That equation can not be solved by what we would normally call "algebraic methods". You

**can** write it as 0.4343x- x log(2x)= -2.825 so that x log(2x)= 2.825- 0.4343x and then log(2x)= 2.825/x- 0.4343. Taking the exponential of both sides, 2x= e^{2.825/x- 0.4343}= e^{-0.4343}e^{2.825/x}. Let y= 2.825/x so x= 2.825/y and 2x= 5.650/y. The equation becomes 5.660/y= e^{-0.4343}e^y. Multiply both sides by ye^{0.4343} to get 5.66e^{0.4343}= ye^y.

**Now** apply the "Lambert W function" to both sides to get y= W(5.66e^{0.4343}). Finally, since x= 2.825/y, x= 2.825/W(5.66e^{0.4343}). The "Lambert W function",

https://en.wikipedia.org/wiki/Lambert_W_function, is defined as the inverse function to f(x)= xe^x.