I can't find second solution to this logarithmic problem!

I kind of got stuck on one step in solving a logarithmic equation.

The equation given was: x^3lnx - 4xlnx = 0

My steps so far:

x^3lnx - 4xlnx = 0

ln((x^x^3)/(x^4x)) = 0

e^ln((x^x^3)/(x^4x)) = e^0

(x^x^3)/(x^4x) = 1

x^x^3 = x^4x

now I would just remove the base to make it x^3=4x. However, this step would also remove the solution x = 1 from the equation. I only got it through guessing and then checking on a graphic calculator.

The second asnwer, x = -2 I do know how to get. I just solved

x^3 - 4x = 0

x(x^2 - 4) = 0

x(x-2)(x+2) = 0 (so the solutions could be +/- 2. By plugging these values back in the original formula I found out that only -2 is the solution.)

THE QUESTION: Can someone please show me how to algebraically find the solution x = 1?

I kind of got stuck on one step in solving a logarithmic equation.

The equation given was: x^3lnx - 4xlnx = 0

My steps so far:

x^3lnx - 4xlnx = 0

ln((x^x^3)/(x^4x)) = 0

e^ln((x^x^3)/(x^4x)) = e^0

(x^x^3)/(x^4x) = 1

x^x^3 = x^4x

now I would just remove the base to make it x^3=4x. However, this step would also remove the solution x = 1 from the equation. I only got it through guessing and then checking on a graphic calculator.

The second asnwer, x = -2 I do know how to get. I just solved

x^3 - 4x = 0

x(x^2 - 4) = 0

x(x-2)(x+2) = 0 (so the solutions could be +/- 2. By plugging these values back in the original formula I found out that only -2 is the solution.)

THE QUESTION: Can someone please show me how to algebraically find the solution x = 1?