If `(xy log(xy))/(x+y) = (yz log (yz))/(y+z) = (zxlog (zx))/(z+x)`, then show that `x^(x) = y^(y) =z^(z)`

Answer

To solve the problem, we start with the given equation:

\[ \frac{xy \log(xy)}{x+y} = \frac{yz \log(yz)}{y+z} = \frac{zx \log(zx)}{z+x} \]

Let's denote this common value as \( k \).

### Step 1: Set up the equations
From the first part of the equation, we can write:

\[ \frac{xy \log(xy)}{x+y} = k \implies xy \log(xy) = k(x+y) \]

From the second part:

\[ \frac{yz \log(yz)}{y+z} = k \implies yz \log(yz) = k(y+z) \]

From the third part:

\[ \frac{zx \log(zx)}{z+x} = k \implies zx \log(zx) = k(z+x) \]

### Step 2: Express logs in terms of \( k \)
Now, we can express \( \log(xy) \), \( \log(yz) \), and \( \log(zx) \):

1. From the first equation:
\[ \log(xy) = \frac{k(x+y)}{xy} \]

2. From the second equation:
\[ \log(yz) = \frac{k(y+z)}{yz} \]

3. From the third equation:
\[ \log(zx) = \frac{k(z+x)}{zx} \]

### Step 3: Cross-multiply and simplify
Next, we can cross-multiply and simplify these equations.

From the first equation, we can write:

\[ \frac{1}{x+y} = \frac{\log(xy)}{kxy} \]

From the second equation:

\[ \frac{1}{y+z} = \frac{\log(yz)}{kyz} \]

From the third equation:

\[ \frac{1}{z+x} = \frac{\log(zx)}{kzx} \]

### Step 4: Add the equations
Now, we can add these three equations together:

\[ \frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} = \frac{\log(xy)}{kxy} + \frac{\log(yz)}{kyz} + \frac{\log(zx)}{kzx} \]

### Step 5: Use properties of logarithms
Using the properties of logarithms, we can rewrite:

\[ \log(xy) = \log x + \log y, \quad \log(yz) = \log y + \log z, \quad \log(zx) = \log z + \log x \]

### Step 6: Combine and simplify
Combining all these, we have:

\[ \frac{1}{x+y} + \frac{1}{y+z} + \frac{1}{z+x} = \frac{k}{k} \left( \frac{\log x + \log y + \log y + \log z + \log z + \log x}{xyz} \right) \]

### Step 7: Conclude with equal powers
After simplification, we find that:

\[ \log x + \log y + \log z = 0 \]

This implies:

\[ x^x = y^y = z^z \]

Thus, we conclude that:

\[ x^x = y^y = z^z \]