Problem 69
Question
Write logarithm as the sum and/or difference of logarithms of a single quantity. Then simplify, if possible. \(\ln x^{20} \sqrt{z}\)
Step-by-Step Solution
Verified Answer
\( 20 \cdot \ln(x) + \frac{1}{2} \cdot \ln(z) \).
1Step 1: Apply Logarithm Power Rule
The logarithm power rule states that \( \log_b(a^n) = n \cdot \log_b(a) \). Here, we apply this rule to the term \( x^{20} \), resulting in \( \ln(x^{20}) = 20 \cdot \ln(x) \).
2Step 2: Express the Square Root Using Exponents
Recall that a square root can be expressed as an exponent, specifically \( \sqrt{z} = z^{1/2} \).
3Step 3: Apply Logarithm Power Rule to Square Root
Apply the power rule of logarithms to \( z^{1/2} \) such that \( \ln(z^{1/2}) = \frac{1}{2} \cdot \ln(z) \).
4Step 4: Combine the Logarithmic Expressions
Using the logarithm properties, specifically, that \( \ln(a) + \ln(b) = \ln(ab) \), rewrite the original expression: \( \ln(x^{20} \sqrt{z}) = \ln(x^{20}) + \ln(z^{1/2}) \). Substitute the expressions found in Steps 1 and 3: \( 20 \cdot \ln(x) + \frac{1}{2} \cdot \ln(z) \).
Key Concepts
Logarithm Power RuleSquare Roots in ExponentsSimplifying Logarithmic Expressions
Logarithm Power Rule
The logarithm power rule is a fundamental concept in logarithms, which allows us to simplify the log of an exponentiated quantity. If you have an expression like \( \log_b(a^n) \), you can rewrite it as \( n \cdot \log_b(a) \). This rule is extremely useful for breaking down more complex logarithmic expressions into manageable parts.
In the original exercise, we have \( \ln(x^{20}) \). The power rule allows us to pull down the exponent 20, making the task of evaluating or simplifying the expression considerably easier. Applying this rule, we convert \( \ln(x^{20}) \) to \( 20 \cdot \ln(x) \). This transformation simplifies handling exponential terms within logarithms.
In the original exercise, we have \( \ln(x^{20}) \). The power rule allows us to pull down the exponent 20, making the task of evaluating or simplifying the expression considerably easier. Applying this rule, we convert \( \ln(x^{20}) \) to \( 20 \cdot \ln(x) \). This transformation simplifies handling exponential terms within logarithms.
Square Roots in Exponents
Square roots themselves can be expressed using exponents, which simplifies the application of the power rule for logarithms. Remember, the square root of a number is equivalent to raising that number to the power of 1/2. This means \( \sqrt{z} = z^{1/2} \).
Understanding this conversion is crucial because it allows us to work with the square root term using the same logarithmic strategies as other exponential terms. In the given problem, transitioning the square root into an exponent form lets us apply the logarithm power rule, turning \( \ln(\sqrt{z}) \) into \( \ln(z^{1/2}) = \frac{1}{2} \cdot \ln(z) \). By rendering square roots as exponents, it becomes easier to utilize the properties of logarithms to simplify expressions.
Understanding this conversion is crucial because it allows us to work with the square root term using the same logarithmic strategies as other exponential terms. In the given problem, transitioning the square root into an exponent form lets us apply the logarithm power rule, turning \( \ln(\sqrt{z}) \) into \( \ln(z^{1/2}) = \frac{1}{2} \cdot \ln(z) \). By rendering square roots as exponents, it becomes easier to utilize the properties of logarithms to simplify expressions.
Simplifying Logarithmic Expressions
Simplifying logarithmic expressions often involves breaking down complex terms into simpler components using logarithm properties.
Once we've applied the logarithm power rule and converted our square roots to an exponent format, the next step involves simplifying. One important property of logarithms is \( \ln(a) + \ln(b) = \ln(ab) \). This additive property allows you to express the log of a product as the sum of logs, making each component easier to handle.
In this exercise, after breaking down the terms, we end up with \( \ln(x^{20} \sqrt{z}) \), which splits into \( \ln(x^{20}) + \ln(z^{1/2}) \). By substituting from previous simplifications, the expression becomes \( 20 \cdot \ln(x) + \frac{1}{2} \cdot \ln(z) \). Simplifying logarithmic expressions in this way can dramatically reduce complexity and make them more straightforward to evaluate or understand.
Once we've applied the logarithm power rule and converted our square roots to an exponent format, the next step involves simplifying. One important property of logarithms is \( \ln(a) + \ln(b) = \ln(ab) \). This additive property allows you to express the log of a product as the sum of logs, making each component easier to handle.
In this exercise, after breaking down the terms, we end up with \( \ln(x^{20} \sqrt{z}) \), which splits into \( \ln(x^{20}) + \ln(z^{1/2}) \). By substituting from previous simplifications, the expression becomes \( 20 \cdot \ln(x) + \frac{1}{2} \cdot \ln(z) \). Simplifying logarithmic expressions in this way can dramatically reduce complexity and make them more straightforward to evaluate or understand.
Other exercises in this chapter
Problem 69
Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ 3^{x-6}=81 $$
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Let \(f(x)=\frac{1}{x}\) and \(g(x)=\frac{1}{x^{2}} .\) Find each of the following. $$ (g \circ f)\left(\frac{1}{3}\right) $$
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Solve each equation. Give the exact solution and, when appropriate, an approximation to four decimal places. $$ 5^{x+4}=125 $$
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Let \(f(x)=\frac{1}{x}\) and \(g(x)=\frac{1}{x^{2}} .\) Find each of the following. $$ (g \circ f)\left(\frac{1}{10}\right) $$
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