Tatiana Cook

2022-10-08

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?

Emmalee Reilly

$\begin{array}{rl}{x}^{3}\mathrm{ln}x-4x\mathrm{ln}x=0& \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}\left({x}^{3}-4x\right)\mathrm{ln}x=0\\ \\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x\left({x}^{2}-4\right)\mathrm{ln}x=0\\ \\ & \phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x\left(x-2\right)\left(x+2\right)\mathrm{ln}x=0\end{array}$

$\phantom{\rule{thickmathspace}{0ex}}⟹\phantom{\rule{thickmathspace}{0ex}}x=0\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{or}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}x=2\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{or}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}x=-2\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}\text{or}\phantom{\rule{thickmathspace}{0ex}}\phantom{\rule{thickmathspace}{0ex}}x=1$

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