Problem 90 - Algebra 2

Question

Let \(u=\log _{b} M\) and \(v=\log _{b} N .\) Prove the Quotient Property of logarithms.

Step-by-Step Solution

Answer

The Quotient Property of logarithms is proven as \( \log_{b}\frac{M}{N} = \log_{b}M - \log_{b}N \).

Step 1: Convert logarithms to exponential form

Since \( u =\log_{b}M \) and \( v =\log_{b}N \), these equalities can be written in exponential form as \( b^{u} = M \) and \( b^{v} = N \) respectively as per the definition of logarithms.

Step 2: Substitution

Substituting the values of M and N in the expression \( \log_{b}\frac{M}{N} \) with \( b^{u} \) and \( b^{v} \) respectively, the expression can be written as \( \log_{b}\frac{b^{u}}{b^{v}} \).

Step 3: Apply the property of exponents

Now, when dividing the same bases, you subtract the exponents according to the properties of exponents. Hence, the expression simplifies to \( \log_{b}(b^{u-v}) \).

Step 4: Simplification

The expression now simplifies to \(u-v\) as \( \log_{b}(b^{n}) = n \).

Step 5: Re-substitute the original values

Re-substituting \( u =\log_{b}M \) and \( v =\log_{b}N \), the expression becomes \( \log_{b}M - \log_{b}N \).

Key Concepts

Logarithm Rules

Logarithm rules are essential tools when working with logarithmic expressions. There's a set of rules that govern the operations on logarithms, making them easier and more straightforward to handle. Some of the key rules include the product rule, quotient rule, and power rule. The quotient rule, as highlighted in the original exercise, states:

• \( \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N \).

This rule shows that the logarithm of a quotient is equivalent to the difference of the logarithms. This can simplify complex calculations and make them more manageable. Remember, these rules only hold when the logarithms share the same base, denoted by \( b \). Understanding these fundamental properties helps in proving more complex theorems and simplifies the process of solving logarithmic equations.
Logarithm rules provide structure and help students understand connections between multiplication, division, and powers in terms of logarithmic expressions. Familiarity with these rules makes it easier to manipulate and solve logarithmic problems efficiently.

Exponential Form

Converting logarithms into their exponential form is a powerful technique used in the original exercise to prove the quotient property of logarithms. The key idea here is understanding that a logarithm is essentially another way to express an exponent.
Consider the expression \( u = \log_b M \). In exponential form, this is expressed as:

• \( b^u = M \)

Here, \( b \) is the base, \( u \) is the exponent, and \( M \) is the result. Similarly, \( v = \log_b N \) translates to \( b^v = N \).
By using the exponential form, it becomes much simpler to handle expressions involving logarithms, especially when substituting values or rearranging equations. This conversion method is crucial when applying other logarithmic rules and properties, allowing for a clearer understanding and simplification of the calculations involved.

Properties of Exponents

The properties of exponents are critical when dealing with exponential and logarithmic expressions. One of the most important properties is how exponents behave during multiplication and division. These properties were applied in the original problem when simplifying \( \frac{b^u}{b^v} \) to \( b^{u-v} \). Here is a brief overview of these properties:

• Product of Powers: \( b^m \cdot b^n = b^{m+n} \)
• Quotient of Powers: \( \frac{b^m}{b^n} = b^{m-n} \)
• Power of a Power: \( (b^m)^n = b^{m \cdot n} \)

The quotient of powers property is showcased in the exercise, where dividing powers with the same base results in subtracting their exponents. This is what allowed the expression \( \log_b \left( \frac{b^u}{b^v} \right) \) to simplify into \( \log_b(b^{u-v}) \).
Knowing how to manipulate exponents is essential in mathematics, especially in algebra and calculus, as it opens doors to simplifying and solving more complex mathematical problems effortlessly.