# I know it's wrong but I don't know WHY

Homework Helper

## Main Question or Discussion Point

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

ln() is only defined over positive values, but you can find solutions like so... where's the error.

## Answers and Replies

anti_crank
ln(x)^-1 = ln(-x) is incorrect. You're probably thinking of ln (1/x) = - ln (x)

Integral
Staff Emeritus
Gold Member
Your trouble is that

$$\ln (-x) \ne (\ln(x))^{-1})$$
This is the correct way to do it.
$$\ln (-x) = \ln(-1 * x) = \ln(-1) + \ln(x)$$

There will be complex solutions to this but none on the Real line.