# Natural Logarithmic Function

## Homework Statement

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

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

mgb_phys
Homework Helper
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!

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
Homework Helper
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.

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Redbelly98
Staff Emeritus