Understanding the Logarithmic Equation

We are given a complex equation involving exponents with natural logarithms:

\( {{\rm{x}}^{{\rm{In}}\left( {\frac{{\rm{y}}}{{\rm{z}}}} \right)}} \cdot {{\rm{y}}^{{\rm{In}}{{\left( {{\rm{xz}} \right)}^2}}} \cdot {{\rm{z}}^{{\rm{In}}\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right)}} = {{\rm{y}}^{4{\rm{\;In\;y}}}} \)

The equation holds for any x > 1, y > 1, and z > 1. Our goal is to find the relationship between \( \ln x \), \( \ln y \), and \( \ln z \). Since x, y, and z are greater than 1, their natural logarithms \( \ln x \), \( \ln y \), and \( \ln z \) are all positive.

Solving the Equation Using Logarithm Properties

To simplify the equation, we can take the natural logarithm of both sides. Remember the logarithm property \( \ln(a \cdot b) = \ln a + \ln b \) and \( \ln(a^p) = p \ln a \).

Taking the natural logarithm of both sides gives:

\( \ln \left( {{\rm{x}}^{{\rm{In}}\left( {\frac{{\rm{y}}}{{\rm{z}}}} \right)}} \cdot {{\rm{y}}^{{\rm{In}}{{\left( {{\rm{xz}} \right)}^2}}} \cdot {{\rm{z}}^{{\rm{In}}\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right)}}} \right) = \ln \left( {{\rm{y}}^{4{\rm{\;In\;y}}}} \right) \)

Using the product rule of logarithms on the left side:

\( \ln\left({\rm{x}}^{{\rm{In}}\left( {\frac{{\rm{y}}}{z}} \right)}\right) + \ln\left({{\rm{y}}^{{\rm{In}}{{\left( {{\rm{xz}} \right)}^2}}}}\right) + \ln\left({{\rm{z}}^{{\rm{In}}\left( {\frac{{\rm{x}}}{y}} \right)}}\right) = \ln\left({{\rm{y}}^{4{\rm{\;In\;y}}}}\right) \)

Now, applying the power rule \( \ln(a^p) = p \ln a \) to each term. Note that the term \( {{\rm{y}}^{{\rm{In}}{{\left( {{\rm{xz}} \right)}^2}}}} \) implies the exponent is \( {\left(\ln(xz)\right)}^2 \) based on the notation \( {\rm{In}}{{\left( {{\rm{xz}} \right)}^2}} \). However, to arrive at one of the given options, the exponent must be interpreted as \( \ln((xz)^2) = 2 \ln(xz) \). We will proceed with this interpretation.

\( \ln\left( {\frac{{\rm{y}}}{{\rm{z}}}} \right) \ln {\rm{x}} + \ln({{\left( {{\rm{xz}} \right)}^2}}) \ln {\rm{y}} + \ln\left( {\frac{{\rm{x}}}{{\rm{y}}}} \right) \ln {\rm{z}} = 4{\rm{\;In\;y}} \ln {\rm{y}} \)

Apply \( \ln(a/b) = \ln a - \ln b \) and \( \ln(ab) = \ln a + \ln b \), and \( \ln(a^2) = 2 \ln a \):

\( (\ln {\rm{y}} - \ln {\rm{z}}) \ln {\rm{x}} + 2(\ln {\rm{x}} + \ln {\rm{z}}) \ln {\rm{y}} + (\ln {\rm{x}} - \ln {\rm{y}}) \ln {\rm{z}} = 4 (\ln {\rm{y}})^2 \)

Expand the terms:

\( \ln {\rm{x}} \ln {\rm{y}} - \ln {\rm{x}} \ln {\rm{z}} + 2 \ln {\rm{x}} \ln {\rm{y}} + 2 \ln {\rm{z}} \ln {\rm{y}} + \ln {\rm{x}} \ln {\rm{z}} - \ln {\rm{y}} \ln {\rm{z}} = 4 (\ln {\rm{y}})^2 \)

Notice that \( -\ln {\rm{x}} \ln {\rm{z}} \) and \( +\ln {\rm{x}} \ln {\rm{z}} \) cancel each other out. The equation simplifies to:

\( \ln {\rm{x}} \ln {\rm{y}} + 2 \ln {\rm{x}} \ln {\rm{y}} + 2 \ln {\rm{z}} \ln {\rm{y}} - \ln {\rm{y}} \ln {\rm{z}} = 4 (\ln {\rm{y}})^2 \)

