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In Exercises \(21-36,\) find the exact value of the logarithmic expression without using a calculator. (If this is not possible, then state the reason.) $$\log _{6} \sqrt[3]{6}$$

Short Answer

Expert verified
The exact value of the logarithmic expression \(\log _{6} \sqrt[3]{6}\) is \(\frac{1}{3}\).

Step by step solution

01

Redefine it using property of exponents

First, convert the cube root into an exponent form as \(\sqrt[3]{6} = 6^\frac{1}{3}\). So, the original expression becomes \(\log _{6} {6^{\frac{1}{3}}}\).
02

Apply properties of logarithms

Next, apply the power rule of logarithms which states that \(\log_b {a^n} = n \cdot \log_b a\), and substitute \(n = \frac{1}{3}\) and \(a = 6\). The expression now simplifies to \(\frac{1}{3} \log _{6} {6}\).
03

Simplify logarithm

The definition of the logarithm states that \(\log_b b = 1\). Using this, simplify further and replace \(\log_{6} 6\) with 1. The final expression becomes \(\frac{1}{3} \cdot 1 = \frac{1}{3}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Properties of Logarithms
The properties of logarithms are mathematical rules that define how logarithms interact with other arithmetic operations. These properties are crucial for simplifying logarithmic expressions and solving logarithmic equations. One fundamental property is the product rule, which states that the logarithm of a product is equal to the sum of the logarithms of its factors, or \( \log_b(xy) = \log_b(x) + \log_b(y) \). On the other hand, the quotient rule says that the logarithm of a quotient is the difference of the logarithms, expressed as \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \).

The property particularly relevant to our exercise is the power rule, which is often used to simplify logarithmic expressions involving exponents, like \( \log_b\left(a^n\right) = n \cdot \log_b(a) \). This property allows one to essentially 'pull down' an exponent from inside a logarithm to the front, making it a coefficient. In terms of simplifying the given logarithmic expression \( \log_{6} \sqrt[3]{6} \), applying the power rule is an essential step towards finding the exact value without the need for a calculator.
Exponent Form
When dealing with roots and logarithms, it's often necessary to convert roots into exponent form. By moving from a radical to an exponent, we can apply other algebraic rules more easily. For instance, the general formula for converting a root into an exponent is \( \sqrt[n]{a} = a^{\frac{1}{n}} \).

In the exercise \( \log_{6} \sqrt[3]{6} \) the cube root of 6 can be rewritten as \( 6^{\frac{1}{3}} \), thus converting the radical expression into the exponent form. This method is essential for simplifying the logarithmic expression using the properties of logarithms. It's a great strategy for any student learning to navigate the complexities of logarithmic expressions, as it makes the subsequent application of logarithm properties more straightforward.
Simplifying Logarithms
Simplifying logarithmic expressions involves using a series of logarithmic properties to write the expression in a simpler or more recognizable form. This process often includes identifying and converting roots to exponents, applying the power rule, and recognizing when the argument of the logarithm and the base are the same number.

As shown in the original solution for \( \log_{6} 6^{\frac{1}{3}} \), after applying the power rule, the expression simplifies to \( \frac{1}{3} \log_{6} 6 \). A crucial simplification step is then to remember that \( \log_b b = 1 \), for any base \( b \). Therefore, in our exercise \( \log_{6} 6 \) simplifies to 1, and the entire expression reduces to \( \frac{1}{3} \).

This result can be reached with the understanding that the logarithm is the inverse operation of exponentiation, and thus \( \log_{b} b \) effectively asks the question, 'To what power must we raise \( b \) to get \( b \) itself?' The answer is always 1, because any number to the power of 1 is itself. Knowing these simplification techniques is vital for students to effectively work with and understand logarithms.

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