/*! 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 66 Simplify the expression. $$ ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 66

Simplify the expression. $$ \ln e^{x^{2}+1} $$

Short Answer

Expert verified
\(x^{2} + 1\)

Step by step solution

01

- Understand the problem

To simplify the expression \(\ln e^{x^{2}+1}\), recognize that it involves both the natural logarithm (\(\ln\)) and the exponential function (\(e^{x}\)).
02

- Applying the properties of logarithms

Use the property \(\ln(e^{a}) = a\). This property tells us that the natural logarithm of \(e\) raised to a power returns that power.
03

- Simplifying the expression

Based on the property, \[ \ln(e^{x^{2}+1}) = x^{2}+1 \].

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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 \(\text{ln}\), is a logarithm to the base e, where e is an irrational and transcendental number approximately equal to 2.71828. The natural logarithm is widely used in mathematics and physics because of its unique properties, especially in relation to exponential functions.
\(\text{ln}(x)\) is the power to which e must be raised to get the number x. More formally, if y = \(\text{ln}(x)\), then \(\text{e}^y = x\).
Some important points about natural logarithms:
  • The natural logarithm of 1 is 0, because \(\text{e}^0 = 1\).
  • \
Exponential Functions
Exponential functions feature a constant base raised to a variable exponent. The general form is \(f(x) = a^{x}\), where a is a positive real number. When the base is e, this specific exponential function becomes very significant in various fields.
Some important characteristics of exponential functions:
  • The function passes through the point (0,1) because any number raised to the power of zero equals one.
  • Exponential growth means the quantity increases quickly, which is why it's used to model population growth, radioactive decay, and interest calculations.
In the problem, we have \(e^{x^2+1}\), where the exponent is \(x^2+1\), demonstrating the typical structure of an exponential function.
Understanding that an exponential function can be inversed by its reflexive logarithmic function is crucial for simplifying the given problem.
Properties of Logarithms
Logarithms have unique properties making them very useful for simplifying complex expressions. Here are a few essential properties:
  • Product Property: \(\text{ln}(a \times b) = \text{ln}(a) + \text{ln}(b)\)
  • Quotient Property: \(\text{ln}(\frac{a}{b}) = \text{ln}(a) - \text{ln}(b)\)
  • Power Property: \(\text{ln}(a^b) = b \times \text{ln}(a)\)
For the given problem \(\text{ln}(e^{x^2+1})\), we applied the power property.
According to the power property, \(\text{ln}(e^{a}) = a\).
So, for our example, \(\text{ln}(e^{x^{2}+1}) = x^{2}+1\). This shows how logarithmic properties allow us to simplify logarithmic expressions efficiently.

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Most popular questions from this chapter

Given the functions defined by \(f(x)=2 x-1\) and \(g(x)=\frac{x+1}{2}\), a. Graph \(y=f(x), y=g(x),\) and the line \(y=x .\) Does the graph suggest that \(f\) and \(g\) are inverses? Why? b. Enter the following functions into the graphing editor. ( $$\mathrm{Y}_{1}=2 x-1$$ \(\mathrm{Y}_{2}=(x+1) / 2\) \(\mathrm{Y}_{3}=\mathrm{Y}_{1}\left(\mathrm{Y}_{2}\right)\) \(\mathrm{Y}_{4}=\mathrm{Y}_{2}\left(\mathrm{Y}_{1}\right)\) c. Create a table of points showing \(Y_{3}\) and \(Y_{4}\) for several values of \(x\). (Hint: Use the right and left arrows to scroll through the table editor to show functions \(Y_{3}\) and \(Y_{4}\).) Does the table suggest that \(f\) and \(g\) are inverses? Why?

(See Example 8 ) a. Estimate the value of the logarithm between two consecutive integers. For example, \(\log _{2} 7\) is between 2 and 3 because \(2^{2}<7<2^{3}\). b. Use the change-of-base formula and a calculator to approximate the logarithm to 4 decimal places. c. Check the result by using the related exponential form. $$ \log _{2} 15 $$

Given \(f(x)=b^{x},\) then \(f^{-1}(x)=\) _____ for \(b>0\) and \(b \neq 1\).

Explain why the product property of logarithms does not apply to the following statement. $$ \begin{array}{l} \log _{5}(-5)+\log _{5}(-25) \\ \quad=\log _{5}[(-5)(-25)] \\ \quad=\log _{5} 125=3 \end{array} $$

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 ). $$ \ln 10=\frac{1}{\log e} $$
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