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

Question

show work if possible

1/2 does not make sense

Answer 1

ln(2x+1) = 2 - ln(x)
Add ln(x) to each side
ln(2x+1) + ln(x) = 2
Recall that the sum of two logs is the product of what you are taking the log of. That is,
ln(a) + ln(b) = ln(ab)
Thus,
ln(x(2x+1)) = 2
Distibute the x.
ln(2x^2 + x) = 2
Apply the exponential e to both sides
e^[ln(2x^2 + x)] = e^2
2x^2 + x = e^2
2x^2 + x - e^2 = 0
Factor
2x^2 + (2e)x - (e)x - e^2 = 0
Factor by grouping.
2x(x+e) - e(x+e) = 0
(x+e)(2x-e) = 0
x+e = 0 or 2x-e = 0
x = -e
or
2x = e
x = e/2
x = -e cannot be a solution because you cannot take the ln of a negative number.
Answer: x = e/2

Answer 2

1/2

Answer 3

1/2

Answer 4

Get the "ln's" on the same side of the equation:

ln(2x+1) = 2 - ln(x) ----> ln(2x+1) + ln(x) = 2

Use a rule of logs that says:

ln a + ln b = ln(ab)

So, ln[(2x + 1)*(x)] = 2
ln[(2(x^2) + x)] = 2

Now, raise "e" to each side:

e^ln[(2(x^2) + x)] = e^2

[(2(x^2) + x)] = e^2 , (by another rule of logs and exponents)

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

Use the quadratice formula to solve for "x"

Answer 5

ln(2x+1)=2-lnx
ln(2x+1)+lnx=2
ln[x(2x+1]=2
2x^2+x=e^2
now you may take the numerical value of e and solve

Answer 6

One half (1/2)