Problem 74
Question
Condensing a Logarithmic Expression In Exercises \(67-82\) , condense the expression to the logarithm of a single quantity. $$2 \ln 8+5 \ln (z-4)$$
Step-by-Step Solution
Verified Answer
The expression can be condensed to \(\ln (64 * (z-4)^5)\)
1Step 1: Apply the properties of logarithms
The expression given is: \(2 \ln 8+5 \ln (z-4)\). First, the Power Rule will be applied to recognize that the coefficient in front of the logarithm can be taken as the power of the argument. Hence, it transforms to \(\ln 8^2 + \ln (z-4)^5\).
2Step 2: Simplify the expression
After the Power Rule is applied, it simplifies to \(\ln 64 + \ln (z-4)^5\). This is because \(8^2 = 64\)
3Step 3: Apply the Product Rule
By applying the Product rule, the expression may be condensed further to \(\ln (64 * (z-4)^5)\)
4Step 4: Final Result
Hence, the expression condenses to \(\ln (64 * (z-4)^5)\). This is the final logarithmic expression
Key Concepts
Properties of LogarithmsPower Rule in LogarithmsProduct Rule in Logarithms
Properties of Logarithms
Understanding the properties of logarithms is essential for condensing logarithmic expressions. Logarithms, the inverse of exponentiation, have unique characteristics that allow us to manipulate and simplify complex logarithmic statements. The three main properties are the Product Rule, the Quotient Rule, and the Power Rule.
The Product Rule states that the log of a product is equal to the sum of the logs of the factors: \[\log_b(mn) = \log_b(m) + \log_b(n)\]. Conversely, the Quotient Rule tells us that the log of a quotient is the difference between the log of the numerator and the log of the denominator: \[\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\]. Lastly, the Power Rule, which we often use when condensing expressions, allows us to move the exponent of the argument to the front of the log: \[\log_b(m^p) = p \cdot \log_b(m)\].
By applying these rules, you can transform a complex logarithmic expression into a simpler equivalent form that makes it easier to evaluate or further manipulate. For students to clearly understand and correctly apply these rules, it's beneficial to practice with various examples that solidify their comprehension.
The Product Rule states that the log of a product is equal to the sum of the logs of the factors: \[\log_b(mn) = \log_b(m) + \log_b(n)\]. Conversely, the Quotient Rule tells us that the log of a quotient is the difference between the log of the numerator and the log of the denominator: \[\log_b\left(\frac{m}{n}\right) = \log_b(m) - \log_b(n)\]. Lastly, the Power Rule, which we often use when condensing expressions, allows us to move the exponent of the argument to the front of the log: \[\log_b(m^p) = p \cdot \log_b(m)\].
By applying these rules, you can transform a complex logarithmic expression into a simpler equivalent form that makes it easier to evaluate or further manipulate. For students to clearly understand and correctly apply these rules, it's beneficial to practice with various examples that solidify their comprehension.
Power Rule in Logarithms
The Power Rule in logarithms is a powerful tool for simplifying expressions with exponents. As seen in our exercise, we can move the coefficient in front of a logarithm to become the exponent of its argument. To apply this rule correctly, identify the coefficient and rewrite the logarithm with the argument raised to the power of this coefficient.
For example, in the expression from the exercise, the Power Rule is applied as follows: \[2 \cdot \ln(8)\text{ becomes }\ln(8^2)\] and \[5 \cdot \ln(z - 4)\text{ becomes }\ln((z - 4)^5)\]. This transformation is essential for condensing the expression, as it allows us to leverage the Product Rule, which requires the logarithms to have the same base without coefficients in front.
For example, in the expression from the exercise, the Power Rule is applied as follows: \[2 \cdot \ln(8)\text{ becomes }\ln(8^2)\] and \[5 \cdot \ln(z - 4)\text{ becomes }\ln((z - 4)^5)\]. This transformation is essential for condensing the expression, as it allows us to leverage the Product Rule, which requires the logarithms to have the same base without coefficients in front.
Exercise Improvement Advice
Students often struggle with the application of the Power Rule when the coefficient is not a whole number or when it's negative. It's beneficial to provide additional exercises incorporating these scenarios to enhance students' understanding and flexibility with the Power Rule.Product Rule in Logarithms
The Product Rule comes into play after utilizing the Power Rule. It's designed to combine multiple logarithms into a single logarithm when they're being added together—as demonstrated in the exercise. This concept hinges upon the fundamental understanding that logarithms convert multiplication into addition.
In practice, we use the Product Rule to combine \[\ln(64)\text{ and }\ln((z - 4)^5)\] by simply multiplying their arguments together, resulting in a single condensed logarithm: \[\ln(64) + \ln((z - 4)^5) = \ln(64 \cdot (z - 4)^5)\]. When students grasp this rule, they find it much easier to work with products inside logarithmic functions.
In practice, we use the Product Rule to combine \[\ln(64)\text{ and }\ln((z - 4)^5)\] by simply multiplying their arguments together, resulting in a single condensed logarithm: \[\ln(64) + \ln((z - 4)^5) = \ln(64 \cdot (z - 4)^5)\]. When students grasp this rule, they find it much easier to work with products inside logarithmic functions.
Key into Conceptual Understanding
While solving exercises, it's important to highlight that the Product Rule only applies to addition of logarithms with the same base. Identifying such opportunities to condense expressions can offer students insight into the interconnectedness of logarithmic and exponential functions and their rules. To promote a deeper understanding, students can be encouraged to explore how changing the order of the factors in the argument does not affect the final value of the logarithm, illustrating the commutative property of multiplication within the logarithmic context.Other exercises in this chapter
Problem 73
Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. \(2 x^{2} e^{2 x}+2 x e^{2 x}=0\)
View solution Problem 73
Use the One-to-One Property to solve the equation for \(x .\) \(\ln (x+4)=\ln 12\)
View solution Problem 74
Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. \(-x^{2} e^{-x}+2 x e^{-x}=0\)
View solution Problem 74
Use the One-to-One Property to solve the equation for \(x .\) \(\ln (x-7)=\ln 7\)
View solution