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In Exercises 29 - 44, find the exact value of the logarithmic expression without using a calculator. (If this is not possible,state the reason.) \( \log_6 \sqrt[3]{6} \)

Short Answer

Expert verified
The exact value of \( \log_6 \sqrt[3]{6} \) is \( \frac{1}{3} \).

Step by step solution

01

Express the cube root in terms of exponents

To express the cube root of 6 in terms of exponents, we can write \( \sqrt[3]{6} \) as \( 6^{1/3} \).
02

Apply the definition of logarithm

A logarithm with base b and argument of b^a is equal to a. Therefore, \( \log_6 (6^{1/3}) \) is equal to \( \frac{1}{3} \).
03

State the final answer

So, the exact value of the logarithm \( \log_6 \sqrt[3]{6} \) is \( \frac{1}{3} \).

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Key Concepts

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

Logarithm Properties
Understanding logarithm properties is crucial for manipulating and simplifying logarithmic expressions effectively. A logarithm, essentially, is the inverse of exponentiation. It asks the question, 'To what power must the base number be raised to produce a certain value?' For instance, the logarithm of a number where the number is the base raised to a power, like in the expression \( \log_b(b^a) \), is simply \( a \). This property is known as the inverse property of logarithms.

Additional key properties include:
  • The Product Rule: \( \log_b(MN) = \log_b(M) + \log_b(N) \), meaning the logarithm of a product is the sum of the logarithms.
  • The Quotient Rule: \( \log_b\left(\frac{M}{N}\right) = \log_b(M) - \log_b(N) \), which states the logarithm of a quotient is the difference of the logarithms.
  • The Power Rule: \( \log_b(M^p) = p \cdot \log_b(M) \), indicating that the logarithm of a number raised to a power is the power times the logarithm of the number.
These properties are integral when simplifying logarithmic expressions, such as the exercise \( \log_6 \sqrt[3]{6} \), where using the Power Rule helps to quickly find the value without calculating the exact root.
Exponents and Radicals
Exponents and radicals are different ways to represent repeated multiplication. An exponent, such as \( b^n \), tells you to multiply the base \( b \) by itself \( n \) times. Conversely, a radical, often presented as a square root or cube root, represents the opposite operation—finding what number multiplied by itself a certain number of times gives you the original value.

Understanding the relationship between exponents and radicals is key to working with expressions involving roots without a calculator. A radical can always be converted to an exponent by remembering that the nth root of a number is the same as that number raised to the \( 1/n \) power. For example, \( \sqrt[3]{6} \) is equivalent to \( 6^{1/3} \). This conversion is particularly useful in combination with logarithm properties, as it allows the original expression to be re-written in a form that is much easier to manage by employing the rules of logarithms.
Solving Logarithms Without Calculator
While calculators are handy, knowing how to solve logarithms without one is a valuable skill, especially when dealing with simple logarithmic values that can be deduced easily. For example, in an expression like \( \log_6 \sqrt[3]{6} \), recognizing that \( \sqrt[3]{6} \) is \( 6^{1/3} \), and that a logarithm essentially asks what exponent is needed to get the value inside the log from the base, allows you to solve the problem straight forward.

Here are several tips for solving logarithms without a calculator:
  • Convert radicals to exponent form to make the relationship more explicit.
  • Use logarithm properties to rewrite the expression into a more manageable form.
  • Look for patterns or familiarity, such as a log's argument being the base raised to an exponent, which simplifies directly to the exponent.
  • If the argument is a known power of the base, the result is simply that known power.
By practicing these strategies and familiarizing yourself with the properties of logarithms, you can confidently tackle a variety of logarithmic expressions without relying on technology.

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Most popular questions from this chapter

Carbon \( 14 \) dating assumes that the carbon dioxide on Earth today has the same radioactive content as it did centuries ago. If this is true,the amount of \( ^{14} C \) absorbed by a tree that grew several centuries ago should be the same as the amount of \( ^{14} C \) absorbed by a tree growing today. A piece of ancient charcoal contains only \( 15\% \) as much radioactive carbon as a piece of modern charcoal. How long ago was the tree burned to make the ancient charcoal if the half-life of \( ^{14} C \) is \( 5715 \) years?

In Exercises 79 - 82, determine whether the statement is true or false. Justify your answer. The graph of a Gaussian model will never have an \( x \)-intercept.

The \( pH \) of a solution is decreased by one unit. The hydrogen ion concentration is increased by what factor?

A cup of water at an initial temperature of \( 78^{\circ}C \) is placed in a room at a constant temperature of \( 21^{\circ}C \). The temperature of the water is measured every 5 minutes during a half-hour period.The results are recorded as ordered pairs of the form \( (t \), \( T) \), where \( t \) is the time (in minutes) and \( T \) is the temperature (in degrees Celsius). \( \left(0, 78.0^{\circ}\right) \), \( \left(5 , 66.0^{\circ}\right) \), \( \left(10, 57.5^{\circ}\right) \), \( \left(15 , 51.2^{\circ}\right) \), \( \left(20 , 46.3^{\circ}\right) \), \( \left(25, 42.4^{\circ}\right) \), \( \left(30 , 39.6^{\circ}\right) \) (a) The graph of the model for the data should be asymptotic with the graph of the temperature of the room. Subtract the room temperature from each of the temperatures in the ordered pairs. Use a graphing utility to plot the data points \( \left(t , T\right) \) and \( \left(t, T - 21\right) \). (b) An exponential model for the data \( \left(t, T - 21\right) \) is given by \( T - 21 = 54.4\left(0.964\right)^t \) Solve for \( T \) and graph the model. Compare the result with the plot of the original data. (c) Take the natural logarithms of the revised temperatures. Use a graphing utility to plot the points \( \left(t, In\left(T - 21\right)\right) \) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form \( In\left(T - 21\right) = at + b \). Solve for \( T \), and verify that the result is equivalent to the model in part (b). (d) Fit a rational model to the data. Take the reciprocals of the \( y \)-coordinates of the revised data points to generate the points \( \dfrac{1}{T - 21} = at + b \). Solve for \( T \), and use a graphing utility to graph the rational function and the original data points. (e) Why did taking the logarithms of the temperatures lead to a linear scatter plot? Why did taking the reciprocals of the temperatures lead to a linear scatter plot?

In Exercises 39 - 44, use a graphing utility to construct a table of values for the function. Then sketch the graph of the function. \( f(x) = e^{-x} \)

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