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
What is the solution of the following equations ?
2x + 3y = 12 and 3x − 2y = 5
Two positive numbers differ by 1280. When the greater number is divided by the smaller number, the quotient is 7 and the remainder is 50. The greater number is:
When 5 children from class A join class B, the number of children in both classes is the same. If 25 children from B, join A, then the number of children in A becomes double the number of children in B. The ratio of the number of children in A to those in B is:
If (x + 6y) = 8, and xy = 2, where x > 0, what is the value of (x 3+ 216y 3)?
If 8k 6+ 15k 3– 2 = 0, then the positive value of \(\left( {{\rm{k}}\,{\rm{ + }}\,\frac{1}{{\rm{k}}}} \right)\) is :
