/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 Simplify each expression. (a) ... [FREE SOLUTION] | 91Ó°ÊÓ

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Simplify each expression. (a) \(\ln e\) (b) \(\ln e^{-2}\) (c) \((\ln e)^{-2}\)

Short Answer

Expert verified
(a) 1; (b) -2; (c) 1.

Step by step solution

01

Simplify \( \ln e \)

The natural logarithm of \( e \), written as \( \ln e \), is equal to 1 because the natural logarithm is the inverse of the exponential function with base \( e \). Thus, we have: \( \ln e = 1 \).
02

Simplify \( \ln e^{-2} \)

Using the properties of logarithms, \( \ln e^{-2} \) can be rewritten as \(-2 \times \ln e \). Since \( \ln e = 1 \) from Step 1, this simplifies to \(-2 \times 1 = -2 \). Thus, \( \ln e^{-2} = -2 \).
03

Simplify \( (\ln e)^{-2} \)

We need to simplify \((\ln e)^{-2}\). From Step 1, we know that \(\ln e = 1\). Therefore, \((\ln e)^{-2} = (1)^{-2}\). Any number to the power of \(-2\), when it is 1, simply remains 1, because \(1^{-2} = 1\). Thus, \((\ln e)^{-2} = 1 \).

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Key Concepts

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

Natural Logarithm
The natural logarithm, denoted as \( \ln \), is a special kind of logarithm that uses the mathematical constant \( e \) as its base. The constant \( e \) is approximately equal to 2.71828 and is a fundamental constant in mathematics, much like \( \pi \). The natural logarithm is known for its unique property: \( \ln e = 1 \). This arises because logarithms count the number of times one number (the base) must be raised to get another number.
  • \( \ln e^x = x \) because \( e^x \) raised to \( e \)'s power results in \( e^x \).
  • \( \ln 1 = 0 \) because any number raised to the power of 0 is 1, including \( e^0 \).
The natural logarithm is particularly useful because it simplifies many expressions in mathematics and serves as a key function in calculus. It transforms the complexity of exponential growth into a simpler, additive process. This powerful tool helps streamline calculations involving growth and decay processes.
Properties of Logarithms
Logarithms, including natural logarithms, have several important properties that make them incredibly useful in mathematics. These properties help simplify expressions and solve equations more easily by transforming multiplicative processes into additive ones.
  • Product Property: \( \ln(ab) = \ln a + \ln b \). This property states that the logarithm of a product is the sum of the logarithms of the factors.
  • Quotient Property: \( \ln\left(\frac{a}{b}\right) = \ln a - \ln b \). The logarithm of a quotient is the difference of the logarithms.
  • Power Property: \( \ln(a^b) = b \cdot \ln a \). This property indicates that the logarithm of a power is the exponent multiplied by the logarithm of the base.
Applying these properties allows us to simplify expressions like \( \ln e^{-2} \) to \( -2 \cdot \ln e \), which becomes easier to compute when you know \( \ln e \) is 1. These properties are fundamental for algebraic manipulations and solving exponential equations.
Exponential Functions
Exponential functions are vital in mathematics because they describe phenomena where the rate of change is proportional to the current value, such as population growth, radioactive decay, or interest calculations. An exponential function is generally expressed as \( f(x) = a \cdot e^{bx} \), where \( a \) is a constant, \( e \) is the base of natural logarithms, and \( bx \) is the exponent.
  • Exponential functions grow extremely fast. As \( x \) increases, \( e^x \) grows exponentially larger.
  • Exponential decay occurs when the exponent is negative, causing the function to decrease rapidly as \( x \) increases.
The relationship between exponential functions and logarithms is significant. Since the logarithm is the inverse of the exponential function, knowing properties of logarithms helps us solve equations involving exponentials. For example, if \( f(x) = e^x \), then \( f^{-1}(x) = \ln x \), showing how logarithms "undo" the action of exponentials. This ties into simplifying expressions such as \( \ln e^{-2} \) or solving equations of the form \( e^x = y \).

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Most popular questions from this chapter

The age of some rocks can be estimated by measuring the ratio of the amounts of certain chemical elements within the rock. The method known as the rubidium-strontium method will be discussed here. This method has been used in dating the moon rocks brought back on the Apollo missions. Rubidium-87 is a radioactive substance with a half-life of \(4.7 \times 10^{10}\) years. Rubidium- 87 decays into the substance strontium- \(87,\) which is stable (nonradioactive). We are going to derive the following formula for the age of a rock: $$T=\frac{\ln \left[\left(\mathcal{N}_{s} / \mathcal{N}_{r}\right)+1\right]}{-k}$$ where \(T\) is the age of the rock, \(k\) is the decay constant for rubidium-87, \(\mathcal{N}_{s}\) is the number of atoms of strontium-87 now present in the rock, and \(\mathcal{N},\) is the number of atoms of rubidium-87 now present in the rock. (a) Assume that initially, when the rock was formed, there were \(\mathcal{N}_{0}\) atoms of rubidium-87 and none of strontium-87. Then, as time goes by, some of the rubidium atoms decay into strontium atoms, but the to tal number of atoms must still be \(\mathcal{N}_{0} .\) Thus, after \(T\) years, we have \(\mathcal{N}_{0}=\mathcal{N}_{r}+\mathcal{N}_{s}\) or, equivalently, $$ \mathcal{N}_{s}=\mathcal{N}_{0}-\mathcal{N}_{r}$$However, according to the law of exponential decay for the rubidium-87, we must have \(\mathcal{N}_{r}=\mathcal{N}_{0} e^{k T} .\) Solve this equation for \(\mathcal{N}_{0}\) and then use the result to eliminate \(\mathcal{N}_{0}\) from equation \((1) .\) Show that the result can be written $$\mathcal{N}_{s}=\mathcal{N}_{r} e^{-k T}-\mathcal{N}_{r}$$ (b) Solve equation (2) for \(T\) to obtain the formula given at the beginning of this exercise.

(a) Let \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t} .\) Show that \([\mathcal{N}(t+1)-\mathcal{N}(t)] / \mathcal{N}(t)=e^{k}-1 .\) (This is actually done in detail in the text. So, ideally, you should look back only if you get stuck or want to check your answer.) (b) Assume as given the following approximation, which was introduced in Exercise 26 of Section 5.2 \(e^{x} \approx x+1 \quad\) provided \(x\) is close to zero Use this approximation to explain why \(e^{k}-1 \approx k\) provided that \(k\) is close to zero. Remark: Combining this result with that in part (a), we conclude that the relative growth rate for the function \(\mathcal{N}(t)=\mathcal{N}_{0} e^{k t}\) is approximately equal to the growth constant \(k .\) As explained in the text, this is one of the reasons why in applications we've not distinguished between the relative growth rate and the decay constant \(k\)

Graph each function and specify the domain, range, intercept(s), and asymptote. $$y=\ln (x+e)$$

[From Example 9(b)] Solve the inequality \(x^{2}-2 x-24 \leq 0 .\) You should find that the solution set is the closed interval [-4,6].

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$10^{-x^{2}} \leq 10^{-12}$$

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