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

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

To solve the problem, we need to determine whether the values \(3\), \(3 \log_{y} x\), \(3 \log_{z} y\), and \(7 \log_{x} z\) are in Arithmetic Progression (AP), Geometric Progression (GP), or Harmonic Progression (HP).

Given that \(x^{18} = y^{21} = z^{28}\), we can start by taking logarithms of all sides.

### Step 1: Take logarithms
We can express the equality in logarithmic form:
\[ \log(x^{18}) = \log(y^{21}) = \log(z^{28}) \]

Using the property of logarithms, \( \log(a^b) = b \log(a) \), we can rewrite this as:
\[ 18 \log x = 21 \log y = 28 \log z \]

### Step 2: Set equal to a common variable
Let \( k = 18 \log x = 21 \log y = 28 \log z \). We can express \(\log x\), \(\log y\), and \(\log z\) in terms of \(k\):
\[ \log x = \frac{k}{18}, \quad \log y = \frac{k}{21}, \quad \log z = \frac{k}{28} \]

### Step 3: Calculate \(3 \log_{y} x\)
Using the change of base formula for logarithms:
\[ \log_{y} x = \frac{\log x}{\log y} = \frac{\frac{k}{18}}{\frac{k}{21}} = \frac{21}{18} = \frac{7}{6} \]
Thus,
\[ 3 \log_{y} x = 3 \cdot \frac{7}{6} = \frac{21}{6} = \frac{7}{2} \]

### Step 4: Calculate \(3 \log_{z} y\)
Similarly,
\[ \log_{z} y = \frac{\log y}{\log z} = \frac{\frac{k}{21}}{\frac{k}{28}} = \frac{28}{21} = \frac{4}{3} \]
Thus,
\[ 3 \log_{z} y = 3 \cdot \frac{4}{3} = 4 \]

### Step 5: Calculate \(7 \log_{x} z\)
Now,
\[ \log_{x} z = \frac{\log z}{\log x} = \frac{\frac{k}{28}}{\frac{k}{18}} = \frac{18}{28} = \frac{9}{14} \]
Thus,
\[ 7 \log_{x} z = 7 \cdot \frac{9}{14} = \frac{63}{14} = \frac{9}{2} \]

### Step 6: List the terms
Now we have:
- \(a = 3\)
- \(b = \frac{7}{2}\)
- \(c = 4\)
- \(d = \frac{9}{2}\)

### Step 7: Check for Arithmetic Progression
To check if these values are in AP, we need to verify if:
\[ 2b = a + c \quad \text{and} \quad 2c = b + d \]

Calculating:
1. \(2b = 2 \cdot \frac{7}{2} = 7\) and \(a + c = 3 + 4 = 7\) (True)
2. \(2c = 2 \cdot 4 = 8\) and \(b + d = \frac{7}{2} + \frac{9}{2} = \frac{16}{2} = 8\) (True)

Since both conditions are satisfied, we conclude that:
\[ 3, 3 \log_{y} x, 3 \log_{z} y, 7 \log_{x} z \text{ are in Arithmetic Progression (AP)}. \]

### Final Answer:
The values \(3\), \(3 \log_{y} x\), \(3 \log_{z} y\), and \(7 \log_{x} z\) are in AP.

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