/*! 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 34 Solve for \(x\) algebraically. ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Solve for \(x\) algebraically. $$\ln (\ln x)=2$$

Short Answer

Expert verified
The solution for \(x\) is \(x = e^{e^2}\)

Step by step solution

01

Change the Perspective of the Equation

To deal with logarithms, first, it's beneficial to re-write the equation in a more friendly way. The equation \(\ln (\ln x)=2\) can be expressed as \(e^{\ln(\ln x)} = e^2\) using properties of logarithms.
02

Simplify the Left Side

The left side simplification leads to \(\ln x = e^2\) because the operation \(e\) to the logarithm is the inverse operation and they can cancel each other out.
03

Change the Perspective Again

To further simplify, the equation \(\ln x = e^2\) can be transformed into \(x = e^{e^2}\) by using the inverse of the natural logarithm function.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Properties of Logarithms
Logarithms have several important properties that make them a powerful tool in mathematics. Understanding these properties is key to solving equations involving logarithms.
  • Product Rule: The logarithm of a product is the sum of the logarithms of the factors. Mathematically, it's expressed as \( \log_b(xy) = \log_b(x) + \log_b(y) \).
  • Quotient Rule: The logarithm of a quotient is the difference of the logarithms. It can be expressed as \( \log_b\left(\frac{x}{y}\right) = \log_b(x) - \log_b(y) \).
  • Power Rule: The logarithm of a power is the exponent times the logarithm of the base. This is written as \( \log_b(x^n) = n \cdot \log_b(x) \).
When solving \( \ln(\ln x) = 2 \), we used the fact that applying the exponential function to a logarithm essentially cancels the logarithm due to their inverse relationship. This manipulation simplifies the equation greatly and is guided by understanding these properties.
Inverse Functions
Inverse functions are integral in understanding equations, especially when dealing with logarithms and exponentials. Two functions are inverses of each other if they "undo" each other's operations.
  • The natural log function \( \ln(x) \) and the exponential function \( e^x \) are classical examples of inverse functions.
This means that if you apply \( e^x \) to \( \ln(x) \) or vice versa, you will get the original input back. For example, in the problem \( \ln(\ln x) = 2 \), after rewriting it as \( e^{\ln(\ln x)} = e^2 \), the \( e \) and \( \ln \) operations cancel each other on \( \ln x \) leading to \( \ln x = e^2 \). This is the power of working with inverses: simplifying complex expressions and easily isolating variables.
Natural Logarithm
The natural logarithm, denoted as \( \ln(x) \), is a logarithm with the base \( e \), where \( e \) is an irrational constant approximately equal to 2.71828.
  • The natural log is commonly used in calculations involving growth and decay processes, and in this context, as seen in mathematical derivations and transformations.
In solving \( \ln(\ln x) = 2 \), the inner \( \ln x \) requires manipulation by understanding its foundational role in mathematics. To find \( x \), recognizing that expressing the natural logarithm in its exponential form \( x = e^{e^2} \) employs the concept of reversing the function. The result is critical in further applications like compound interest calculations, natural growth modeling, and logarithmic scales.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Explain how to use the graph of the first function \(f\) to produce the graph of the second function \(F\). $$f(x)=4^{x}, F(x)=4^{x}-3$$

Use a graphing utility to graph each function. If the function has a horizontal asymptote, state the equation of the horizontal asymptote. $$f(x)=\frac{e^{x}+e^{-x}}{2}$$

Assuming that air resistance is proportional to velocity, the velocity \(v,\) in feet per second, of a falling object after \(t\) seconds is given by \(v=64\left(1-e^{-t / 2}\right)\) a. Graph this equation for \(t \geq 0\) b. Determine algebraically, to the nearest 0.1 second, when the velocity is 50 feet per second. c. Determine the horizontal asymptote of the graph of \(v\). d. Write a sentence that explains the meaning of the horizontal asymptote in the context of this application.

Explain how to use the graph of the first function \(f\) to produce the graph of the second function \(F\). $$f(x)=\left(\frac{5}{2}\right)^{x}, F(x)=-\left[\left(\frac{5}{2}\right)^{x}\right]$$

A medical care package is air lifted and dropped to a disaster area. During the free-fall portion of the drop, the time, in seconds, required for the package to obtain a velocity of \(v\) feet per second is given by the function $$t=2.43 \ln \frac{150+v}{150-v}, \quad 0 \leq v<150$$ a. Determine the velocity of the package 5 seconds after it is dropped. Round to the nearest foot per second. b. Determine the vertical asymptote of the function. c. Write a sentence that explains the meaning of the vertical asymptote in the context of this application.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.