Solving the Exponential Equation

The problem asks us to find the value of \(n\) given the equation \(\frac{a^{n+2}+b^{n+2}}{a^n+b^n} = ab\), where \(a \ne b\) and \(a, b > 0\).

We start by rearranging the given equation:

\[\frac{a^{n+2}+b^{n+2}}{a^n+b^n} = ab\]

Multiply both sides by \((a^n+b^n)\) to clear the denominator:

\[a^{n+2}+b^{n+2} = ab(a^n+b^n)\]

Distribute \(ab\) on the right side:

\[a^{n+2}+b^{n+2} = a \cdot a^n \cdot b + a \cdot b \cdot b^n\]

Using the exponent rule \(x^m \cdot x^p = x^{m+p}\), we get:

\[a^{n+2}+b^{n+2} = a^{n+1}b + ab^{n+1}\]

Rearranging Terms

Now, let's move terms to one side to group similar powers of \(a\) and \(b\):

\[a^{n+2} - a^{n+1}b = ab^{n+1} - b^{n+2}\]

Factoring the Equation

Factor out common terms from each side of the equation. On the left side, \(a^{n+1}\) is common. On the right side, \(b^{n+1}\) is common:

\[a^{n+1}(a - b) = b^{n+1}(a - b)\]

Solving for n

We are given that \(a \ne b\), which means \(a - b \ne 0\). Since \(a - b\) is not zero, we can divide both sides of the equation by \((a - b)\):

\[\frac{a^{n+1}(a - b)}{a - b} = \frac{b^{n+1}(a - b)}{a - b}\]

This simplifies to:

\[a^{n+1} = b^{n+1}\]

To solve for \(n\), we can rearrange this equation:

\[\frac{a^{n+1}}{b^{n+1}} = 1\]

Using the exponent rule \(\frac{x^m}{y^m} = \left(\frac{x}{y}\right)^m\), we get:

\[\left(\frac{a}{b}\right)^{n+1} = 1\]

We are given that \(a > 0\), \(b > 0\), and \(a \ne b\). This means the base \(\frac{a}{b}\) is a positive number and is not equal to 1.

For any positive base \(x\) where \(x \ne 1\), the equation \(x^y = 1\) is true only if the exponent \(y\) is 0.

In our equation, the base is \(\frac{a}{b}\) and the exponent is \(n+1\).

Therefore, we must have:

\[n+1 = 0\]

Solving for \(n\):

\[n = -1\]

We can verify this solution by substituting \(n=-1\) back into the original equation:

\[\frac{a^{-1+2}+b^{-1+2}}{a^{-1}+b^{-1}} = \frac{a^{1}+b^{1}}{\frac{1}{a}+\frac{1}{b}} = \frac{a+b}{\frac{b+a}{ab}} = (a+b) \times \frac{ab}{a+b} = ab\]

This matches the right side of the original equation.

Thus, the value of \(n\) is -1.