/*! 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 85 You are given an equation and a ... [FREE SOLUTION] | 91Ó°ÊÓ

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You are given an equation and a root that was obtained in an example in the text. In each case: (a) verify (algebraically) that the root indeed satisfies the equation; and (b) use a calculator to check that the root satisfies the equation. [From Example \(3(a)] \quad \ln (\ln x)=2 ; x=e^{e^{2}}\)

Short Answer

Expert verified
The root \( x = e^{e^2} \) satisfies the equation \( \ln(\ln x) = 2 \) both algebraically and using a calculator.

Step by step solution

01

Understand the Given Equation and Root

The equation is \( \ln(\ln x) = 2 \), and the proposed root is \( x = e^{e^2} \). We need to verify if this value of \( x \) satisfies the equation.
02

Substitute the Root into the Equation

Substitute \( x = e^{e^2} \) into the equation \( \ln(\ln x) = 2 \) to verify:1. Calculate \( \ln x = \ln(e^{e^2}) = e^2 \), because the natural logarithm is the inverse of the exponential function.2. Substitute \( \ln x = e^2 \) into \( \ln(\ln x) = 2 \), which becomes \( \ln(e^2) \).
03

Simplify and Confirm the Equation Holds

Simplify \( \ln(e^2) = 2 \). This is true because the natural logarithm of \( e^2 \) is \( 2 \) due to the property \( \ln(e^a) = a \). Hence, \( \ln(\ln x) = 2 \) holds true, verifying the root algebraically.
04

Use a Calculator to Verify the Root

1. Calculate \( \ln x = \ln(e^{e^2}) \) using a calculator to get \( e^2 \).2. Then, calculate \( \ln(\ln x) = \ln(e^2) \) to check if it equals 2.3. The calculator should confirm that \( \ln(e^2) = 2 \), thereby verifying the solution numerically.

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Key Concepts

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

Exponential Functions
Exponential functions are fundamental in mathematics, and they are defined by expressions where the variable is in the exponent. An exponential function has the form \( f(x) = a^x \), where \( a \) is the base. These functions grow very quickly, especially as the exponent increases.
One special case of the exponential function is when the base is the mathematical constant \( e \), approximately equal to 2.718. The function \( e^x \) is known as the natural exponential function. This function is particularly important because it describes continuous growth or decay in natural processes.
In solving the given equation \( \, \ln (\ln x) = 2 \), we need to recognize that \( e^{e^2} \) is built on the principle of exponential function. Here, \( e^{e^2} \) means \( e \) raised to the power \( e^2 \); thus, it heavily relies on the properties and operations of exponential functions to solve or simplify the expressions.
Logarithmic Functions
Logarithmic functions are the inverse of exponential functions, and they are essential for solving equations that involve exponentials. The logarithm function with base \( a \) is commonly denoted as \( \log_a(x) \). In mathematics, we often use the natural logarithm, written as \( \ln(x) \), which has a base of \( e \).
The expression \( \ln( ext{something}) \) tells us what power we have to raise \( e \) to get the 'something'. Hence, \( \ln(e^2) = 2 \) because \( e \) raised to the power of 2 gives us \( e^2 \). This property is crucial when solving equations like \( \ln(\ln x) = 2 \).
Logarithmic functions make it possible to turn multiplicative problems into additive ones, facilitating a simpler approach to solving complex equations. This transformation simplifies the process of working with very large or very small numbers.
Equation Verification
Equation verification, in mathematical terms, involves confirming that a proposed solution satisfies a given equation. This is done through both algebraic manipulation and numerical calculation.
In the problem \( \ln(\ln x) = 2 \), verifying that \( x = e^{e^2} \) satisfies the equation involves substituting the value back into the equation. We check that all properties and transformations hold true throughout the calculation. First, substitute \( x = e^{e^2} \) into \( \ln(\ln x) \). By calculating each step, you ensure mathematical accuracy.
If the final results equate correctly to the original equation's right-hand side, you have verified the solution. This process is supported by using calculators to confirm that the arithmetic agrees with the theory.
Roots of Equations
Finding the roots of an equation means finding the values for which the equation holds true (is equal to zero in the general case). The roots are the solutions to the equation.
For the equation \( \ln(\ln x) = 2 \), the root given is \( x = e^{e^2} \). The procedure of finding this root involves transforming the equation using inverse functions or algebraic manipulation.
The root is a critical solution that satisfies the equation under all mathematical conditions applied during its derivation. Once you find a root, it should make the original equation true when substituted back. In practical applications, verifying roots can involve additional methods, including graphical or numerical techniques, to further confirm the found solutions.

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Most popular questions from this chapter

[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].

(a) Use a graphing utility to estimate the root(s) of the equation to the nearest one-tenth (as in Example 6). (b) Solve the given equation algebraically by first rewriting it in logarithmic form. Give two forms for each answer: an exact expression and a calculator approximation rounded to three decimal places. Check to see that each result is consistent with the graphical estimate obtained in part (a). $$10^{x^{2}}=40$$

The intensity of the sounds that the human ear can detect varies over a very wide range of values. For instance, a whisper from 1 meter away has an intensity of approximately \(10^{-10}\) watts per square meter \(\left(\mathrm{W} / \mathrm{m}^{2}\right)\), whereas, from a distance of 50 meters, the intensity of a launch of the Space Shuttle is approximately \(10^{8} \mathrm{W} / \mathrm{m}^{2} .\) For a sound with intensity \(I\), the sound level \(\beta\) is defined by $$ \beta=10 \log _{10}\left(I / I_{0}\right) $$ where the constant \(I_{0}\) is the sound intensity of a barely audible sound at the threshold of hearing. The units for the sound level \(\beta\) are decibels, abbreviated dB. (a) Solve the equation \(\beta=10 \log _{10}\left(1 / I_{0}\right)\) for \(I\) by first dividing by 10 and then converting to exponential form. (b) The sound level for a power lawnmower is \(\beta=100 \mathrm{db}\). and that for a cat purring is \(\beta=10\) db. Use your result in part (a) to determine how many times more intense is the power mower sound than the cat's purring.

If \(a^{2}+b^{2}=7 a b,\) where \(a\) and \(b\) are positive, show that $$\log \left[\frac{1}{3}(a+b)\right]=\frac{1}{2}(\log a+\log b)$$ no matter which base is used for the logarithms (but it understood that the same base is used throughout).

The population figures in Exercise 20 are actually for mid-2000. As of that time, the population of Mexico was 99.6 million, with a relative growth rate of \(2.0 \% /\) year. Let \(t=0\) correspond to June 2000 and use the exponential growth model \(\mathcal{N}(t)=99.6 e^{0.02 t}\) to verify that \(\mathcal{N}(0.5)>100\) Interpret the result.

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