I can't find second solution to this logarithmic problem!The equation given was: x^3lnx - 4xlnx = 0

Tatiana Cook 2022-10-08 Answered
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?
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Answers (1)

Emmalee Reilly
Answered 2022-10-09 Author has 6 answers
x 3 ln x 4 x ln x = 0 ( x 3 4 x ) ln x = 0 x ( x 2 4 ) ln x = 0 x ( x 2 ) ( x + 2 ) ln x = 0
x = 0  or  x 2 = 0  or  x + 2 = 0  or  ln x = 0 ,
x = 0 or x = 2 or x = 2 or x = 1
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