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If x≠0, express x in the following equation with y

The equation given is:

\[
5 + 5^{2x + y} - 5^{x + 1} - 5^{x + y} = 0
\]

To solve this equation for \(x\) in terms of \(y\), let's follow these steps:

1. **Factor the equation where possible:**
   We can factor out common terms to simplify the equation.

   Let's first rewrite the equation:

   \[
   5 + 5^{2x + y} - 5 \cdot 5^{x} - 5^{x + y} = 0
   \]

   Notice that \(5 \cdot 5^{x} = 5^{x+1}\), so the equation remains the same. Now let's factor out common factors:

   \[
   5(1 - 5^x) + 5^{x+y}(5^{x} - 1) = 0
   \]

2. **Isolate terms involving \(x\):**

   \[
   5(1 - 5^x) = -5^{x+y}(5^x - 1)
   \]

   Divide both sides by 5:

   \[
   1 - 5^x = -5^{x+y-1}(5^x - 1)
   \]

   Now, express it in a more manageable form:

   \[
   \frac{1 - 5^x}{5^x - 1} = -5^{x+y-1}
   \]

3. **Simplify the equation:**

   Since \(1 - 5^x\) and \(5^x - 1\) are negatives of each other, the equation simplifies to:

   \[
   \frac{-5^x + 1}{5^x - 1} = -5^{x+y-1}
   \]

   Simplifying the fraction:

   \[
   \frac{-(5^x - 1)}{5^x - 1} = -5^{x+y-1}
   \]

   This further simplifies to:

   \[
   -1 = -5^{x+y-1}
   \]

   Therefore:

   \[
   5^{x+y-1} = 1
   \]

4. **Solve for \(x\):**

   Since \(5^{x+y-1} = 1\), the exponent must be zero because \(5^0 = 1\). Therefore:

   \[
   x + y - 1 = 0
   \]

   Solving for \(x\), we get:

   \[
   x = 1 - y
   \]

So, the solution is \(x = 1 - y\).

Would you like any further details or have any questions?

### Related Questions:
1. How do you solve exponential equations with different bases?
2. What are the properties of exponential functions that allow us to solve such equations?
3. How do you graph \(y = 5^{x}\) and \(y = 5^{1-x}\)?
4. What is the importance of factoring in solving algebraic equations?
5. Can this method be applied to equations with more than one exponential term with different bases?
6. How does changing the base of the exponent affect the solution of the equation?
7. What are the general steps to isolate \(x\) in equations involving exponential functions?
8. How can logarithms be used to solve exponential equations?

### Tip:
When solving exponential equations, always look for ways to factor or simplify expressions to make the exponents more manageable.

Learn how to solve the exponential equation 5 + 5^(2x + y) - 5^(x + 1) - 5^(x + y) = 0 for x with a detailed step-by-step solution. This problem covers key concepts in exponential equations, factoring, and solving equations. Suitable for advanced high school students.

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