Question

How do you solve `ln x = 1 - ln (x+8)`?

Answer 1

I found: `x=-4+sqrt(16+e)`

Explanation:

Write it rearranging as:
`ln(x)+ln(x+8)=1`
use the fact that `lnx+lny=ln(xy)`
so:
`ln[x(x+8)]=1`
use the definition of log:
`lnx=a -> x=e^a`
`x(x+8)=e^1`
`x^2+8x-e=0`
using the Quadratic Formula:
`x_(1,2)=(-8+-sqrt(64+4e))/2=(-8+-2sqrt(16+e))/2=`
`=-4+-sqrt(16+e)`
so you get two solutions but one, `-4-sqrt(16+e)`, doesn't work when substituted into the original equation (it is a negative number, try it!) and you can discard it keeping only:
`x=-4+sqrt(16+e)`

Answer 2

x=`-4+-sqrt(16+e)`

Explanation:

Rewriting the equation , it is ln x + ln(x+8) =1

ln x(x+8)=1

`x^2 +8x= e`

`x^2 +8x -e=0`

x= `(-8+-sqrt (64+4e))/2`

x=`-4+-sqrt(16+e)`