Find the value of `log_y"x"xxlog_zyxxlog_xz`

Options

A. 0
B. 1
C. logx
D. xyz

Answer

To find the value of \( \log_y x \cdot \log_z y \cdot \log_x z \), we can use the change of base formula for logarithms. The change of base formula states that:

\[ \log_a b = \frac{1}{\log_b a} \]

Let's solve the problem step by step.

### Step 1: Rewrite the logarithms using the change of base formula

We can express each logarithm in terms of a common base. Let's choose base \( z \):

\[ \log_y x = \frac{\log_z x}{\log_z y} \]
\[ \log_z y = \log_z y \quad (\text{remains the same}) \]
\[ \log_x z = \frac{\log_z z}{\log_z x} = \frac{1}{\log_z x} \quad (\text{since } \log_z z = 1) \]

### Step 2: Substitute these into the original expression

Now substituting these into the original expression:

\[ \log_y x \cdot \log_z y \cdot \log_x z = \left(\frac{\log_z x}{\log_z y}\right) \cdot \log_z y \cdot \left(\frac{1}{\log_z x}\right) \]

### Step 3: Simplify the expression

Now, we can simplify the expression:

\[ = \frac{\log_z x}{\log_z y} \cdot \log_z y \cdot \frac{1}{\log_z x} \]

Notice that \( \log_z x \) in the numerator and denominator cancels out, and \( \log_z y \) also cancels out:

\[ = 1 \]

### Final Answer

Thus, the value of \( \log_y x \cdot \log_z y \cdot \log_x z \) is \( 1 \).

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