`2xln(x) + x = 0` Solve the equation algebraically. Round your result to three decimal places. Verify your answer using a graphing utility.

Textbook Question

Chapter 3, 3.4 - Problem 77 - Precalculus (3rd Edition, Ron Larson).
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jmfj's profile picture

jmfj | (Level 1) Adjunct Educator

Posted on

Start by factoring out the x from each of the two terms on the left side of the equation. This gets you:

x(2lnx + 1) = 0 which implies that x = 0 as well as 2lnx + 1 = 0.

Now let's solve 2lnx + 1 = 0 by moving the 1 from the left to the right, getting 2lnx = -1.  Divide off the 2 getting lnx = -1/2.  Un-natural log the left side by using "e".  It will look like: e^(lnx) = e^(-1/2).  The "e" and the ln cancel leaving just x.  Therefore, x = e^(-1/2).  If you punch this into your calculator you get 0.607.  Don't forget, you started this process out with x = 0 and now you have x = 0.607.  Hopefully you remember that it is impossible to find the ln of 0, so you must exclude x = 0 from your answer.  This leaves x = 0.607 as the only answer.

kspcr111's profile picture

kspcr111 | In Training Educator

Posted on

Given

2xln(x) + x = 0

=> 2x ln(x) = -x

cancelling x on both sides we get

=> 2 ln(x) = -1

=> ln(x) = (-1/2)

=> x = e^(-1/2) = 0.607

the graph of the equations

2x ln(x) +x =0

and

y= 2x ln(x) +x

are as follows

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