Solving Logarithm Equation: Finding the Relation Between x, y, and z

The problem provides us with three variables, x, y, and z, defined in terms of logarithms with different bases:

• \( x = \log_{c}(ab) \)
• \( y = \log_{a}(bc) \)
• \( z = \log_{b}(ca) \)

We need to determine which of the given options represents the correct relationship between x, y, and z.

Calculating (1 + x), (1 + y), and (1 + z)

Let's examine the form of the options, particularly the one involving inverse terms like \( (1+x)^{-1} \). It is often helpful in such problems to calculate the terms (1 + x), (1 + y), and (1 + z) first.

Recall the logarithm property: \( \log_m(m) = 1 \). We can use this to express the number '1' with the same base as the given logarithm.

Calculating (1 + x)

For x = \( \log_{c}(ab) \):

\( 1 + x = 1 + \log_{c}(ab) \)

We can rewrite 1 as \( \log_{c}(c) \). So,

\( 1 + x = \log_{c}(c) + \log_{c}(ab) \)

Using the logarithm property \( \log_m(M) + \log_m(N) = \log_m(MN) \), we combine the terms:

\( 1 + x = \log_{c}(c \cdot ab) \)

\( 1 + x = \log_{c}(abc) \)

Calculating (1 + y)

For y = \( \log_{a}(bc) \):

\( 1 + y = 1 + \log_{a}(bc) \)

Rewrite 1 as \( \log_{a}(a) \):

\( 1 + y = \log_{a}(a) + \log_{a}(bc) \)

Combine using the logarithm property:

\( 1 + y = \log_{a}(a \cdot bc) \)

\( 1 + y = \log_{a}(abc) \)

Calculating (1 + z)

For z = \( \log_{b}(ca) \):

\( 1 + z = 1 + \log_{b}(ca) \)

Rewrite 1 as \( \log_{b}(b) \):

\( 1 + z = \log_{b}(b) + \log_{b}(ca) \)

Combine using the logarithm property:

\( 1 + z = \log_{b}(b \cdot ca) \)

\( 1 + z = \log_{b}(abc) \)

Calculating the Inverse Terms: (1 + x)^-1, (1 + y)^-1, (1 + z)^-1

Now, let's find the inverse of each of these terms:

• \( (1 + x)^{-1} = \frac{1}{1 + x} = \frac{1}{\log_{c}(abc)} \)
• \( (1 + y)^{-1} = \frac{1}{1 + y} = \frac{1}{\log_{a}(abc)} \)
• \( (1 + z)^{-1} = \frac{1}{1 + z} = \frac{1}{\log_{b}(abc)} \)

Recall the change of base formula property: \( \frac{1}{\log_m(n)} = \log_n(m) \). Applying this property to each term:

• \( (1 + x)^{-1} = \frac{1}{\log_{c}(abc)} = \log_{abc}(c) \)
• \( (1 + y)^{-1} = \frac{1}{\log_{a}(abc)} = \log_{abc}(a) \)
• \( (1 + z)^{-1} = \frac{1}{\log_{b}(abc)} = \log_{abc}(b) \)

Summing the Inverse Terms

Let's find the sum of these inverse terms:

\( (1 + x)^{-1} + (1 + y)^{-1} + (1 + z)^{-1} = \log_{abc}(c) + \log_{abc}(a) + \log_{abc}(b) \)

Using the logarithm property \( \log_m(M) + \log_m(N) = \log_m(MN) \) again, we can combine the terms on the right side, as they all have the same base (abc):

\( \log_{abc}(c) + \log_{abc}(a) + \log_{abc}(b) = \log_{abc}(c \cdot a \cdot b) \)

\( = \log_{abc}(abc) \)

Finally, using the property \( \log_m(m) = 1 \), we get:

\( \log_{abc}(abc) = 1 \)

Thus, we have found that:

\( (1 + x)^{-1} + (1 + y)^{-1} + (1 + z)^{-1} = 1 \)

This matches one of the provided options.

Conclusion

Based on our calculations using logarithm properties, the correct relationship between x, y, and z is \( (1 + x)^{-1} + (1 + y)^{-1} + (1 + z)^{-1} = 1 \).

Revision Table: Key Logarithm Properties Used

Additional Information: Understanding Logarithms

A logarithm is the exponent to which a base must be raised to produce a given number. For example, \( \log_{10}(100) = 2 \) because \( 10^2 = 100 \). In the expression \( \log_b(n) = x \), b is the base, n is the number (or argument), and x is the logarithm (or exponent).

Logarithms are powerful tools used to simplify complex calculations involving multiplication, division, and exponentiation by converting them into addition, subtraction, and multiplication, respectively.

The properties used in this problem are fundamental to manipulating logarithmic expressions and solving equations involving logarithms. Understanding how to change the base and how the product rule works is crucial for simplifying expressions like the ones for x, y, and z.