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Chapter 3: Problem 77

Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility. $$2 x \ln x+x=0$$

Short Answer

Expert verified
The solution to the equation \(2 x \ln x+x=0\) is \(x = 0.183\).

Step by step solution

01

Rewrite Equations

Rewrite the equation as: \(2 x \ln x = -x\).
02

Divide by \(x\)

Next, divide through the entire equation by \(x\) to simplify, yielding: \(2\ln x = -1\).
03

Solve for \(x\)

To solve for \(x\), we can convert the equation to exponential form. We know that if \(\ln a = b\), then \(e^b = a\), therefore \(e^{-1} = 2\). Now compute the value of \(x = \frac{e^{-1}}{2}\).
04

Verification Using a Graphing Utility

To verify the answer, you can plot the function \(y=2x\ln x+x\) and the line \(y=0\) in the same graph. The intersection points of the plotted function and the line \(y=0\) represent the solution to the equation. Verifying our calculated solution around \(x = 0.183\), we can visually observe that the function does indeed intersect the line \(y=0\) at this value.

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Most popular questions from this chapter

Use a graphing utility to graph and solve the equation. Approximate the result to three decimal places. Verify your result algebraically. $$2 \ln (x+3)=3$$

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$\log (3 x+4)=\log (x-10)$$

Function \(\quad\) Value \(\begin{array}{ll}\text { 57. } f(x)=\ln x & x=18.42 \\ \text { 58. } f(x)=3 \ln x & x=0.74\end{array}\) \(\begin{array}{lll}y & f(x)=3 \ln x & x=0.74 \\ \text { 59. } g(x)=8 \ln x & x=0.05\end{array}\) 60\. \(g(x)=-\ln x \quad x=\frac{1}{2}\)

Solve the logarithmic equation algebraically. Approximate the result to three decimal places. $$6 \log _{3}(0.5 x)=11$$

Writing Use your school's library, the Internet, or some other reference source to write a paper describing John Napier's work with logarithms.
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