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Chapter 4: Problem 24

In Exercises \(21-30,\) write each logarithm as a sum and\or difference of logarithmic expressions. Eliminate exponents and radicals and evaluate logarithms wherever possible. Assume that \(a, x, y\) \(z>0\) and \(a \neq 1\). $$\ln \frac{\sqrt[4]{y^{3}}}{e^{5}}$$

Short Answer

Expert verified
The simplified logarithmic expression is \(\frac{3}{4}\) \(ln(y)\) - 5.

Step by step solution

01

Apply the Quotient Rule

The quotient rule in logarithms says that the logarithm of a quotient can be written as the difference of the logarithms. So, the logarithm expression is broken down into difference of two logarithms as follows: \(ln(\frac{\sqrt[4]{y^{3}}}{e^{5}})\) = \(ln(\sqrt[4]{y^{3}})\) - \(ln(e^{5})\).
02

Evaluate the Logarithm of Base e

Note that the natural logarithm \(ln(e^{5})\) is just 5, since the natural logarithm has a base of e. Therefore, the expression can be further simplified as follows: \(ln(\sqrt[4]{y^{3}})\) - 5.
03

Apply the Rule of Logarithm of a Root

The rule of logarithm of a root says that the logarithm of a root can be written as a fraction. Let's apply this rule to simplify the term \(ln(\sqrt[4]{y^{3}})\). We get: \(\frac{3}{4}\) \(ln(y)\) - 5.

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Key Concepts

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

Quotient Rule
The quotient rule in logarithms is a handy tool that helps us break down complex expressions. When you have a logarithm of a quotient, such as \( \ln \left( \frac{A}{B} \right) \), it can be simplified to \( \ln(A) - \ln(B) \). This conversion makes it easier to handle and solve logarithmic expressions. In practical terms:
  • It simplifies the evaluation of logarithmic expressions.
  • Helps identify parts of the expression that can be solved individually.
  • Makes complex expressions more manageable by separating them into simpler components.
In the given exercise, we used the quotient rule to simplify \( \ln \left( \frac{\sqrt[4]{y^{3}}}{e^{5}} \right) \) to \( \ln(\sqrt[4]{y^{3}}) - \ln(e^{5}) \). Splitting complex fractions into differences makes the next steps much clearer.
Natural Logarithm
A natural logarithm, denoted as \( \ln(x) \), has a special base known as 'e', which is approximately 2.718. The main purpose of natural logarithms is to simplify expressions where this base is involved.Understanding the properties:
  • The inverse of natural logarithms is the exponential function \( e^x \).
  • With base 'e', \( \ln(e^n) \) is simplified directly to 'n'.
  • It appears frequently in calculus and applied mathematics because of its natural growth patterns.
In our problem, \( \ln(e^{5}) \) simplifies to 5 because the effect of the base 'e' and the logarithm cancel each other out entirely. This property made it easy for us to reduce the expression to \( \ln(\sqrt[4]{y^{3}}) - 5 \).
Logarithm of a Root
When dealing with logarithms of roots, understanding and applying the rule for roots helps simplify expressions substantially. The rule states that the logarithm of a root, such as \( \sqrt[n]{A} \), can be expressed as a fraction: \( \frac{1}{n} \ln(A) \).Key points to remember:
  • Converting roots into fractional exponents aids in simplification.
  • Allows for the transformation of complex numbers into more easily handled terms.
  • This principle holds whether the root is square, cube, fourth, or any other kind.
In the example, \( \ln(\sqrt[4]{y^{3}}) \) was rewritten as \( \frac{3}{4} \ln(y) \), because the exponent 3 on \/( quantity \( y \) is multiplied by the fraction \( \frac{1}{4} \) from the fourth root. Applying this rule enables us to further simplify and evaluate very complex-looking logarithms.

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Most popular questions from this chapter

Evaluate the given quantity by referring to the function \(f\) given in the following table. $$\begin{array}{cc}x & f(x) \\\\-2 & 1 \\\\-1 & 2 \\\0 & 0 \\\1 & -1 \\\2 & -2\end{array}$$ $$f^{-1}(2)$$

The spread of a disease can be modcled by a logistic function. For example, in carly 2003 there was an outbreak of an illness called SARS (Severe Acute Respiratory Syndrome) in many parts of the world. The following table gives the total momber of cases in Canada for the wecks following March 20,2003 (Source: World Health Organization) (Note: The total number of cases dropped from 149 to 140 between weeks 3 and 4 because some of the cases thought to be SARS were reclassified as other discases.) $$\begin{array}{|c|c|}\hline\text { Weeks since } & \\\\\text { March } 20,2003 & \text { Total Cases } \\\0 & 9 \\\1 & 62 \\\2 & 132 \\\3 & 149 \\\4 & 140 \\\5 & 216 \\\6 & 245 \\\7 & 252 \\\8 & 250 \\\\\hline\end{array}$$ (a) Explain why a logistic function would suit this data well. (b) Make a scatter plot of the data and find the logistic function of the form \(f(x)=\frac{\epsilon}{1+a \varepsilon^{-1}}\) that best fits the data. (c) What docs \(c\) signify in your model? (d) The World Health Organization declared in July 2003 that SARS no longer posed a threat in Canada. By analyring this data, explain why that would be so.

The cost of removing chemicals from drinking water depends on how much of the chemical can safcly be left behind in the water. The following table lists the annual removal costs for arsenic in terms of the concentration of arsenic in the drinking water. (Source: Environmental Protection Agency) $$\begin{array}{|c|c|}\hline\text { Arsenic Concentration } & \text { Annual Cost } \\\\\text { (micrograms per liter) } & \text { (millions of dollars) } \\\\\hline 3 & 645 \\\5 & 379 \\\10 & 166 \\\20 & 65\\\ \hline\end{array}$$ (a) Interpret the data in the table. What is the relation between the amount of arsenic left behind in the removal process and the annual cost? (One microgram is equal to \(10^{-6}\) gram.) (b) Make a scatter plot of the data and find the exponential function of the form \(C(x)=C a^{*}\) that best fits the data. Here, \(x\) is the arscnic concentration. (c) Why must \(a\) be less than 1 in your model? (d) Using your model, what is the annual cost to obtain an arsenic concentration of 12 micrograms per liter? (e) It would be best to have the smallest possible amount of arsenic in the drinking water, but the cost may be prohibitive. Use your model to calculate the annual cost of processing such that the concentration of arsenic is only 2 micrograms per liter of water. Interpret your result.

Find the interest rate \(r\) if the interest on the initial deposit is compounded continuously and no withdrawals or further deposits are made. Initial amount: \(\$ 4000 ;\) Amount in 8 years: \(\$ 6000\)

Solve the logarithmic equation and eliminate any extraneous solutions. If there are no solutions, so state. $$\log (x+5)-\log \left(4 x^{2}+5\right)=0$$
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