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Chapter 3: Problem 32

Use properties of logarithms to expand each logarithmic expression as much as possible. Where possible, evaluate logarithmic expressions without using a calculator. $$\log \sqrt[5]{\frac{x}{y}}$$

Short Answer

Expert verified
\(\frac{1}{5} \log x - \frac{1}{5} \log y\)

Step by step solution

01

Apply the power rule

Begin by recognizing that the expression inside the logarithm is a fifth root. The power rule for logarithms states that \(\log a^n = n \log a\). We can apply this rule to the given expression. The fifth root of an expression is the same as raising that expression to the power of \(\frac{1}{5}\). Hence, we rewrite the given expression as: \[ \frac{1}{5} \log \left(\frac{x}{y}\right) \]
02

Apply the quotient rule

The quotient rule for logarithms states that \(\log \frac{a}{b} = \log a - \log b\). We can apply this rule to the fraction \(\frac{x}{y}\) inside the logarithm. Hence, we rewrite the expression as: \[ \frac{1}{5} (\log x - \log y) \]
03

Apply the properties of real numbers

By applying the properties of real numbers, we use the distributive law to multiply each term by \(\frac{1}{5}\). The exercise is then expanded to: \[ \frac{1}{5} \log x - \frac{1}{5} \log y \]

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Key Concepts

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

Logarithmic Properties
If you're studying logarithms, understanding their properties is essential. These properties simplify complex expressions and make calculations easier. Three fundamental properties are the product rule, the quotient rule, and the power rule. Each of these rules help break down complex logarithmic expressions.

  • The product rule states that the log of a product equals the sum of the logs: \( \log(ab) = \log a + \log b \).
  • The quotient rule states that the log of a quotient equals the difference of the logs: \( \log\frac{a}{b} = \log a - \log b \).
  • The power rule states that the log of a power is the exponent times the log of the base: \( \log(a^n) = n \log a \).
These properties can transform difficult logarithmic expressions into simpler terms. Consider them essential tools in your math toolbox. They are not only applicable to numeric values but are widely used in calculations involving algebraic variables.
Power Rule
The power rule for logarithms helps to simplify expressions involving exponents. Understanding this rule can make working with powers in log expressions much easier.

Imagine you have an expression like \( \log(a^n) \). According to the power rule, you can simplify this to \( n \log a \). This is particularly useful when you have roots involved because roots can be expressed as fractional exponents.

Let's relate this back to the original exercise where the expression is \( \log \sqrt[5]{\frac{x}{y}} \). The fifth root can be rewritten as raising the expression to the power of \( \frac{1}{5} \). Applying the power rule, the expression becomes \( \frac{1}{5} \log(\frac{x}{y}) \). This gives a simpler form that is easier to work with in subsequent steps.
Quotient Rule
The quotient rule is another powerful property of logarithms. It tells us how to deal with logs of fractions, which is really handy in simplifying expressions.

The rule is simple: for a fraction \( \frac{a}{b} \), the log of this fraction equals the log of the numerator minus the log of the denominator. So, \( \log \frac{a}{b} = \log a - \log b \). This rule is key when you need to break down expressions where one variable is divided by another.

In the original exercise, we used the quotient rule to transform \( \log \left(\frac{x}{y}\right) \) into \( \log x - \log y \). After this transformation, we then applied the power rule which brought the fractional power to the front, creating a completely expanded and simplified expression: \( \frac{1}{5}(\log x - \log y) \). As a result, it becomes \( \frac{1}{5} \log x - \frac{1}{5} \log y \), showcasing the simplicity that logarithmic properties bring to complex expressions.

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Most popular questions from this chapter

Find the inverse of \(f(x)=x^{2}+4, x \geq 0\) (Section \(1.8, \text { Example } 7)\).

In parts (a)-(c), graph \(f\) and \(g\) in the same viewing rectangle. a. \(f(x)=\ln (3 x), g(x)=\ln 3+\ln x\) b. \(f(x)=\log \left(5 x^{2}\right), g(x)=\log 5+\log x^{2}\) c. \(f(x)=\ln \left(2 x^{3}\right), g(x)=\ln 2+\ln x^{3}\) d. Describe what you observe in parts (a)-(c). Generalize this observation by writing an equivalent expression for \(\log _{b}(M N),\) where \(M>0\) and \(N>0\) e. Complete this statement: The logarithm of a product is equal to____.

Hurricanes are one of nature's most destructive forces. These low-pressure areas often have diameters of over 500 miles. The function \(f(x)=0.48 \ln (x+1)+27\) models the barometric air pressure, \(f(x),\) in inches of mercury, at a distance of \(x\) miles from the eye of a hurricane. Use this function to solve Exercises \(133-134\) Graph the function in a [0,500,50] by [27,30,1] viewing rectangle. What does the shape of the graph indicate about barometric air pressure as the distance from the eye increases?

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. As the number of compounding periods increases on a fixed investment, the amount of money in the account over a fixed interval of time will increase without bound.

Exercises \(153-155\) will help you prepare for the material covered in the next section. U.S. soldiers fight Russian troops who have invaded New York City. Incoming missiles from Russian submarines and warships ravage the Manhattan skyline. It's just another scenario for the multi-billion-dollar video games Call of Duty, which have sold more than 100 million games since the franchise's birth in 2003 The table shows the annual retail sales for Call of Duty video games from 2004 through 2010 . Create a scatter plot for the data. Based on the shape of the scatter plot, would a logarithmic function, an exponential function, or a linear function be the best choice for modeling the data? $$\begin{array}{cc} \hline \text { Year } & \begin{array}{c} \text { Retail Sales } \\ \text { (millions of dollars) } \end{array} \\ \hline 2004 & 56 \\ 2005 & 101 \\ 2006 & 196 \\ 2007 & 352 \\ 2008 & 436 \\ 2009 & 778 \\ 2010 & 980 \end{array}$$
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