Problem 19
Question
Use properties of logarithms to expand logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. \(\ln \sqrt[5]{x}\)
Step-by-Step Solution
Verified Answer
The expanded form of \(\ln \sqrt[5]{x}\) is \(\frac{1}{5} \ln x\).
1Step 1: Identify the Expression
We need to evaluate or simplify: \(\ln \sqrt[5]{x}\).
2Step 2: Apply Order of Operations
Following the order of operations (PEMDAS/BODMAS), we simplify the expression step by step, handling parentheses, exponents, multiplication/division, then addition/subtraction.
3Step 3: Compute the Result
Carrying out the operations systematically, we arrive at the result.
4Step 4: State the Result
The expanded form of \(\ln \sqrt[5]{x}\) is \(\frac{1}{5} \ln x\).
Key Concepts
Logarithmic ExpressionsNatural LogarithmExponentiation and LogarithmsAlgebraic Manipulation
Logarithmic Expressions
Understanding logarithmic expressions is crucial for students tackling advanced math topics. Logarithms are the inverses of exponentiation. This means, if you have an expression like \( b^y = x \), the logarithm of \( x \) with base \( b \) is \( y \) or \( \text{log}_b(x) = y \). The expression \( \text{log}_b(x) \) is read as 'log base \( b \) of \( x \)'.
When we use logarithmic expressions, we often apply certain rules or properties to simplify or manipulate these expressions. For example, the power rule, which allows us to move exponents outside the log as a multiplier, as seen in the provided exercise. This makes complex logarithmic expressions easier to tackle, as it breaks down a more complicated problem into simpler parts that are more manageable to solve. A solid grasp of these rules simplifies a student's approach to logarithms considerably.
When we use logarithmic expressions, we often apply certain rules or properties to simplify or manipulate these expressions. For example, the power rule, which allows us to move exponents outside the log as a multiplier, as seen in the provided exercise. This makes complex logarithmic expressions easier to tackle, as it breaks down a more complicated problem into simpler parts that are more manageable to solve. A solid grasp of these rules simplifies a student's approach to logarithms considerably.
Natural Logarithm
The natural logarithm, denoted as \( \text{ln}(x) \), is a specific type of logarithm where the base is the transcendental number \( e \), approximately equal to 2.71828. This natural base is fundamental to various mathematical and physical phenomena, especially in calculus and growth processes.
Why is it 'natural'?
The term 'natural' comes from the natural growth rates and compound interest where \( e \) continuously pops up. Thus, \( \text{ln} \) is not just a passing concept but is deeply rooted in the way we understand exponential growth and decay. Students will see \( \text{ln} \) in various contexts ranging from calculating the half-life of a radioactive substance to understanding continuous compounding in finance.Exponentiation and Logarithms
Exponentiation and logarithms are mathematical operations that are inverses of each other. Exponentiation is the process of raising a number to a power, which is a compact way of writing multiplication repeated several times. For example, \( 2^3 = 2 \times 2 \times 2 \).
On the flip side, logarithms help us figure out what power we need to raise a base to get a certain result. Continuing our previous example, \( \text{log}_2(8) = 3 \) because \( 2^3 = 8 \). It's much like asking, '2 raised to what power gives me 8?' Understanding this inverse relationship is crucial because it allows us to switch between the two operations. This proves invaluable when solving equations where the variable is part of an exponent—a standard scenario in exponential growth or decay problems.
On the flip side, logarithms help us figure out what power we need to raise a base to get a certain result. Continuing our previous example, \( \text{log}_2(8) = 3 \) because \( 2^3 = 8 \). It's much like asking, '2 raised to what power gives me 8?' Understanding this inverse relationship is crucial because it allows us to switch between the two operations. This proves invaluable when solving equations where the variable is part of an exponent—a standard scenario in exponential growth or decay problems.
Algebraic Manipulation
Algebraic manipulation is the skill of rearranging and simplifying mathematical expressions and equations. It involves a range of operations, including expanding, factoring, and using properties of arithmetic to combine like terms.
When you're dealing with logarithms, algebraic manipulation can include actions such as applying the distributive property, combining or breaking up logarithms using the product, quotient, or power rules, and even switching between logarithmic and exponential forms to find a solution. Mastering algebraic manipulation is essential for a student not just for the sake of solving an equation, but for understanding the underlying structure of mathematical relationships. This skill set highlights the idea that mathematics is not just about calculating but also about understanding and is of immense value in making students better problem solvers across all science, technology, engineering, and mathematics (STEM) fields.
When you're dealing with logarithms, algebraic manipulation can include actions such as applying the distributive property, combining or breaking up logarithms using the product, quotient, or power rules, and even switching between logarithmic and exponential forms to find a solution. Mastering algebraic manipulation is essential for a student not just for the sake of solving an equation, but for understanding the underlying structure of mathematical relationships. This skill set highlights the idea that mathematics is not just about calculating but also about understanding and is of immense value in making students better problem solvers across all science, technology, engineering, and mathematics (STEM) fields.
Other exercises in this chapter
Problem 19
Solve each exponential equation by expressing each side as a power of the same base and then equating exponents. $$8^{x+3}=16^{x-1}$$
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Write each equation in its equivalent logarithmic form. $$7^{y}=200$$
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Use the exponential decay model for carbon- \(14, A=A_{0} e^{-0.000121 t}\) to solve Exercises \(19-20\) Skeletons were found at a construction site in San Fran
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Solve each exponential equation by expressing each side as a power of the same base and then equating exponents. $$8^{1-x}=4^{x+2}$$
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