Natural Logarithmic Function

1. Jun 9, 2008

Rossinole

1. The problem statement, all variables and given/known data

Solve ln(2x+1)=2-ln(x) for x.

2. Relevant equations

3. The attempt at a solution

e^(ln(2x+1)) = e^((2-ln(x))

2x + 1 = e^(2/x)

2x^2 + x = e^2

2x^2+x-e^2 = 0

At this point, I know you're supposed to use the quadratic equation. But, my problem is with how to treat e^2. Should I multiply it out in the discriminant where c = e^2? or just e?

Thanks for any help.

2. Jun 9, 2008

nicksauce

e^2 is just a number... you can call it c if you like.

3. Jun 9, 2008

mgb_phys

You can either just calculate it out (it's only a number!) or include it in the quadratic formula and quote your answers in terms of e.
It depends wether this is for home work or real work!

4. Jun 9, 2008

BrendanH

There is an error in the second part

e^(ln(2x+1)) = e^((2-ln(x))

2x + 1 = e^(2/x)

These are NOT equivalent. Remember that you can only turn logarithmic expressions
into quotients of logarithms in a case like this: ln z - ln x = ln (z/x)

Try again using that info.

5. Jun 9, 2008

HallsofIvy

Staff Emeritus
I suspect that was a typo because in the next line he has
2x^2+ x= e^2 which is correct.

Added Later: In fact, I would have been inclined to solve the problem as follows:
ln(2x+1)=2-ln(x) so, adding ln(x) to both sides, ln(2x+1)+ ln(x)= ln(x(2x+1))= 2.
Now take the exponetial of both sides: x(2x+1)= 2x2+ x= e2 and you can solve that quadratic equation for x.

Last edited: Jun 10, 2008
6. Jun 9, 2008

Redbelly98

Staff Emeritus
BrendanH, Rossinole meant (e^2)/x, not e^(2/x). The next line in the derivation,
2x^2 + x = e^2
is correct.