Question

solve 2x+2lnx=0

Answer 1

for better or worse, and though it seems weird, this kind of equation cannot be solved by simple algebra! even if you play around with it:
\[x+lnx=0\]\[lnx=-x\]\[e^{-x}=x\]\[xe^{x}=1\]the equation still never "gets anywhere"--you always get stuck with x's that can never be combined or factored or whatever.

so, there are methods that approximate answers by starting with a guess and narrowing down. dr math has already put together a great explanation!:
http://mathforum.org/library/drmath/view/68497.html
good luck.

Answer 2

please, use the fact that the function f(x)=2x+2lnx realizes a bijection from ]0, infinity[ to R , then think that you can use the Intermediate Values Theorem by chosing an interval like [a, b] such that f(a).f(b)<0. By resolving the exercise, we'll find xe^x=1, then a first interval to chose is easy to view, for example [0.25, 1], after we'll try to reduce this interval as small we can, like [0.4 0.6], we have f(0.4)=-1.03 and f(0.6)=0.17, then we can give an approximation about the solution of the equation, this solution is single, x≈0.55!

Answer 3

These type of equations are best solved by first graphing. If you draw both the line y=-2x and the curve y=2lnx you find if they have any intersections but also how many there are.

Then if the values of x are not immiately clear you can use Newton's method to solve with very good first guesses from the graph.