Problem 81
Question
Write each logarithmic expression as one logarithm. See Example 7. $$ \frac{1}{3}\left[\log _{b}\left(M^{2}-9\right)-\log _{b}(M+3)\right] $$
Step-by-Step Solution
Verified Answer
The expression simplifies to \( \log_b\left((M - 3)^{1/3}\right) \).
1Step 1: Expand the Expression
The given expression is \( \frac{1}{3}\left[\log _{b}\left(M^{2}-9\right)-\log _{b}(M+3)\right] \). The expression inside the brackets involves the difference of two logs, which can be combined using the logarithm property \( \log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right) \). Therefore, express this as \( \log_b\left(\frac{M^2 - 9}{M + 3}\right) \).
2Step 2: Apply Logarithmic Simplification
The expression now reads \( \frac{1}{3}\log_b\left(\frac{M^2 - 9}{M + 3}\right) \). According to the power rule of logarithms, \( c \log_b(A) = \log_b(A^c) \), you can simplify this further. Hence, write it as \( \log_b\left(\left(\frac{M^2 - 9}{M + 3}\right)^{1/3}\right) \).
3Step 3: Simplify the Radicand
The radicand \( M^2 - 9 \) can be factored since it is a difference of squares: \( M^2 - 9 = (M + 3)(M - 3) \). Substituting this back into the expression gives: \( \log_b\left(\left(\frac{(M + 3)(M - 3)}{M + 3}\right)^{1/3}\right) \).
4Step 4: Simplify the Fraction
Within the logarithm, simplify the fraction by cancelling the \( M + 3 \) in the numerator and denominator, leaving \( \log_b\left(\left(M - 3\right)^{1/3}\right) \).
5Step 5: Final Expression as One Logarithm
Since we expressed everything as a single logarithm, the final simplified logarithmic expression is \( \log_b\left((M - 3)^{1/3}\right) \).
Key Concepts
Logarithm PropertiesLogarithmic SimplificationDifference of Squares
Logarithm Properties
Logarithm properties are key tools in simplifying complex logarithmic expressions. Logarithms have several properties that allow us to manipulate and simplify expressions, such as the product, quotient, and power rules.
In the original exercise, we use the difference property of logarithms, which states that \( \log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right) \). This property lets us combine two logarithms into a single log expression.
By applying this rule, the problem simplifies from holding two separate logs to just one. It's essentially saying that subtracting logs means you're dividing their internal values. Knowing and applying these properties makes it easier to condense complex expressions into a simple form that's easier to work with.
In the original exercise, we use the difference property of logarithms, which states that \( \log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right) \). This property lets us combine two logarithms into a single log expression.
By applying this rule, the problem simplifies from holding two separate logs to just one. It's essentially saying that subtracting logs means you're dividing their internal values. Knowing and applying these properties makes it easier to condense complex expressions into a simple form that's easier to work with.
Logarithmic Simplification
Logarithmic simplification involves the use of properties of logarithms to rewrite expressions in a more compact form. After combining logarithms using properties, further simplifications can often be made.
One crucial rule at play is the power rule of logarithms: \( c \log_b(A) = \log_b(A^c) \). This rule allows multiplication inside a logarithmic expression to become a power. When the original expression was \( \frac{1}{3} \log_b\left(\frac{M^2 - 9}{M + 3}\right) \), the power rule rewrote it as a single logarithm with an exponent of \( \frac{1}{3} \).
This step is about transforming the form, making sure the entire expression is as neat and compact as possible. Remember, simplification helps to not only make expressions neater but also easier to interpret when solving logarithmic equations.
One crucial rule at play is the power rule of logarithms: \( c \log_b(A) = \log_b(A^c) \). This rule allows multiplication inside a logarithmic expression to become a power. When the original expression was \( \frac{1}{3} \log_b\left(\frac{M^2 - 9}{M + 3}\right) \), the power rule rewrote it as a single logarithm with an exponent of \( \frac{1}{3} \).
This step is about transforming the form, making sure the entire expression is as neat and compact as possible. Remember, simplification helps to not only make expressions neater but also easier to interpret when solving logarithmic equations.
Difference of Squares
The difference of squares is a specific algebraic identity that expresses a difference between two square numbers. It is given by the formula \( a^2 - b^2 = (a + b)(a - b) \). Recognizing and applying this pattern helps simplify expressions further.
In our problem, the expression \( M^2 - 9 \) is recognized as a difference of squares. Here, \( a = M \) and \( b = 3 \), making it \( (M^2 - 9) = (M + 3)(M - 3) \). Factoring expressions using the difference of squares quickly breaks them down into products of linear factors.
Understanding this concept is foundational when working with polynomials, as recognizing this pattern helps in factoring and reducing expressions which can lead to solving equations more easily. This step allowed us to simplify the entire logarithmic expression further, as it offered a pathway to cancel terms in the numerator and the denominator.
In our problem, the expression \( M^2 - 9 \) is recognized as a difference of squares. Here, \( a = M \) and \( b = 3 \), making it \( (M^2 - 9) = (M + 3)(M - 3) \). Factoring expressions using the difference of squares quickly breaks them down into products of linear factors.
Understanding this concept is foundational when working with polynomials, as recognizing this pattern helps in factoring and reducing expressions which can lead to solving equations more easily. This step allowed us to simplify the entire logarithmic expression further, as it offered a pathway to cancel terms in the numerator and the denominator.
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Problem 81
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