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

Thanks in advance

2 Answers
May 4, 2018

x=sqrt(1+e)-1~~0.928

Explanation:

Add ln(x+2) to both sides to get:
lnx+ln(x+2)=1

Using the addition rule of logs we get:
ln(x(x+2))=1

Then by e"^" each term we get:
x(x+2)=e

x^2+2x-e=0

x=(-2+-sqrt(2^2+4e))/2

x=(-2+-sqrt(4+4e))/2

x=(-2+-sqrt(4(1+e)))/2

x=(-2+-2sqrt(1+e))/2

x=-1+-sqrt(1+e)

However, with the ln()s, we can only have positive values, so sqrt(1+e)-1 can be taken.

May 4, 2018

x = sqrt(e+1) - 1

Explanation:

lnx=1−ln(x+2)
As 1 = ln e
implies ln x = ln e -ln(x+2)

ln x = ln(e/(x+2))
Taking the antilog on both sides,
x = e/(x+2)
implies x^2 + 2x = e
Complete the squares.
implies (x+1)^2 = e + 1
implies x + 1 = +-sqrt(e + 1)
implies x = sqrt(e + 1) - 1 or x = -sqrt(e +1 ) - 1

We neglect the second value as it would be negative, and the logarithm of a negative number is undefined.