Solve Logarithmic Equation: 3 Solutions Explained

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The discussion focuses on solving the logarithmic equation (ln x)^3 = 3 ln x, which has three solutions. The initial attempt yielded only one solution, ln x = sqrt(3), but overlooked others due to dividing by ln x, which can eliminate valid solutions. The equation can be transformed into y^3 - 3y = 0, allowing for the identification of all solutions. The complete solutions are x = 1, e^sqrt(3), and e^-sqrt(3). Understanding the implications of division in logarithmic equations is crucial for finding all potential solutions.
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Homework Statement


I'm trying to solve this logarithmic equation, which has 3 possible solutions. However, I can only get to one of them. What are other ways I can solve it so I get to these other solutions?


Homework Equations


(ln x)^3 = 3 ln x


The Attempt at a Solution


(ln x)^3 = 3 ln x
(ln x)^2 = 3
ln x = sqrt(3)
e^sqrt(3) = x
 
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MAke the substitution y=lnx. Then your equation becomes y3-3y=0. Can you solve this for y?

(Note: You don't have to make the substitution; you could do it using your method. Your problems are:
(ln x)^3 = 3 ln x
(ln x)^2 = 3
Here you lost a solution by dividing by lnx
ln x = sqrt(3)
Here you only took one value of the square root)
 
Last edited:
Thanks a lot.

x = 1, e^sqrt(3), e^-sqrt(3)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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