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Using Properties of Logarithms In Exercises \(15-20\) , use the properties of logarithms to rewrite and simplify the logarithmic expression. $$\ln \frac{6}{e^{2}}$$

Short Answer

Expert verified
The simplified version of the given logarithmic expression \( \ln \frac{6}{e^{2}} \) is \( \ln(6) - 2 \) .

Step by step solution

01

Apply the rule \( \ln(a/b) = \ln(a) - \ln(b) \)

Split the given logarithmic expression \( \ln \frac{6}{e^{2}} \) into two, so it becomes \( \ln(6) - \ln(e^{2}) \) . This rule allows us to simplify complex logarithmic expressions containing division.
02

Apply the rule \( \ln(a^n) = n* \ln(a) \)

When we simplify \( \ln(e^{2}) \) using the rule, it becomes \( 2*\ln(e) \). This rule is useful when we want to simplify logarithmic expressions containing exponents.
03

Reduce \( \ln(e) \)

Now we know that \( \ln(e) \) equals 1 because logarithm base \( e \) of \( e \) is one. So, the expression becomes \( \ln(6) - 2*1 \) or \( \ln(6) - 2\) .

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Key Concepts

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

Natural Logarithm Simplification
Simplifying natural logarithms can be quite straightforward once you grasp the fundamental rules governing their manipulation. The natural logarithm, represented as \( \ln(x) \), is the logarithm to the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. When simplifying \( \ln \), it's essential to remember that \( \ln(e^x) = x \) and \( \ln(1) = 0 \).

Using these properties, many complex expressions become simpler. For example, \( \ln(e^3) \) can be quickly simplified to 3 since the power to which \( e \) must be raised to obtain \( e^3 \) is just 3. This direct simplification is invaluable for working through higher-level mathematics where natural logs frequently appear.
Logarithmic Expression
A logarithmic expression involves finding the power to which a number, called the base, must be raised to produce another number. For instance, the expression \( \ln(x) \) asks at what power must \( e \) be raised to result in \( x \). Understanding how to manipulate these expressions is a cornerstone in mathematics, especially when dealing with exponential functions and growth models.

When you're confronted with a complex logarithmic expression involving division or multiplication, the properties of logarithms allow you to break down the expression into a series of simpler terms. This is particularly useful not only in simplification but also in solving equations where the variable you seek is within a logarithm.
Logarithmic Division Rule
Mastering the logarithmic division rule is key to breaking down complex expressions into something more manageable. The rule states that the logarithm of a quotient is equal to the difference of the logarithms. In mathematical terms, \( \ln(\frac{a}{b}) = \ln(a) - \ln(b) \).

This is especially helpful when you have a fraction inside a logarithm. For example, if you're given \( \ln(\frac{x}{y}) \), you can simplify it to \( \ln(x) - \ln(y) \) using the division rule. This rule is a vital piece of the puzzle for simplifying and solving logarithmic problems efficiently and effectively.
Logarithmic Exponent Rule
The logarithmic exponent rule can be a powerhouse tool for simplification. It tells us that the logarithm of a number raised to a power is equal to that power times the logarithm of the number itself. Mathematically, we express this as \( \ln(a^n) = n* \ln(a) \).

This rule is pivotal when dealing with expressions where the argument of the logarithm has an exponent. To put this into action, consider the term \( \ln(x^3) \); it simplifies to \( 3*\ln(x) \) by applying the logarithmic exponent rule. This fundamental concept is often applied in calculus and algebra, and it's a major asset when working through problems involving growth processes, decay, and many natural phenomena modeled by exponential functions.

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Most popular questions from this chapter

Forensics At 8: 30 A.M., a coroner went to the home of a person who had died during the night. In order to estimate the time of death, the coroner took the person's temperature twice. At 9: 00 A.M. the temperature was \(85.7^{\circ} \mathrm{F},\) and at 11: 00 A.M. the temperature was \(82.8^{\circ} \mathrm{F} .\) From these two temperatures, the coroner was able to determine that the time elapsed since death and the body temperature were related by the formula \(t=-10 \ln \frac{T-70}{98.6-70}\) where \(t\) is the time in hours elapsed since the person died and \(T\) is the temperature (in degrees Fahrenheit) of the person's body. (This formula comes from a general cooling principle called Newton's Law of Cooling. It uses the assumptions that the person had a normal body temperature of \(98.6^{\circ} \mathrm{F}\) at death and that the room temperature was a constant \(70^{\circ} \mathrm{F}\).) Use the formula to estimate the time of death of the person.

Think About It For how many integers between 1 and 20 can you approximate natural logarithms, given the values \(\ln 2 \approx 0.6931,\) ln \(3 \approx 1.0986,\) and ln 5\(\approx 1.6094 ?\) Approximate these logarithms (do not use a calculator).

Due to the installation of noise suppression materials, the noise level in an auditorium decreased from 93 to 80 decibels. Find the percent decrease in the intensity level of the noise as a result of the installation of these materials.

Expanding a Logarithmic Expression In Exercises \(37-58\) , use the properties of logarithms to expand the expression as a sum, difference, and or constant multiple of logarithms. (Assume all variables are positive.) $$\ln x y z^{2}$$

Due to the installation of a muffler, the noise level of an engine decreased from 88 to 72 decibels. Find the percent decrease in the intensity level of the noise as a result of the installation of the muffler.

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