If `x^(y)= y^(z)= z^(x) and zx= y^(2)`, then which of the following is correct?

Options

A. `z=(2xy)/(x+y)`
B. `y= (x-z)/(x+z)`
C. `x= (y-z)/(yz)`
D. `xyz = (x-z+y)/(x+z-y)`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
\[ x^y = y^z = z^x \]
and
\[ zx = y^2 \]

2. **Assuming a Common Value**:
Let us assume that \( x^y = y^z = z^x = k \).

3. **Expressing x, y, and z in terms of k**:
- From \( x^y = k \), we can express \( x \) as:
\[ x = k^{\frac{1}{y}} \]
- From \( y^z = k \), we can express \( y \) as:
\[ y = k^{\frac{1}{z}} \]
- From \( z^x = k \), we can express \( z \) as:
\[ z = k^{\frac{1}{x}} \]

4. **Substituting into the second equation**:
Now we substitute \( z \) and \( x \) into the equation \( zx = y^2 \):
\[ (k^{\frac{1}{x}})(k^{\frac{1}{y}}) = (k^{\frac{1}{z}})^2 \]
This simplifies to:
\[ k^{\frac{1}{x} + \frac{1}{y}} = k^{\frac{2}{z}} \]

5. **Equating the exponents**:
Since the bases are the same, we can equate the exponents:
\[ \frac{1}{x} + \frac{1}{y} = \frac{2}{z} \]

6. **Finding a common expression**:
Rearranging gives us:
\[ \frac{1}{x} + \frac{1}{y} = \frac{2}{z} \]
This can be rewritten as:
\[ \frac{x + y}{xy} = \frac{2}{z} \]

7. **Cross-multiplying**:
Cross-multiplying gives:
\[ z(x + y) = 2xy \]

8. **Rearranging to find z**:
Thus, we can express \( z \) as:
\[ z = \frac{2xy}{x + y} \]

9. **Conclusion**:
Therefore, we have found that:
\[ z = \frac{2xy}{x + y} \]