Solve for X: ln(x) + ln(x+1) = 1

  • Thread starter Thread starter r_swayze
  • Start date Start date
AI Thread Summary
The equation ln(x) + ln(x+1) = 1 can be simplified to x^2 + x - e = 0 after applying properties of logarithms and exponentials. The constant e represents e^1, which is crucial for solving the quadratic equation. The quadratic formula is recommended for finding the roots, as the equation does not factor easily. Participants confirm that using the quadratic formula is the appropriate method to solve for x. The discussion emphasizes the importance of correctly manipulating the equation to reach a solvable form.
r_swayze
Messages
65
Reaction score
0
Im having trouble finding the solution to this problem, can anyone walk me through this?

So far I have:

ln(x) + ln(x+1) = 1

ln(x)(x+1) = 1

e^1 = x(x+1)

e^1 = x^2 + x

This is where I get stuck.
Am I on the right track?
 
Physics news on Phys.org
r_swayze said:
e^1 = x^2 + x

Yes this is correct. Write e^1 as just e, which is constant.
If you move it to the other side you'll get:
x2+x-e =0

Now how do you solve the equation ax2+bx+c=0?
 
rock.freak667 said:
Yes this is correct. Write e^1 as just e, which is constant.
If you move it to the other side you'll get:
x2+x-e =0

Now how do you solve the equation ax2+bx+c=0?

I don't think x^2 + x - e = 0 can factor out, at least not without using the quadratic formula
 
r_swayze said:
I don't think x^2 + x - e = 0 can factor out, at least not without using the quadratic formula

Then use the wquadratic equation formula and you'd solve for x.
 
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
Back
Top