If `x^(18)=y^(21)=z^(28)` then `3,3log_(y)x,3log_(z)y,7log_(x)z` are in

Options

A. A.P.
B. G.P.
C. H.P.
D. None

Answer

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

1. \( x^{18} = y^{21} = z^{28} = k \) (where \( k \) is a constant)

From this, we can express \( x \), \( y \), and \( z \) in terms of \( k \):

\[ x = k^{\frac{1}{18}}, \quad y = k^{\frac{1}{21}}, \quad z = k^{\frac{1}{28}} \]

Next, we need to find the logarithms:

### Step 1: Calculate \( \log_y x \)

Using the change of base formula, we have:

\[ \log_y x = \frac{\log x}{\log y} \]

Substituting the expressions for \( x \) and \( y \):

\[ \log_y x = \frac{\log(k^{\frac{1}{18}})}{\log(k^{\frac{1}{21}})} = \frac{\frac{1}{18} \log k}{\frac{1}{21} \log k} = \frac{21}{18} = \frac{7}{6} \]

### Step 2: Calculate \( \log_z y \)

Similarly, we find:

\[ \log_z y = \frac{\log y}{\log z} \]

Substituting the expressions for \( y \) and \( z \):

\[ \log_z y = \frac{\log(k^{\frac{1}{21}})}{\log(k^{\frac{1}{28}})} = \frac{\frac{1}{21} \log k}{\frac{1}{28} \log k} = \frac{28}{21} = \frac{4}{3} \]

### Step 3: Calculate \( \log_x z \)

Now we calculate:

\[ \log_x z = \frac{\log z}{\log x} \]

Substituting the expressions for \( z \) and \( x \):

\[ \log_x z = \frac{\log(k^{\frac{1}{28}})}{\log(k^{\frac{1}{18}})} = \frac{\frac{1}{28} \log k}{\frac{1}{18} \log k} = \frac{18}{28} = \frac{9}{14} \]

### Step 4: Formulate the expressions

Now we can express the values we need to check:

1. \( 3 \)
2. \( 3 \log_y x = 3 \cdot \frac{7}{6} = \frac{21}{6} = \frac{7}{2} \)
3. \( 3 \log_z y = 3 \cdot \frac{4}{3} = 4 \)
4. \( 7 \log_x z = 7 \cdot \frac{9}{14} = \frac{63}{14} = \frac{9}{2} \)

### Step 5: Check if they are in Arithmetic Progression (AP)

To check if the four numbers \( 3, \frac{7}{2}, 4, \frac{9}{2} \) are in AP, we can check if:

\[ 2 \cdot \frac{7}{2} = 3 + 4 \]

Calculating:

\[ 2 \cdot \frac{7}{2} = 7 \quad \text{and} \quad 3 + 4 = 7 \]

Next, we check:

\[ 2 \cdot 4 = \frac{7}{2} + \frac{9}{2} \]

Calculating:

\[ 2 \cdot 4 = 8 \quad \text{and} \quad \frac{7}{2} + \frac{9}{2} = \frac{16}{2} = 8 \]

Both conditions are satisfied, hence:

### Conclusion

The numbers \( 3, 3 \log_y x, 3 \log_z y, 7 \log_x z \) are in Arithmetic Progression.