If `(x(y+z-x))/(logx)=(y(z+x-y))/(logy)=(z(x+y-z))/(logz)` then `x^(y).y^(x)=z^(y).y^(z)=x^(z).z^(x)`

Answer

To solve the problem, we need to prove that if

\[ \frac{x(y+z-x)}{\log x} = \frac{y(z+x-y)}{\log y} = \frac{z(x+y-z)}{\log z} \]

then

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

### Step 1: Set the common value
Let

\[ \frac{x(y+z-x)}{\log x} = \frac{y(z+x-y)}{\log y} = \frac{z(x+y-z)}{\log z} = \frac{1}{t} \]

for some \( t \).

### Step 2: Express logs in terms of \( t \)
From the first equation, we have:

\[ \log x = t \cdot x(y + z - x). \]

From the second equation:

\[ \log y = t \cdot y(z + x - y). \]

From the third equation:

\[ \log z = t \cdot z(x + y - z). \]

### Step 3: Use the logarithmic properties
We will use the properties of logarithms to express \( x^y y^x \), \( z^y y^z \), and \( x^z z^x \).

1. For \( x^y y^x \):

\[ \log(x^y y^x) = y \log x + x \log y. \]

Substituting the values from Step 2:

\[ = y(t x (y + z - x)) + x(t y (z + x - y)). \]

### Step 4: Simplify the expression
Combine the terms:

\[ = t \left( yx(y + z - x) + xy(z + x - y) \right). \]

### Step 5: Factor out common terms
Notice that both terms have \( xy \):

\[ = txy \left( (y + z - x) + (z + x - y) \right). \]

### Step 6: Simplify the expression further
Combine the terms inside the parentheses:

\[ = txy \left( 2z \right). \]

Thus,

\[ \log(x^y y^x) = 2xyz t. \]

### Step 7: Repeat for \( z^y y^z \) and \( x^z z^x \)
Following similar steps for \( z^y y^z \) and \( x^z z^x \):

- For \( z^y y^z \):

\[ \log(z^y y^z) = y \log z + z \log y = 2xyz t. \]

- For \( x^z z^x \):

\[ \log(x^z z^x) = z \log x + x \log z = 2xyz t. \]

### Step 8: Conclude the proof
Since all three expressions equal \( e^{2xyz t} \), we conclude that:

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

Thus, we have proved the required statement.