Problem 21
Question
For the following exercises, condense each expression to a single logarithm using the properties of logarithms. \(\ln \left(6 x^{9}\right)-\ln \left(3 x^{2}\right)\)
Step-by-Step Solution
Verified Answer
The condensed expression is \( \ln(2x^7) \).
1Step 1: Apply the Quotient Rule
The expression given is \( \ln(6x^9) - \ln(3x^2) \). According to the properties of logarithms, specifically the quotient rule, \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). Apply this property here: \( \ln\left(\frac{6x^9}{3x^2}\right) \).
2Step 2: Simplify the Fraction
Now simplify the fraction inside the logarithm. When simplifying \( \frac{6x^9}{3x^2} \), first simplify the coefficients: \( \frac{6}{3} = 2 \). Then simplify the exponents: \( \frac{x^9}{x^2} = x^{9-2} = x^7 \). This gives us: \( \ln(2x^7) \).
3Step 3: Conclusion: Final Expression
The condensed expression as a single logarithm is \( \ln(2x^7) \).
Key Concepts
Properties of LogarithmsLogarithmic SimplificationQuotient Rule in Logarithms
Properties of Logarithms
Logarithms are extremely useful mathematical tools that help us simplify logarithmic expressions. Essential properties of logarithms include:
- The Product Rule: \( \ln(a) + \ln(b) = \ln(ab) \). This is used when dealing with the sum of logarithms.
- The Quotient Rule: \( \ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right) \). This property is used when we have a difference of logarithms, as shown in the original exercise.
- The Power Rule: \( \ln(a^b) = b \cdot \ln(a) \). This is useful when you have a logarithm of an expression raised to a power.
Logarithmic Simplification
Simplifying expressions involving logarithms often involves using their properties to reduce complex forms to simpler, single-logarithm expressions. The primary goal here is to condense multiple logs into one, making the expression more straightforward to interpret or solve.
In the given exercise, we started with the expression \( \ln(6x^9) - \ln(3x^2) \). The process of logarithmic simplification entails several steps:
In the given exercise, we started with the expression \( \ln(6x^9) - \ln(3x^2) \). The process of logarithmic simplification entails several steps:
- Recognize the type of logarithmic operation involved. Here, it’s a subtraction, which calls for the quotient rule.
- Use the properties intelligently, applying one or a combination as needed to simplify the expression. For example, simplifying \( \ln\left(\frac{6x^9}{3x^2}\right) \) by reducing the fraction inside the logarithm.
- Always check to see if additional simplification is possible, like reducing the expression \( \frac{6}{3} \) to 2 and combining the powers of x to get \( x^7 \).
Quotient Rule in Logarithms
The quotient rule for logarithms is an invaluable tool when you need to simplify the difference between two logarithms with the same base. This rule states that if you have a logarithm subtraction, like \( \ln(a) - \ln(b) \), you can combine them into a single logarithm of a quotient: \( \ln\left(\frac{a}{b}\right) \).
In the example provided, we applied the quotient rule directly to \( \ln(6x^9) - \ln(3x^2) \). This transformed it into:\[ \ln\left(\frac{6x^9}{3x^2}\right) \].
After applying the rule, further simplification is often necessary. Break down the fraction by dividing both coefficients and powers separately. The operation results in \( \frac{6}{3} = 2 \) and \( x^{9-2} = x^7 \), yielding \( \ln(2x^7) \).
Using the quotient rule simplifies calculations and aids in arriving at more manageable logarithmic expressions.
In the example provided, we applied the quotient rule directly to \( \ln(6x^9) - \ln(3x^2) \). This transformed it into:\[ \ln\left(\frac{6x^9}{3x^2}\right) \].
After applying the rule, further simplification is often necessary. Break down the fraction by dividing both coefficients and powers separately. The operation results in \( \frac{6}{3} = 2 \) and \( x^{9-2} = x^7 \), yielding \( \ln(2x^7) \).
Using the quotient rule simplifies calculations and aids in arriving at more manageable logarithmic expressions.
Other exercises in this chapter
Problem 21
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