solve this for me please: |ln x| = 1?

|ln(x)| = 1
ln(x) = ±1

ln(x) = 1
x = e

ln(x) = -1
x = 1/e

x = {1/e, e}

if | ln x | = 1, then either ln x = 1 or ln x = -1

if ln x = 1, then e^1 = x ==> x = e
if ln x = -1, then e^-1 = x => x = 1/e

two solutions: x = e or x = 1/e

careful Kahsel... logs can be negative (the range is (-inf , inf)); what you're thinking of is that you can't take the log of a negative number (or 0, for that matter)-- the domain of logs is positive numbers only; however, the range is all real numbers

x = e

Those bars mean 'the magnitude of', which basically means how much do you have, regardless of the sign.
But it doesn't really matter, since logs can never be negative.

Ln x is just shorthand for Log base e of x, or Log e (x) - and a log is equal to 1 when the base and the result are the same.

So basically, when x = e (or about 2.718), |ln x| = 1!

Edit: - yeah, disregard my negative bit thing. Made a mistake - x can't be negative, sorry. Not the answer. So could be e^-1, or 1/e, not just e. ...Yep.
Peh.

lnx=1
x=e^1=e=2.7183
lnx=-1
x=e^-1=0.36788

|ln(x)| = 1
ln(x) = ±1
x = e^1, e^-1
x = e, 1/e