Help solving log (natural number) equation

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To solve the equation 2lnX = 2 + ln(X-1), the first step involves raising both sides by e to eliminate the logarithm. This leads to the equation X^2/(X-1) = e^2. By rearranging, it simplifies to X^2 = e^2(X-1), resulting in the quadratic equation X^2 - e^2X + e^2 = 0. The next steps involve applying the quadratic formula to find the values of X. This method provides a clear path to the solution of the logarithmic equation.
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I am stuck on finding the solution to this problem.

2lnX = 2 + ln(X-1)

Please show all steps for the solution.

Thank you.
 
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Try raising both sides by e. Use algebra to simplify.

- Warren
 
I get this far.
X^2/X-1 = e^2
X^2 = e^2(X-1)
x^2 = e^2 X - e^2
 
You now have a quadraic equation in x. Remember e2 is just a number.
 
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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