Understanding the Problem: Logarithms, GP, and AP

The question provides us with three conditions involving three variables, x, y, and z:

1. The terms \( \log_x y \), \( \log_z x \), and \( \log_y z \) are in a Geometric Progression (GP).
2. The product of the variables is \( xyz = 64 \).
3. The terms \( x^3 \), \( y^3 \), and \( z^3 \) are in an Arithmetic Progression (AP).

We need to determine the relationship between the terms \( xy \), \( yz \), and \( zx \): are they in AP, GP, both, or neither?

Analyzing the Geometric Progression Condition

If three terms \( a, b, c \) are in GP, then the square of the middle term is equal to the product of the other two terms, i.e., \( b^2 = ac \).

Applying this to the given terms \( \log_x y, \log_z x, \log_y z \):

$$ (\log_z x)^2 = (\log_x y)(\log_y z) $$

We can use the change of base formula for logarithms, which states that \( \log_b a = \frac{\log_c a}{\log_c b} \). Let's use the natural logarithm (\( \ln \)) as the base for convenience.

$$ \left(\frac{\ln x}{\ln z}\right)^2 = \left(\frac{\ln y}{\ln x}\right)\left(\frac{\ln z}{\ln y}\right) $$

Assuming \( \ln y \neq 0 \) (which implies \( y \neq 1 \)), we can cancel \( \ln y \) from the right side:

$$ \left(\frac{\ln x}{\ln z}\right)^2 = \frac{\ln z}{\ln x} $$

Now, multiply both sides by \( (\ln z)^2 \cdot (\ln x) \):

$$ (\ln x)^2 \cdot (\ln x) = (\ln z) \cdot (\ln z)^2 $$

$$ (\ln x)^3 = (\ln z)^3 $$

Taking the cube root of both sides gives:

$$ \ln x = \ln z $$

Since the natural logarithm function is one-to-one, this implies:

$$ x = z $$

This is a crucial finding. The GP condition implies that \( x \) and \( z \) must be equal.

Analyzing the Arithmetic Progression Condition

If three terms \( a, b, c \) are in AP, then twice the middle term is equal to the sum of the other two terms, i.e., \( 2b = a + c \).

Applying this to the given terms \( x^3, y^3, z^3 \):

$$ 2y^3 = x^3 + z^3 $$

We already found that \( x = z \). Substituting \( z = x \) into the AP condition:

$$ 2y^3 = x^3 + x^3 $$

$$ 2y^3 = 2x^3 $$

Dividing both sides by 2:

$$ y^3 = x^3 $$

Assuming \( x, y, z \) are positive (as they are bases/arguments of logarithms), we can take the cube root:

$$ y = x $$

This implies that \( y \) must be equal to \( x \).

Combining the Conditions and Finding the Values of x, y, z

From the GP condition, we got \( x = z \). From the AP condition, we got \( y = x \). Combining these, we conclude that:

$$ x = y = z $$

Now, we use the third given condition, \( xyz = 64 \).

Substitute \( x = y = z \) into this equation:

$$ x \cdot x \cdot x = 64 $$

$$ x^3 = 64 $$

Taking the cube root of both sides:

$$ x = \sqrt[3]{64} $$

$$ x = 4 $$

Since \( x = y = z \), we have:

$$ x = 4, y = 4, z = 4 $$

Determining the Nature of xy, yz, and zx

Now we need to find the values of \( xy \), \( yz \), and \( zx \) using \( x=4, y=4, z=4 \).

• \( xy = 4 \times 4 = 16 \)
• \( yz = 4 \times 4 = 16 \)
• \( zx = 4 \times 4 = 16 \)

The sequence of terms is \( 16, 16, 16 \).

Checking if 16, 16, 16 is in AP

A sequence \( a, b, c \) is in AP if \( b - a = c - b \). Here, \( a=16, b=16, c=16 \).

• Difference between the second and first term: \( 16 - 16 = 0 \)
• Difference between the third and second term: \( 16 - 16 = 0 \)

Since the common difference is 0, the terms \( 16, 16, 16 \) are in AP.

Checking if 16, 16, 16 is in GP

A sequence \( a, b, c \) is in GP if \( \frac{b}{a} = \frac{c}{b} \). Here, \( a=16, b=16, c=16 \).

• Ratio of the second term to the first term: \( \frac{16}{16} = 1 \)
• Ratio of the third term to the second term: \( \frac{16}{16} = 1 \)

Since the common ratio is 1, the terms \( 16, 16, 16 \) are in GP.

Since the sequence \( 16, 16, 16 \) satisfies both the condition for AP (common difference 0) and the condition for GP (common ratio 1), the terms \( xy, yz, zx \) are in both AP and GP.

Conclusion

Based on the analysis of the given conditions, we found that \( x=y=z=4 \). This leads to \( xy=16 \), \( yz=16 \), and \( zx=16 \). The sequence \( 16, 16, 16 \) is confirmed to be in both Arithmetic Progression and Geometric Progression.

Therefore, \( x=y=z=4 \).

Revision Table: Key Concepts in Progressions

Additional Information: Special Cases of AP and GP

A sequence consisting of identical terms, like \( k, k, k, \ldots \), represents a special case that is simultaneously an AP and a GP.

• It is an AP with a common difference of 0.
• It is a GP with a common ratio of 1 (provided \( k \neq 0 \)). If \( k=0 \), the definition of GP requires careful consideration, but in typical problems involving positive numbers like log bases, this is not an issue.

In this problem, since \( x, y, z \) are bases of logarithms, they must be positive and not equal to 1. Our solution \( x=y=z=4 \) satisfies these conditions, and the resulting sequence \( 16, 16, 16 \) is indeed a constant sequence (non-zero), which is both AP and GP.