/*! 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 69 (a) Find the domain of the funct... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Find the domain of the function \(f\) defined by \(f(x)=\ln (\ln x)\) (b) Find \(f^{-1}(x)\) for the function \(f\) in part (a).

Short Answer

Expert verified
The domain of \( f(x) = \ln(\ln x) \) is \( x > e \). The inverse is \( f^{-1}(x) = e^{(e^x)} \).

Step by step solution

01

Understand the Function's Requirements

The function given is \( f(x) = \ln(\ln x) \). This function applies the natural logarithm twice. First, it calculates \( \ln x \) and then applies the natural logarithm \( \ln (\ln x) \). For each of these log functions to be defined, their arguments must be greater than zero.
02

Find Domain of Inner Logarithm

The innermost function is \( \ln x \). The natural logarithm is only defined for positive numbers \( x > 0 \). However, since the argument of \( \ln x \) is itself within another logarithm, it must also be greater than 1. Thus, \( x > 1 \).
03

Find Domain of Outer Logarithm

For the outer logarithm \( \ln (\ln x) \) to be defined, \( \ln x \) must be greater than 0. Solving \( \ln x > 0 \) gives \( x > e^0 = 1 \). Thus, the domain of the function \( f(x) = \ln(\ln x) \) is \( x > e \).
04

Define the Inverse Function Requirement

To find \( f^{-1}(x) \), we start from the equation \( y = \ln(\ln x) \) and aim to express \( x \) in terms of \( y \). This involves reversing the operations of \( f \), essentially solving \( y = \ln(\ln x) \) for \( x \).
05

Solve for \( x \)

Starting with the equation \( y = \ln(\ln x) \), exponentiate both sides to get \( e^y = \ln x \). Then, exponentiate again to solve for \( x \). This yields \( x = e^{(e^y)} \). Thus, \( f^{-1}(x) = e^{(e^x)} \).

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Key Concepts

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

Inverse Function
An inverse function essentially reverses the effect of the function it is associated with. Given a function \( f(x) = \ln(\ln x) \), its inverse function \( f^{-1}(x) \) will take an output value from \( f \) and return the original input \( x \). To determine this inverse, we start by rewriting the function equation \( y = \ln(\ln x) \).

To isolate \( x \), you first undo the outer logarithm by exponentiating both sides, resulting in \( e^y = \ln x \). Then, undo the inner logarithm in a similar way: \( x = e^{(e^y)} \).

This outcome \( f^{-1}(x) = e^{(e^x)} \) confirms how the inverse reverses \( f \). With this formula, if you apply \( f \) to a value, applying \( f^{-1} \) retrieves the original input, completing the cycle.
Natural Logarithm
The natural logarithm, often notated as \( \ln(x) \), is a logarithmic function to the base \( e \). The constant \( e \) (approximately 2.718) is a mathematical constant used widely in growth calculations, particularly in natural processes. The function \( \ln(x) \) gives the power that \( e \) needs to be raised to, to get \( x \). For example, \( \ln(e) = 1 \), because \( e^1 = e \).

When calculating \( \ln(\ln x) \), you are essentially applying the natural log twice. The inner \( \ln x \) must be positive, meaning \( x > 1 \). Then, for the outer \( \ln \) to be defined, \( \ln x \) itself must be greater than zero, resulting in \( x > e \). This makes the domain of \( f(x) = \ln(\ln x) \) be \( x > e \). The natural log's properties like base \( e \) and positivity are key in defining the allowable inputs.
Function Composition
Function composition involves combining two or more functions such that the output of one function becomes the input of another. In the context of \( f(x) = \ln(\ln x) \), the composition starts with \( g(x) = \ln x \), then applies \( h(y) = \ln(y) \) to \( g(x) \). This composition is represented as \( f(x) = h(g(x)) \).

The process can be thought of in layers: \( g(x) = \ln x \) first processes the input \( x \), transforming it into \( \ln x \). Then, the function \( h(y) = \ln(y) \) further processes the output of \( g(x) \), resulting in \( \ln(\ln x) \).

Function composition is notated as \( (h \circ g)(x) \), indicating \( h \) is applied after \( g \). Understanding function composition helps in analyzing how complex functions are built from simpler ones, as seen in this exercise.

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Most popular questions from this chapter

In 1969 the United States National Academy of Sciences issued a report entitled 91Ó°ÊÓ and Man. One conclusion in the report is that a world population of 10 billion "is close to (if not above) the maximum that an intensively managed world might hope to support with some degree of comfort and individual choice." (The figure "10 billion" is sometimes referred to as the carrying capacity of the Earth.) (a) When the report was issued in \(1969,\) the world population was about 3.6 billion, with a relative growth rate of \(2 \%\) per year. Assuming continued exponential growth at this rate, estimate the year in which the Earth's carrying capacity of 10 billion might be reached. (b) Repeat the calculations in part (a) using the following more recent data: In 2000 the world population was about 6.0 billion, with a relative growth rate of \(1.4 \%\) per year. How does your answer compare with that in part (a)?

Solve the inequalities. Where appropriate, give an exact answer as well as a decimal approximation. $$e^{x^{2}-4 x} \geq e^{5}$$

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(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). $$10^{x^{2}}=40$$

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