Problem 73
Question
Write each logarithm as the sum and/or difference of logarithms of a single quantity. Then simplify, if possible. See Example 6. $$ \ln \sqrt{e x} $$
Step-by-Step Solution
Verified Answer
\(\ln \sqrt{e x} = \frac{1}{2} + \frac{1}{2} \ln x\).
1Step 1: Rewrite the Square Root
The expression given is \( \ln \sqrt{e x} \). Recall the property that \( \sqrt{a} = a^{1/2} \). So, rewrite the expression as \( \ln (e x)^{1/2} \).
2Step 2: Apply the Power Rule for Logarithms
The power rule states that \( \ln(a^b) = b \ln a \). Applying this to \( \ln (e x)^{1/2} \), we get \( \frac{1}{2} \ln (e x) \).
3Step 3: Apply the Product Rule for Logarithms
The product rule for logarithms is \( \ln(ab) = \ln a + \ln b \). Thus, apply it to \( \frac{1}{2} \ln (e x) \) to get \( \frac{1}{2} (\ln e + \ln x) \).
4Step 4: Simplify the Expression
Since \( \ln e = 1 \), replace it in the expression. This simplifies to \( \frac{1}{2} (1 + \ln x) \). Distributing the \( \frac{1}{2} \) yields \( \frac{1}{2} + \frac{1}{2} \ln x \).
Key Concepts
Power Rule for LogarithmsProduct Rule for LogarithmsSimplification of Logarithms
Power Rule for Logarithms
The power rule for logarithms is a fundamental principle that is incredibly helpful when dealing with exponents inside a logarithmic expression.
Simply put, the rule states that for any real number \( a \) and any non-zero real number \( b \), the logarithm of \( a^b \) can be expressed as \( b \ln a \).
Simply put, the rule states that for any real number \( a \) and any non-zero real number \( b \), the logarithm of \( a^b \) can be expressed as \( b \ln a \).
- This means you can "bring down" the exponent as a multiple in front of the logarithm.
- For example, the expression \( \ln(x^2) \) can be rewritten as \( 2 \ln x \).
- The power rule helps simplify situations where the argument of the logarithm is raised to a power.
Product Rule for Logarithms
The product rule for logarithms is another essential tool needed to break down complex logarithmic expressions.
According to this rule, the logarithm of a product can be expressed as the sum of the logarithms of the multiplicands.
It is a staple property when working with logarithmic equations, allowing you to break down multiplicative components into additive ones.
According to this rule, the logarithm of a product can be expressed as the sum of the logarithms of the multiplicands.
- Mathematically, this is stated as \( \ln(ab) = \ln a + \ln b \).
- This rule splits a single logarithm into multiple terms, each with its own logarithmic expression.
- For example, if you have \( \ln(2x) \), it can be split into \( \ln 2 + \ln x \).
It is a staple property when working with logarithmic equations, allowing you to break down multiplicative components into additive ones.
Simplification of Logarithms
Simplifying logarithmic expressions is crucial, as it transforms complex problems into simpler, more solvable ones. Using the properties of logarithms such as the power and product rules, we can streamline these expressions significantly.
- Rewriting the square root as a power: \( \ln (e x)^{1/2} \). - Using the power rule: \( \frac{1}{2} \ln (e x) \). - Applying the product rule: \( \frac{1}{2}(\ln e + \ln x) \). - And finally simplifying the known logarithm \( \ln e = 1 \): \( \frac{1}{2} + \frac{1}{2} \ln x \).
Through each of these steps, the expression becomes far more digestible, demonstrating the power of logarithmic simplification.
- Start by rewriting any roots or powers using the power rule.
- Next, apply the product or quotient rules to separate out multiplicative or divisive components.
- Afterwards, substitute known identities, such as \( \ln e = 1 \), to simplify further.
- Rewriting the square root as a power: \( \ln (e x)^{1/2} \). - Using the power rule: \( \frac{1}{2} \ln (e x) \). - Applying the product rule: \( \frac{1}{2}(\ln e + \ln x) \). - And finally simplifying the known logarithm \( \ln e = 1 \): \( \frac{1}{2} + \frac{1}{2} \ln x \).
Through each of these steps, the expression becomes far more digestible, demonstrating the power of logarithmic simplification.
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Problem 73
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