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.
Tuesday, June 23, 2015
Precalculus, Chapter 3, 3.4, Section 3.4, Problem 77
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