# What is the inverse of f(x) = 2ln(x-1)-ix-x?

Jul 5, 2017

${f}^{- 1} \left(x\right) = {e}^{i x} - x + 1$#

#### Explanation:

If f"$f \left(x\right) = 2 \text{ln} \left(x + 1\right) - i x - x$, then ${f}^{- 1} \left(x\right)$is the inverse, or the reflection of $f \left(x\right)$ in the line $y = x$.

So, let's make $f \left(x\right) = 2 \text{ln} \left(x + 1\right) - i x - x = 0$, then $2 \text{ln} \left(x + 1\right) = i x + x = i \left(2 x\right)$, dividing both sides by 2 gives $f \left(x\right) = \text{ln} \left(x + 1\right) = \frac{i \left(2 x\right)}{2} = i x$.

As ${e}^{\text{ln} \left(a\right)} = a$, ${e}^{\ln \left(x + 1\right)} = x + 1 = {e}^{i x}$.

Now take away $\left(x + 1\right)$ from both sides to make it equal 0, $0 = {e}^{i x} - x + 1$.