/*! 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 95 Solve each equation. $$ 2|\l... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 95

Solve each equation. $$ 2|\ln x|-6=0 $$

Short Answer

Expert verified
The solutions to the equation are \(x = e^3\) and \(x = e^{-3}\).

Step by step solution

01

Isolate the Absolute Value

The equation is \(2|\ln x| - 6 = 0\). The first task is to isolate the absolute value on one side. This can be done by adding 6 to both sides and then dividing by 2. \(2|\ln x| = 6\) \(|\ln x| = 3\)
02

Remove the Absolute Value

With absolute values, consider that either the positive or negative equivalent could equal to 3. That results in two separate equations to solve: \(\ln x = 3\) and \(\ln x = -3\).
03

Solve for x

Next, solve each equation. Since the base of natural logarithm is e, take e to the power of each side. For the first equation: \(e^{ln x} = e^3\), resulting in \(x = e^3\). For the second equation: \(e^{ln x} = e^{-3}\), resulting in \(x = e^{-3}\). However, avoid solutions that would make the original log undefined or negative in the original equation. Thus, \(x = e^3\) and \(x = e^{-3}\) are the two solutions as \(e^3\) and \(e^{-3}\) are both positive.

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Most popular questions from this chapter

Determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$ \text { If } \log (x+3)=2, \text { then } e^{2}=x+3 $$

Graph: \(f(x)=\frac{4 x^{2}}{x^{2}-9}\) (Section 3.5, Example 6)

In Exercises 139鈥�142, determine whether each statement is true or false. If the statement is false, make the necessary change(s) to produce a true statement. $$ \frac{\log _{2} 8}{\log _{2} 4}=\frac{8}{4} $$

Find \(\ln 2\) using a calculator. Then calculate each of the following: \(1-\frac{1}{2} ; \quad 1-\frac{1}{2}+\frac{1}{3} ; \quad 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\) \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5} ; \ldots\) Describe what you observe.

You overhear a student talking about a property of logarithms in which division becomes subtraction. Explain what the student means by this.
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