If `"log"_(y) x = "log"_(z)y = "log"_(x)z`, then

Options

A. `x lt y lt z`
B. `x gt y ge z`
C. `x lt y le z`
D. `x = y =z`

Answer

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

\[ \log_y x = \log_z y = \log_x z \]

Let's denote this common value as \( k \). Therefore, we can write:

1. \(\log_y x = k\)
2. \(\log_z y = k\)
3. \(\log_x z = k\)

From the properties of logarithms, we can rewrite these equations in exponential form:

1. From \(\log_y x = k\), we have:
\[ x = y^k \tag{1} \]

2. From \(\log_z y = k\), we have:
\[ y = z^k \tag{2} \]

3. From \(\log_x z = k\), we have:
\[ z = x^k \tag{3} \]

Now, we can substitute the expressions from equations (1) and (2) into equation (3).

Substituting equation (2) into equation (1):
\[ x = (z^k)^k = z^{k^2} \tag{4} \]

Now substituting equation (4) into equation (3):
\[ z = (z^{k^2})^k = z^{k^3} \]

For this equation to hold true, we can analyze two cases:

1. If \( z \neq 0 \), we can divide both sides by \( z \) (assuming \( z \neq 1 \)):
\[ 1 = z^{k^3 - 1} \]
This implies that \( k^3 - 1 = 0 \) or \( z = 1 \).

Therefore, \( k^3 = 1 \) leads to \( k = 1 \).

2. If \( k = 1 \), we can substitute back into our earlier equations:
- From (1): \( x = y^1 \) implies \( x = y \)
- From (2): \( y = z^1 \) implies \( y = z \)
- From (3): \( z = x^1 \) implies \( z = x \)

Thus, we have:
\[ x = y = z \]

This means that all three variables are equal.

Now, we can check the options provided in the question:

1. \( x < y < z \)
2. \( x > y \geq z \)
3. \( x < y \leq z \)
4. \( x = y = z \)

The only option that matches our conclusion is option 4.

**Final Answer:**
\[ x = y = z \]

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