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Chapter 4: Problem 59

Write the logarithmic expression as a single logarithm with coefficient \(1,\) and simplify as much as possible. (See Exercises \(5-6)\) $$ \frac{1}{2} \ln (x+1)-\frac{1}{2} \ln (x-1) $$

Short Answer

Expert verified
The expression simplifies to \( \frac{1}{2} \ln \left( \frac{x+1}{x-1} \right) \).

Step by step solution

01

Apply Logarithm Power Rule

Recall that the logarithm power rule states that for any logarithm \( \log_{b}(a^c) = c \log_{b}(a)\). Apply this rule to each term in the expression: \(\frac{1}{2} \ln (x+1) = \ln ((x+1)^{1/2})\) and \(\frac{1}{2} \ln (x-1) = \ln ((x-1)^{1/2})\). The expression becomes: \ln((x+1)^{1/2}) - \ln((x-1)^{1/2})\.
02

Apply Logarithm Quotient Rule

Use the logarithm quotient rule which states that \( \log_{b}(a) - \log_{b}(c) = \log_{b}\left( \frac{a}{c} \right) \). Applying this rule to the current expression yields: \ln \left( \frac{(x+1)^{1/2}}{(x-1)^{1/2}} \right)\.
03

Simplify the Expression

Simplify the expression inside the logarithm. The term \( \frac{(x+1)^{1/2}}{(x-1)^{1/2}} \) can be rewritten as \( \left( \frac{x+1}{x-1} \right)^{1/2} \). Thus, the expression becomes: \ln \left( \left( \frac{x+1}{x-1} \right)^{1/2} \right)\.
04

Apply Logarithm Power Rule Again

Apply the logarithm power rule one more time to simplify further. The rule states that \( \log_{b}(a^c) = c \log_{b}(a)\). So, \ln \left( \left( \frac{x+1}{x-1} \right)^{1/2} \right) = \frac{1}{2} \ln \left( \frac{x+1}{x-1} \right)\.

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Key Concepts

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

logarithm power rule
The logarithm power rule allows us to simplify expressions where an exponent is involved in a logarithm. The rule states that for any logarithm, if you have an exponent inside the logarithm, you can bring the exponent in front: \[\text{log}_{b}(a^c) = c \text{log}_{b}(a)\].
In our problem, we had the terms \$\frac{1}{2} \log_{b}(x+1)\$ and \$\frac{1}{2} \log_{b}(x-1)\$.
This can be simplified by applying the logarithm power rule as follows:
\[ \frac{1}{2} \log_{b}(x+1) = \log_{b}((x+1)^{1/2}) \]
\[ \frac{1}{2} \log_{b}(x-1) = \log_{b}((x-1)^{1/2}) \]

This step helps to prepare the expression for further simplification.
logarithm quotient rule
The logarithm quotient rule helps us manage the subtraction of logarithms. This rule states that the difference between two logarithms is the logarithm of the quotient of their arguments: \[\text{log}_{b}(a) - \text{log}_{b}(c) = \text{log}_{b}(a/c)\].
In our exercise, after applying the power rule, we had:
\$\log_{b}((x+1)^{1/2}) - \log_{b}((x-1)^{1/2})\$.
Using the quotient rule, it becomes:
\[\log_{b}\left( \frac{(x+1)^{1/2}}{(x-1)^{1/2}} \right)\]

This step simplifies our expression significantly, preparing it for final simplification.
expression simplification
Expression simplification refers to the process of making the mathematical expression as concise and simple as possible. After applying both the power and quotient rules, we need to further simplify the expression:
The term \[\frac{(x+1)^{1/2}}{(x-1)^{1/2}}\] can be written as \[\frac{(x+1)}{(x-1)}^{1/2}\].
Thus, the expression becomes:
\[\log_{b}\left(\frac{x+1}{x-1}\right)^{1/2}\]
Applying the logarithm power rule one last time, we get:
\[\left(\frac{1}{2}\right) \text{log}_{b}\left( \frac{x+1}{x-1}\right)\]
This is our final and simplified expression with a single logarithm:

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Most popular questions from this chapter

Solve the equation. Write the solution set with the exact values given in terms of common or natural logarithms. Also give approximate solutions to 4 decimal places. \(2^{1-6 x}=7^{3 x+4}\)

Determine if the statement is true or false. For each false statement, provide a counterexample. For example, \(\log (x+y) \neq \log x+\log y\) because \(\log (2+8) \neq \log 2+\log 8\) (the left side is 1 and the right side is approximately 1.204 ). $$ \log _{8}\left(\frac{1}{w}\right)=-\log _{8} w $$

Solve the equation. Write the solution set with the exact solutions. Also give approximate solutions to 4 decimal places if necessary. \(\log _{6}(7 x-2)=1+\log _{6}(x+5)\)

a. The populations of two countries are given for January 1,2000 , and for January 1,2010 . Write a function of the form \(P(t)=P_{0} e^{k t}\) to model each population \(P(t)\) (in millions) \(t\) years after January 1, 2000.$$ \begin{array}{|l|c|c|c|} \hline & \begin{array}{c} \text { Population } \\ \text { in 2000 } \\ \text { (millions) } \end{array} & \begin{array}{c} \text { Population } \\ \text { in 2010 } \\ \text { (millions) } \end{array} & \boldsymbol{P}(t)=\boldsymbol{P}_{0} e^{k t} \\ \hline \text { Switzerland } & 7.3 & 7.8 & \\ \hline \text { Israel } & 6.7 & 7.7 & \\ \hline \end{array}$$ b. Use the models from part (a) to predict the population on January \(1,2020,\) for each country. Round to the nearest hundred thousand. c. Israel had fewer people than Switzerland in the year 2000 , yet from the result of part (b), Israel will have more people in the year \(2020 ?\) Why? d. Use the models from part (a) to predict the year during which each population will reach 10 million if this trend continues.

Determine whether the two functions are inverses. \(h(x)=7 x-3\) and \(k(x)=\frac{x+3}{7}\)
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