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Chapter 4: Problem 46

Write the expression as the logarithm of a single quantity. $$ \frac{1}{3}\left[2 \ln (x+3)+\ln x-\ln \left(x^{2}-1\right)\right] $$

Short Answer

Expert verified
The expression simplifies to \( \ln [\frac{(x^2+6x+9)x}{(x^{2}-1)^{1/3}}] \)

Step by step solution

01

Identify the logarithmic properties applicable to the expression

The expression given is a combination of the logarithm properties, namely, the product rule, quotient rule, and the power rule. They must be applied accordingly where each term is added or subtracted. The overall product will be multiplied by the factor \(\frac{1}{3}\) at the end.
02

Apply the product rule

Combine the two terms with addition towards the middle of the expression, \(2 \ln (x+3) + \ln x\). According to the product rule, the sum of two logarithms can be expressed as the logarithm of their product. This gives \( \ln [(x+3)^2 \cdot x] = \ln [(x^2+6x+9)x]\).
03

Apply the quotient rule

Now, combine the previously found term, \(\ln [(x^2+6x+9)x]\) with the third term \(\ln (x^{2}-1)\), which is to be subtracted. According to the quotient rule, the difference between two logarithms can be expressed as the logarithm of the division of the terms. This gives us \( \ln \frac{(x^2+6x+9)x}{x^{2}-1} \).
04

Apply the power rule

The overall expression is multiplied by \(\frac{1}{3}\). We can introduce this into the expression by using the power rule. This transforms the expression to \( \ln [\frac{(x^2+6x+9)x}{(x^{2}-1)^{1/3}}] \).

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Most popular questions from this chapter

Population Growth The number of a certain type of bacteria increases continuously at a rate proportional to the number present. There are 150 present at a given time and 450 present 5 hours later. $$ \begin{array}{l}{\text { (a) How many will there be } 10 \text { hours after the initial time? }} \\ {\text { (b) How long will it take for the population to double? }} \\ {\text { (c) Does the answer to part (b) depend on the starting }} \\ {\text { time? Explain your reasoning. }}\end{array} $$

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