Solving a Logarithm

[Dorian Gray]

Greetings Mathematicians,


I am having some issues with this problem: ln x + ln (x-1) = 1

I attached my work. Any comments, suggestions, or info is always appreciated.

Thanks

 

[Deleted member 4993]

Greetings Mathematicians,


I am having some issues with this problem: ln x + ln (x-1) = 1

I attached my work. Any comments, suggestions, or info is always appreciated.View attachment 1622

Thanks

ln x + ln (x-1) = 1

ln[x(x-1)] = 1

ln[x(x-1)] = ln(e)

x(x-1) = e

Now continue...

 

[Dorian Gray]

Thanks for the info! However, isn't my step 4 the same as your last step? I am not sure where to take it after x(x-1)= e OR x² -x = e

 

[srmichael]

Thanks for the info! However, isn't my step 4 the same as your last step? I am not sure where to take it after x(x-1)= e OR x² -x = e

Yes, they are the same.

Do you see that \(\displaystyle \displaystyle x^2-x=e\) is a quadratic? Move the "e" to the left hand side and then use the quadratic formula to solve.

 

[Dorian Gray]

thank you

Thank you SrMichael.I used the quadratic formula and got the correct answer.

 

[lookagain]

Thank you SrMichael.I used the quadratic formula and got the correct answer.

Dorian Gray,

how did you determine that you had the right answer

when solving for x in the quadratic equation \(\displaystyle \ \ x^2 - x - e = 0\)

if using the Quadratic formula (where here a = 1, b = -1, and c = -e),

that the candidates for solutions in the original logarithmic equation

turn out to be


\(\displaystyle x \ = \ \dfrac{1 - \sqrt{1 + 4e}}{2}, \ \ and \ \ x \ = \ \dfrac{1 + \sqrt{1 + 4e}}{2} \ \ ?\)



Did you know that it must be that x > 1 for solutions to the original equation,

and upon checking these candidate solutions from the solving of the

quadratic equation, that you had to discard the first potential solution?


Or, just how did you know that you got the correct answer?