Question

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

Thanks in advance

Answer 1

`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.

Answer 2

`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.