Question

Prove Corollary \(4.8\) by showing that for any \(x, y\), and \(z\), each greater than \(1, x^{\log _{y} z}=z^{\log _{y} x}\).

Short answer

We proved that \(x^{\log_{y} z} = z^{\log_{y} x}\) by simplifying both expressions to a common form, utilizing the change of base formula and logarithmic properties.

Step-by-step solution

Understand the Problem

We are given the expression \(x^{\log_{y} z}\) and we need to prove that it is equal to \(z^{\log_{y} x}\). Recall that \(\log_{y} z\) means the logarithm of \(z\) to the base \(y\), and similarly for \(\log_{y} x\).

Apply the Change of Base Formula

By applying the change of base formula, we can express \(\log_{y} z\) as \(\frac{\log_{b} z}{\log_{b} y}\) and \(\log_{y} x\) as \(\frac{\log_{b} x}{\log_{b} y}\), where \(b\) is any positive number. For simplicity, we usually take \(b=e\) or \(b=10\).

Rewrite the Expression Using the Change of Base

We rewrite \(x^{\log_{y} z} = x^{\frac{\log_{b} z}{\log_{b} y}}\) and \(z^{\log_{y} x} = z^{\frac{\log_{b} x}{\log_{b} y}}\).

Simplify the Expression

Utilize the exponent rule \(a^{\frac{m}{n}} = (a^m)^{\frac{1}{n}}\). Simplify both sides: both \(x^{\frac{\log_{b} z}{\log_{b} y}}\) and \(z^{\frac{\log_{b} x}{\log_{b} y}}\) simplify to \(\exp\left(\frac{\log_{b} z \cdot \log_{b} x}{\log_{b} y}\right)\) by properties of exponents and logarithms.

Conclude Equality

Because both expressions simplify to the same form, we conclude \(x^{\log_{y} z} = z^{\log_{y} x}\). Thus, we have proven the corollary using the properties of logarithms and exponents.

Key concepts

Logarithmic Identities

Logarithmic identities are essential tools in mathematics, especially when dealing with exponent and logarithm operations. These identities help simplify and solve complex expressions involving logs. A fundamental property of logarithms is their ability to translate multiplication into addition, as described by the product rule:

• Product rule: \(\log_b(m \cdot n) = \log_b(m) + \log_b(n) \)
• Quotient rule: \(\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n) \)
• Power rule: \(\log_b(m^n) = n \cdot \log_b(m) \)

These identities allow us to break down and manage expressions in a manageable form. The exercise relies on such identities to showcase the inherent relationships between logarithmic and exponential forms.

Change of Base Formula

The change of base formula is a vital technique for manipulating logarithms, enabling conversion between different logarithmic bases. This is especially useful when the base required is not readily available on calculators or when specific base conversions are needed in theoretical proofs. The formula itself is:\[\log_b(a) = \frac{\log_k(a)}{\log_k(b)}\]Here, you can choose any base \(k\) that suits your needs. Common choices for \(k\) are either the natural logarithm, denoted \(e\), or the common logarithm, base 10. The change of base formula plays a pivotal role in the solution by allowing us to express complex exponential forms in simpler ratios, as seen in the exercise where both \(x^{\log_y z}\) and \(z^{\log_y x}\) are rewritten using this formula for ease of simplification.

Exponential and Logarithmic Relationships

In mathematics, exponential and logarithmic functions have an intertwined relationship. An exponential function in one form can be translated into a logarithmic form and vice versa. This connection is fundamental in solving equations where these operations are present. Consider that:

• A number raised to a logarithm exponent, such as \(a^{\log_b(c)}\), can be imagined as asking "to what power must \(b\) be raised, to achieve \(c\)?" but reformulated with \(a\) as an intermediary.
• Natural exponential properties, with expressions like \(e^{\ln(x)} = x\), underpin key transformations in expression proving.

The problem exemplifies these relationships by transforming and equating expressions into a common simplified form, using logarithms and exponent rules. The identity in the exercise \(x^{\log_{y} z} = z^{\log_{y} x}\) highlights this silky transition between different forms and exemplifies the strength of paired logarithmic-exponential manipulation.