Problem 383 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Find the positive solutions of the system of equations } x^{x+y}=y^{n}, y^{x+y}=x^{2 n} y^{n}, \text { where } n>0 \text { . } $$

Step-by-Step Solution

Answer

The positive solutions of this system of equations are \(x = 1\) and \(y = 1\).

Step 1: Convert the equations to logarithmic form

We should convert the equations \(x^{x+y}=y^{n}\) and \(y^{x+y}=x^{2n} y^{n}\) to logarithmic form using logarithmic properties. The first equation becomes \((x+y) \log(x) = n \log(y)\), and the second equation is converted to \((x+y) \log(y) = (2n+n) \log(x) + n \log(y)\).

Step 2: Simplify the long equations

Next, we must simplify the obtained equations. Subtract \(n \log(y)\) from both sides of the first equation \((x+y) \log(x) - n \log(y) = 0\). Then, from the second equation subtract \((x+y) \log(y)\) to get \(2n \log(x) = 0\).

Step 3: Observe the equations and derive solutions

Look at the simplified equations and derive solutions for \(x\) and \(y\). If \(2n \log(x) = 0\), then \(\log(x) = 0\) and therefore \(x = 1\). Then, with \(x=1\), the first equation simplifies as \((1+y) \log(1) - n \log(y) = 0\). Because \(\log(1) = 0\), the equation simplifies to \(- n \log(y) = 0\), which implies \(\log(y) = 0\), so \(y = 1\).

Key Concepts

Logarithmic Form

Starting with a system of exponential equations like in the given exercise, it is often necessary to convert the expressions into logarithmic form to solve them. Logarithmic form is preferred because it allows one to untangle the variables intertwined in the exponents, making them more manageable.

For example, an exponential equation of the form \(a^{b}=c\) can be transformed into its logarithmic counterpart \(b \log(a)=\log(c)\). In this exercise, we had \(x^{x+y}=y^{n}\) converting to \( (x+y) \log(x) = n \log(y)\), following the logarithmic principle that operations in the exponent can be brought down to multiply the logarithm of the base.

Logarithmic Properties

Understanding the properties of logarithms is crucial while handling equations involving them. These properties can include the product rule, quotient rule, power rule, and others, which help simplify the complex equations into a more solvable form.

Specifically for simplifying equations, the product rule states that \(\log(a) + \log(b) = \log(ab)\), and the power rule says that \(n \log(a) = \log(a^n)\). In the context of our problem, properties like \(\log(1) = 0\) and that coefficients can be brought in as powers (like \(2n \log(x) = \log(x^{2n})\)) are used to simplify the equations to a point where the variables can be isolated and solved.

Simplifying Equations

When simplifying equations, the goal is to reduce the complexity while maintaining equivalency, making the solution process less daunting. This involves various algebraic manipulations such as adding, subtracting, factorizing or expanding expressions, and using logarithmic properties.

In this exercise, after transforming to logarithmic form, further simplification was required. Terms were grouped, and like terms were eliminated by subtraction. This process further reduced the equations from their logarithmic form to a simple linear form, allowing us to deduce the value of variables. Such simplification is a key step as it reveals the core relationship between the variables that must be true for the system to hold, eventually leading to the solution of \(x = 1\) and \(y = 1\).