3}{5}\right)^{4 - 6x} \]

### Step 5: Express both sides with the same base
We can rewrite \( \frac{5}{3} \) as \( \left(\frac{3}{5}\right)^{-1} \):
\[ \left(\frac{3}{5}\right)^{-1} = \left(\frac{3}{5}\right)^{4 - 6x} \]

### Step 6: Set the exponents equal
Since the bases are the same, we can set the exponents equal to each other:
\[ -1 = 4 - 6x \]

### Step 7: Solve for \( x \)
Now, we will solve for \( x \):
\[ -1 - 4 = -6x \\ -5 = -6x \\ x = \frac{5}{6} \]

### Final Answer
Thus, the solution for \( x \) is:
\[ x = \frac{5}{6} \]