Question

Show that \(\log _{a}\left(\frac{M}{N}\right)=\log _{a} M-\log _{a} N,\) where \(a, M,\) and \(N\) are positive real numbers and \(a \neq 1\).

Short answer

\(\log _{a}\left(\frac{M}{N}\right) = \log _{a} M - \log _{a} N\)

Step-by-step solution

Understanding the Properties of Logarithms

Recall that logarithms have specific properties. One of these is the quotient rule: \ For any positive real numbers and any positive base different from 1, the logarithm of a quotient is equal to the difference of the logarithms: \ \ \ \ \[ \log_{a} \left( \frac{M}{N} \right) = \log_{a} M - \log_{a} N \] \ \ In this exercise, we must prove that using the properties of logarithms.

Define the Logarithms

Let \( x = \log_{a} M \) and \( y = \log_{a} N \). This means that: \ \ \ \[ a^x = M \] \ and \ \ \[ a^y = N \] \ \ Our goal is now to express \( \log_{a} \left( \frac{M}{N} \right) \) in terms of \( x \) and \( y \).

Express the Quotient Inside the Logarithm

Using the definitions we have: \ \[ \frac{M}{N} = \frac{a^x}{a^y} \] \ This can be simplified further by applying the properties of exponents: \ \[ \frac{a^x}{a^y} = a^{x-y} \]

Apply the Logarithm to the Simplified Expression

Take the logarithm base \(a\) of both sides: \ \[ \log_{a} \left( \frac{M}{N} \right) = \log_{a} \left( a^{x-y} \right) \] \ Using the logarithm power rule, which states \( \log_{a}(a^k) = k \), we have: \ \[ \log_{a} \left( a^{x-y} \right) = x - y \] \ Substitute back \( \log_{a} M \) for \( x \) and \( \log_{a} N \) for \( y \): \ \[ x - y = \log_{a} M - \log_{a} N \]

Conclude the Proof

Therefore, we have shown that: \ \[ \log_{a} \left( \frac{M}{N} \right) = \log_{a} M - \log_{a} N \] \ \ This completes our proof.

Key concepts

logarithm rules

Logarithm rules are a set of properties that make it easier to work with logarithms. They help in simplifying logarithmic expressions and solving equations. Here are some common rules:

• Product Rule: \(\u{2061}\text{log}\big(b \times c\big) = \text{log}(b) + \text{log}(c)\)
  
• Quotient Rule: \(\text{\text{log}}\big(\frac{b}{c}\big) = \text{log}(b) - \text{log}(c)\)
  
• Power Rule: \(\text{\text{log}}\big(b^c\big) = c \times \text{log}(b)\)
  

These rules rely heavily on the properties of exponents. For instance, to use the quotient rule, one must understand that \(\text{\text{log}}\big(\frac{M}{N}\big)\) translates to the difference of individual logarithms, as shown in the step-by-step solution. Mastering these rules provides a toolkit for effectively navigating any logarithmic problem.

quotient rule in logarithms

The quotient rule in logarithms states: \(\text{\text{log}}_{a}\big(\frac{M}{N}\big) = \text{\text{log}}_{a}(M) - \text{\text{log}}_{a}(N)\).
This rule simplifies the logarithm of a division into a subtraction problem, making calculations easier. In the provided exercise, this rule is proven using the properties of exponents.
Here's a breakdown:

• Define individual logarithms: Let \(\text{\text{log}}_{a}(M) = x\) and \(\text{\text{log}}_{a}(N) = y\).
  
• Express the quotient: \(\frac{M}{N} = \frac{a^x}{a^y}\).
  
• Simplify using exponents: \(\frac{a^x}{a^y} = a^{x-y}\).
  
• Apply the logarithm: \(\text{\text{log}}_{a}(a^{x-y}) = x - y\), leading to the rule: \(\text{\text{log}}_{a}(M) - \text{\text{log}}_{a}(N)\).
  

With practice, understanding and using the quotient rule becomes intuitive and straightforward.

proof of logarithmic properties

Proving logarithmic properties often involves using their underlying definitions and related properties of exponents. For example, to prove the quotient rule: \(\text{\text{log}}_{a}\big(\frac{M}{N}\big) = \text{\text{log}}_{a}(M) - \text{\text{log}}_{a}(N)\),
we applied steps that logically follow from the definition of logarithms. First, we equate the logarithm to a variable, then use the properties of exponents to simplify the function within the logarithm.
Steps:

• Define: Let \(\text{\text{log}}_{a}(M) = x\) and \(\text{\text{log}}_{a}(N) = y\).
  
• Express quotient: \(\frac{M}{N} = \frac{a^x}{a^y}\).
  
• Simplify exponent: \(\frac{a^x}{a^y} = a^{x-y}\).
  
• Logarithm property: \(\text{\text{log}}_{a}(a^{x-y}) = x - y\).
  
• Rewriting: Replace \(\text{\text{log}}_{a}(M)\) and \(\text{\text{log}}_{a}(N)\) to confirm the rule.
  

This systematic approach reassures us that the properties of logarithms rest on solid mathematical foundations.

exponent properties in logarithms

Exponent properties are fundamental to understanding and simplifying logarithms. Two key properties are:

• Power of a Quotient: This property helps when dealing with a fraction where the numerator and denominator are both in exponential form: \(\frac{a^x}{a^y} = a^{x-y}\).
  
• Logarithm of an Exponential: Applying the logarithm to an exponent allows further simplification: \(\text{\text{log}}_{a}(a^{x-y}) = x - y\).
  

In our provided proof, we used these properties to transform and simplify expressions, ensuring we could apply the logarithmic rules smoothly.
By expressing \(\frac{M}{N}\) in terms of exponents and then taking the logarithm, we get an equation that confirms the quotient rule in logarithms. Understanding these exponent properties is crucial as it bridges the gap between logarithmic and exponential expressions, making the manipulation and simplification of such terms more intuitive.

logarithms base change

The base change formula for logarithms allows us to convert logarithms from one base to another. This is particularly useful because some bases are easier to work with than others. The formula is:
\(\text{\text{log}}_{b}(M) = \frac{\text{\text{log}}_{a}(M)}{\text{\text{log}}_{a}(b)}\)
Using this formula can simplify calculations, especially when dealing with unfamiliar bases. For instance, if you have a logarithm base that is not convenient, you can convert it to base 10 or base e (natural logarithm), which are more common.
Steps:

• Identify the original logarithm: \(\text{\text{log}}_{a}(M)\).
  
• Choose a new base: Often 10 or e.
  
• Apply the base change formula: \(\text{\text{log}}_{a}(M) = \frac{\text{\text{log}}_{b}(M)}{\text{\text{log}}_{b}(a)}\).
  

This conversion is powerful in solving problems involving different bases, allowing flexibility in choosing the most convenient base for a given problem. Understanding how to change bases and applying the base change formula is fundamental in advanced logarithmic operations.