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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?

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  • $\begingroup$ Factor the equation: $x^3 \cdot \ln x - 4 x \cdot \ln x = (x^3 - 4x) \ln x = 0$. If a product is zero, one of its factors must be zero. And ... $x = -2$ is not a solution, since $\ln (-2)$ is undefined. $\endgroup$ – Hans Engler Apr 4 '14 at 15:44
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$$\begin{align}x^3 \ln x - 4x\ln x = 0 &\iff (x^3 - 4x)\ln x = 0 \\ \\ &\iff x(x^2 - 4)\ln x = 0\\ \\ & \iff x(x - 2)(x+2)\ln x = 0 \\ \end{align}$$

$$\implies x = 0 \;\text{ or } \;x-2 = 0 \;\text{ or }\; x+2 = 0 \;\text{ or }\;\ln x = 0, $$

$$\implies x = 0 \;\;\text{or}\;\; x = 2\;\;\text{or}\;\;x = -2\;\;\text{or} \;\;x = 1$$

Hence, there are four solutions to the given equation.

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  • $\begingroup$ Wow, I've been doing it wrong :)... $\endgroup$ – Juraj Apr 4 '14 at 18:02
  • $\begingroup$ You're welcome, Juraj! $\endgroup$ – Namaste Apr 4 '14 at 18:06

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