Find the solution set (x,y)if x power y+1 = 1 divide by x power 2-2y  =256

The given equation is:

\[
\frac{x^{y+1}}{x^{2 - 2y}} = 256
\]

To solve for \( x \) and \( y \), let's simplify this expression step-by-step.

### Step 1: Simplify the Expression
We can combine the powers in the numerator and the denominator:

\[
\frac{x^{y+1}}{x^{2 - 2y}} = x^{(y+1) - (2 - 2y)}
\]

Simplify the exponent:

\[
x^{(y+1) - (2 - 2y)} = x^{y + 1 - 2 + 2y} = x^{3y - 1}
\]

So, the equation becomes:

\[
x^{3y - 1} = 256
\]

### Step 2: Express \( 256 \) as a Power of 2
Notice that \( 256 = 2^8 \), so we can rewrite the equation as:

\[
x^{3y - 1} = 2^8
\]

### Step 3: Analyze Possible Values of \( x \) and \( y \)
There are two cases to consider:

1. **Case 1**: \( x = 2 \)

   If \( x = 2 \), then:

   \[
   2^{3y - 1} = 2^8
   \]

   Since the bases are the same, we can equate the exponents:

   \[
   3y - 1 = 8
   \]

   Solving for \( y \):

   \[
   3y = 9 \Rightarrow y = 3
   \]

   So, one solution is \( (x, y) = (2, 3) \).

2. **Case 2**: \( x = -2 \)

   If \( x = -2 \), then we must examine whether \( x^{3y - 1} = 256 \) can hold, as the left side would be a power of a negative base. For real solutions, \( y \) would need to be chosen carefully, and typically integer values of \( y \) may not work.

   Therefore, we discard this case for real solutions.

### Conclusion
The only solution in real numbers is:

\[
(x, y) = (2, 3)
\]

Would you like further details or have any questions?

---

Here are five related questions:
1. What if \( 256 \) were a different power of 2?
2. How would this change if we changed the base from \( x \) to another variable?
3. Can this approach be used for non-integer exponents?
4. How do we solve similar equations if we include complex numbers?
5. What if we change the value of the right side from \( 256 \) to something else?

**Tip:** For equations involving exponents, try to rewrite numbers as powers of common bases, like 2, 3, or 5, to simplify comparisons.

Find the solution set (x,y) if x^y+1 = 1 / x^2-2y = 256

Explore how to find the solution set for (x, y) in the exponential equation x^(y+1) / x^(2-2y) = 256. This problem involves exponent rules and simplification techniques, with a detailed step-by-step solution. Suitable for students in Grades 10-12.

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