Combine the terms on the left side that contain \( \ln {\rm{y}} \):

\( ( \ln {\rm{x}} + 2 \ln {\rm{x}} + 2 \ln {\rm{z}} - \ln {\rm{z}} ) \ln {\rm{y}} = 4 (\ln {\rm{y}})^2 \)

\( ( 3 \ln {\rm{x}} + \ln {\rm{z}} ) \ln {\rm{y}} = 4 (\ln {\rm{y}})^2 \)

Since y > 1, \( \ln {\rm{y}} \ne 0 \). We can divide both sides by \( \ln {\rm{y}} \):

\( 3 \ln {\rm{x}} + \ln {\rm{z}} = 4 \ln {\rm{y}} \)

Rearranging the equation to isolate \( \ln {\rm{y}} \):

\( \ln {\rm{y}} = \frac{3 \ln {\rm{x}} + \ln {\rm{z}}}{4} \)

Relating the Result to Arithmetic Mean

The arithmetic mean (AM) of a set of numbers is the sum of the numbers divided by the count of the numbers. For example, the AM of a, b, c, and d is \( \frac{a+b+c+d}{4} \).

The equation we derived is \( \ln {\rm{y}} = \frac{3 \ln {\rm{x}} + \ln {\rm{z}}}{4} \). This can be written as:

\( \ln {\rm{y}} = \frac{\ln {\rm{x}} + \ln {\rm{x}} + \ln {\rm{x}} + \ln {\rm{z}}}{4} \)

This clearly shows that \( \ln {\rm{y}} \) is the arithmetic mean of the four terms: \( \ln {\rm{x}} \), \( \ln {\rm{x}} \), \( \ln {\rm{x}} \), and \( \ln {\rm{z}} \).

Conclusion

Based on the step-by-step simplification of the given logarithmic equation, we found that \( \ln {\rm{y}} = \frac{3 \ln {\rm{x}} + \ln {\rm{z}}}{4} \). This relationship corresponds exactly to the definition of the arithmetic mean of the terms \( \ln {\rm{x}}, \ln {\rm{x}}, \ln {\rm{x}}, \) and \( \ln {\rm{z}} \).

Therefore, the correct statement is that \( \ln {\rm{y}} \) is the AM of \( \ln {\rm{x}}, \ln {\rm{x}}, \ln {\rm{x}}, \) and \( \ln {\rm{z}} \).

Revision Table: Key Concepts Reviewed

This problem involved applying several fundamental concepts:

• Properties of exponents in equations.
• Properties of natural logarithms (ln):
  • \( \ln(ab) = \ln a + \ln b \)
  • \( \ln(a/b) = \ln a - \ln b \)
  • \( \ln(a^p) = p \ln a \)
• Solving algebraic equations involving logarithmic terms.
• Definition of Arithmetic Mean (AM).

Additional Information: Means and Logarithms

Besides the arithmetic mean (AM), other common types of means are the Geometric Mean (GM) and Harmonic Mean (HM). These means have interesting properties and relationships, especially in the context of logarithms.

• Arithmetic Mean (AM): For positive numbers \( a_1, a_2, \ldots, a_n \), AM \( = \frac{a_1 + a_2 + \ldots + a_n}{n} \).
• Geometric Mean (GM): For positive numbers \( a_1, a_2, \ldots, a_n \), GM \( = \sqrt[n]{a_1 \cdot a_2 \cdot \ldots \cdot a_n} \). Taking the natural logarithm of the GM gives \( \ln(\text{GM}) = \ln\left(\sqrt[n]{a_1 \cdot \ldots \cdot a_n}\right) = \frac{1}{n} \ln(a_1 \cdot \ldots \cdot a_n) = \frac{\ln a_1 + \ldots + \ln a_n}{n} \). This shows that the logarithm of the geometric mean is the arithmetic mean of the logarithms.
• Harmonic Mean (HM): For positive numbers \( a_1, a_2, \ldots, a_n \), HM \( = \frac{n}{\frac{1}{a_1} + \frac{1}{a_2} + \ldots + \frac{1}{a_n}} \). The reciprocal of the harmonic mean is the arithmetic mean of the reciprocals.

In this problem, we found that \( \ln y \) is the AM of \( \ln x, \ln x, \ln x, \ln z \). This is a direct AM relationship involving the logarithms themselves, not the logarithm of a geometric mean.