Problem 27
Question
Evaluate each expression without using a calculator. $$\log _{2} \sqrt{2}$$
Step-by-Step Solution
Verified Answer
The evaluation of the expression \( \log_{2} \sqrt{2} \) without using a calculator results in \(1/2\).
1Step 1: Convert the square root to an exponent
Start by expressing the square root of 2 as an exponent. The square root of any number x is equivalent to raising that number to the power of \(1/2\). So, \( \sqrt{2} \) can be written as \(2^{1/2}\). This gives us the new expression \( \log_{2} 2^{1/2} \).
2Step 2: Apply the power rule of logarithms
The power rule of logarithms states that \( \log_{b} a^{n} \) is equivalent to \( n \log_{b} a\). Applying this rule to our expression, we can bring down the exponent \(1/2\) in front of the logarithm to get \( (1/2) \log_{2} 2 \).
3Step 3: Evaluate the logarithm
We know that the logarithm base 2 of 2 is 1, as any log base b of b equals 1. So the expression simplifies to \(1/2 * 1\) or simply \(1/2\). This is our final answer.
Key Concepts
Exponent ConversionPower Rule of LogarithmsLogarithmic Identities
Exponent Conversion
When faced with logarithmic expressions, understanding how to handle exponents is crucial. Exponent conversion simplifies the process of evaluating logarithms, especially when dealing with roots.
For instance, the square root of a number can be expressed as that number raised to the power of \( \frac{1}{2} \). This exponent form is derived from the definition of roots in terms of exponents: \( x^{\frac{1}{n}} \) is the n-th root of \(x\). So, \(\sqrt{2}\) becomes \(2^{\frac{1}{2}}\) through exponent conversion. It's the initial step that often sets the stage for further simplification using logarithmic rules.
Understanding exponent conversion allows students to handle more complex expressions involving roots by converting them into a form more amenable to logarithmic manipulation. This step is frequent in mathematical evaluations and is the cornerstone for solving problems that involve logarithms and radical expressions.
For instance, the square root of a number can be expressed as that number raised to the power of \( \frac{1}{2} \). This exponent form is derived from the definition of roots in terms of exponents: \( x^{\frac{1}{n}} \) is the n-th root of \(x\). So, \(\sqrt{2}\) becomes \(2^{\frac{1}{2}}\) through exponent conversion. It's the initial step that often sets the stage for further simplification using logarithmic rules.
Understanding exponent conversion allows students to handle more complex expressions involving roots by converting them into a form more amenable to logarithmic manipulation. This step is frequent in mathematical evaluations and is the cornerstone for solving problems that involve logarithms and radical expressions.
Power Rule of Logarithms
The power rule of logarithms is a transformative tool in the world of logarithmic evaluation. It states that for any positive real number \(a\), base \(b\) (where \(b > 0\) and \(b e 1\)), and exponent \(n\), the logarithm of a power is the exponent times the logarithm of the base:
\[ \log_{b} a^{n} = n \cdot \log_{b} a \]
Using this rule, expressions like \( \log_{2} 2^{\frac{1}{2}} \) can be simplified by bringing the exponent in front of the logarithm. This deceptively simple move can turn a seemingly complex expression into an easily solvable one, reducing \( \log_{2} 2^{\frac{1}{2}} \) to \( \frac{1}{2} \cdot \log_{2} 2 \) with ease.
This rule is incredibly useful because it not only simplifies the logarithmic representation but also sets the stage for applying basic logarithmic identities, further simplifying the expression. The power rule is often a student's first step in breaking down and solving logarithmic equations.
\[ \log_{b} a^{n} = n \cdot \log_{b} a \]
Using this rule, expressions like \( \log_{2} 2^{\frac{1}{2}} \) can be simplified by bringing the exponent in front of the logarithm. This deceptively simple move can turn a seemingly complex expression into an easily solvable one, reducing \( \log_{2} 2^{\frac{1}{2}} \) to \( \frac{1}{2} \cdot \log_{2} 2 \) with ease.
This rule is incredibly useful because it not only simplifies the logarithmic representation but also sets the stage for applying basic logarithmic identities, further simplifying the expression. The power rule is often a student's first step in breaking down and solving logarithmic equations.
Logarithmic Identities
Logarithmic identities are the formulas and properties that govern the operation of logarithms. One of the most commonly used logarithmic identities is that the logarithm of any number at its own base is 1:
\[ \log_{b} b = 1 \]
Applying this to the expression \( \frac{1}{2} \cdot \log_{2} 2 \), we can quickly determine that \( \log_{2} 2 \) is equal to 1 since the base and the number are the same. With this identity, the expression simplifies to \( \frac{1}{2} \cdot 1 \), which equals \( \frac{1}{2}\).
Understanding these identities is essential as they are frequently the key to unlocking and simplifying logarithmic expressions. Identities like \( \log_{b} 1 = 0 \) (since any number to the power of 0 is 1) and the product, quotient, and change-of-base rules for logarithms provide a comprehensive toolkit for evaluating logarithmic expressions without a calculator. Mastering these identities ensures that students can confidently tackle a wide array of logarithmic problems.
\[ \log_{b} b = 1 \]
Applying this to the expression \( \frac{1}{2} \cdot \log_{2} 2 \), we can quickly determine that \( \log_{2} 2 \) is equal to 1 since the base and the number are the same. With this identity, the expression simplifies to \( \frac{1}{2} \cdot 1 \), which equals \( \frac{1}{2}\).
Understanding these identities is essential as they are frequently the key to unlocking and simplifying logarithmic expressions. Identities like \( \log_{b} 1 = 0 \) (since any number to the power of 0 is 1) and the product, quotient, and change-of-base rules for logarithms provide a comprehensive toolkit for evaluating logarithmic expressions without a calculator. Mastering these identities ensures that students can confidently tackle a wide array of logarithmic problems.
Other exercises in this chapter
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