Given that a ≠ b and a, b > 0, and \(\rm \frac{a^{n+2}+b^{n+2}}{a^n+b^n}\) = ab then n =
-1
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.
Important Questions from Linear Equation in 2 Variable
The cost of 6 pens and 19 pencils is ₹275. If the cost of a pen decreases by ₹7 and the cost of a pencil increases by ₹7, then the cost of 15 pens and 4 pencils is 132. What is the original cost of 6 pens and 2 pencils?
- The cost of 12 pens and 7 pencils is ₹285. If the cost of a pen decreases by ₹6 and the cost of a pencil increases by ₹5, then the cost of 11 pens and 3 pencils is ₹118. What is the original cost of 2 pens and 2 pencils?
- The cost of 7 pens and 17 pencils is ₹92. If the cost of a pen decreases by ₹0.50 and the cost of a pencil increases by ₹8, then the cost of 8 pens and 14 pencils is 186. What is the original cost of 9 pens and 6 pencils?
- The cost of 19 pens and 13 pencils is ₹254. If the cost of a pen decreases by ₹6 and the cost of a pencil increases by ₹9, then the cost of 14 pens and 2 pencils is ₹106. What is the original cost of 12 pens and 15 pencils?
- The price of 4 pens and 5 notebooks is ₹130. If the price of a pen increases by 25% and the price of a notebook decreases by 20%, then the price of 2 pens and 4 notebooks becomes ₹82. What is the original price of 1 pen and 2 notebooks?
