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Chapter 4: Problem 65

Use properties of logarithms to condense logarithmic expression. Write the expression as a single logarithm whose coefficient is 1. Where possible, evaluate logarithmic expressions without using a calculator. \(\frac{1}{2}\left(\log _{5} x+\log _{5} y\right)-2 \log _{5}(x+1)\)

Short Answer

Expert verified
The single logarithmic expression is \(\log_5(\sqrt{xy} / (x + 1)^2)\).

Step by step solution

01

Apply log rule for the sum of logs

First, we apply the rule that the log of a product is the sum of the logs of its factors. This rule recognizes that \(\log_b(xy) = \log_b(x) + \log_b(y)\). Applying this rule to the first term in the given expression, we get \(\frac{1}{2}\log_5(xy)\).
02

Use properties of logarithms for multiplication by a constant

Next, we apply the property that a coefficient in front of a logarithm can be rewritten as an exponent of the argument of the logarithm. That means \(\frac{1}{2}\log_5(xy)\) becomes \(\log_5((xy)^{1/2})\), or \(\log_5(\sqrt{xy})\).
03

Apply the property of logs for power

Now we apply the rule \(\log_b(a^n) = n\log_b(a)\) to the second term in the expression, such that -2 \(\log_5(x+1)\) becomes \(\log_5((x+1)^{-2})\).
04

Simplify expression

Lastly, we simplify the expression by combining the logs. The result is \(\log_5(\sqrt{xy} / (x + 1)^2)\). Since the division corresponds to a subtraction in the logarithms, this step is also justified by the logarithm rules.

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Key Concepts

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

Understanding Logarithmic Expressions
Logarithmic expressions are the written form of logarithms, which are essentially the inverse operations of exponential expressions. A basic understanding of logarithms is paramount to dealing with more complex mathematical concepts. For instance, the logarithmic expression \(\log_b x\) asks the question: 'To what power must we raise the base \(b\) to receive the number \(x\)?' The importance of understanding comes into play when we need to condense or expand these expressions using various properties of logarithms.

As showcased in the original exercise, condensing a logarithmic expression involves combining multiple log terms into a single term. The process typically includes the use of several logarithm properties, but the key is to remember that each operation has a specific purpose and a strict set of rules governing it. The more comfortable you become with recognizing and applying these properties, the easier it will be to condense or expand logarithmic expressions accurately. It's like putting together a puzzle where each piece must fit perfectly based on the rules of logarithms.
Applying the Logarithm Product Rule
One crucial property that allows us to manipulate logarithms is the logarithm product rule. It states that the logarithm of a product is equal to the sum of the logarithms of each factor: \(\log_b(xy) = \log_b(x) + \log_b(y)\). In simple terms, if you have two numbers being multiplied, you can break down their logarithm into two separate terms.

When applying the logarithm product rule, as seen in the exercise, a term such as \(\frac{1}{2}(\log_5 x + \log_5 y)\) can be rewritten by combining the logs into a single log representing the square root of the product of \(x\) and \(y\), due to the coefficient \(\frac{1}{2}\). This is a transformative step that can simplify complex expressions and is often a prerequisite for solving logarithmic equations or further condensing the expressions.
Mastering the Logarithm Power Rule
The logarithm power rule is another significant tool in handling logarithms. It helps simplify expressions where a logarithm has an exponent. Formally, the power rule can be expressed as: \(\log_b(a^n) = n \log_b(a)\). This means you can take the exponent on the argument of a logarithm and 鈥榤ove鈥� it in front of the log.

Within the exercise context, the power rule is applied in reverse to move the coefficient in front of a log so that \( -2 \log_5(x+1) \) becomes \(\log_5((x+1)^{-2})\). This step is often used in conjunction with the product rule for further simplification. Understandably, it鈥檚 essential to gain fluency in switching between these forms through the power rule as it provides a pathway to simplify logarithmic expressions further or solve equations that would otherwise be too complicated.

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Most popular questions from this chapter

Students in a mathematics class took a final examination. They took equivalent forms of the exam in monthly intervals thereafter. The average score, \(f(t),\) for the group after \(t\) months was modeled by the human memory function \(f(t)=75-10 \log (t+1), \quad\) where \(\quad 0 \leq t \leq 12 . \quad\) Use \(\quad\) a graphing utility to graph the function. Then determine how many months elapsed before the average score fell below 65.

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. If \(x=\frac{1}{k} \ln y,\) then \(y=e^{k x}\)

Complete the table for a savings account subject to contimuous compounding ( \(A=P e^{n}\) ). Round answers to one decimal place. Amount Invested 8000 dollar Annual Interest Rate 20.3% Accumulated Amount 12,000 dollar Time \(t\) in Years _______

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 an equation to answer this question: How far from the eye of a hurricane is the barometric air pressure 29 inches of mercury? Use the \([\mathrm{TRACE}]\) and \([\mathrm{ZOOM}]\) features or the intersect command of your graphing utility to verify your answer.

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. The function \(W(t)=2600\left(1-0.51 e^{-0.075 t}\right)^{3}\) models the weight, \(W(t),\) in kilograms, of a female African elephant at age \(t\) years. (1 kilogram \(=2.2\) pounds) Use a graphing utility to graph the function. Then \([\mathrm{TRACE}]\) along the curve to estimate the age of an adult female elephant weighing 1800 kilograms.
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