Solving Natural Logarithmic Equations

[Rossinole]

Homework Statement

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

Homework Equations

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.

 

[nicksauce]

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

 

[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!

 

[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.

 

[HallsofIvy]

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.

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.

 

[Redbelly98]

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

edit added:
Halls beat me to it.