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?
$$\begin{align}x^3 \ln x - 4x\ln x = 0 &\iff (x^3 - 4x)\ln x = 0 \\ \\ &\iff x(x^2 - 4)\ln x = 0\\ \\ & \iff x(x - 2)(x+2)\ln x = 0 \\ \end{align}$$
$$\implies x = 0 \;\text{ or } \;x-2 = 0 \;\text{ or }\; x+2 = 0 \;\text{ or }\;\ln x = 0, $$
$$\implies x = 0 \;\;\text{or}\;\; x = 2\;\;\text{or}\;\;x = -2\;\;\text{or} \;\;x = 1$$
Hence, there are four solutions to the given equation.