solve for x

2=lnx+2x the answer is x=1 but i dont see who they got that

\(2=lnx+2x\)

\(lnx=2(1-x)\)

\(\frac{lnx}{2}=1-x\)

\(ln\sqrt x=1-x\)

\(x+ln\sqrt x=1\)

\(f(x)=x+ln\sqrt x-1=0\)

\(\LARGE x=1\)

  !

2=lnx+2x    

I think I would graph  y=lnx+2x-2   using a graphing calculator and see where y=0

There is no super straight forward algebraic way of doing it, although someone may come up with some complicated method.  One must exist. :)

You can see from the graph that x=1 is the only answer.

Rewrite the equation as

2 - 2x = ln(x),

and sketch the two graphs y = 2 - 2x and y = ln(x).

(You shouldn't need a graph plotter to do this).

Solutions to the equation will be the x co-ordinates of their points of intersection. Clearly there is just one, at x = 1.