How to Solve an Exponential Equation with a Mistake in the Second Step?

[Rectifier]

The problem
Solve $e^x-e^{-x} = 6$ .

The attempt
$$ e^x-e^{-x} = 6 \\ e^x(1-e^{-1}) = 6 \\ e^x = \frac{6}{(1-e^{-1})} \\ x = \ln \left( \frac{6}{1-e^{-1}} \right) \\ $$

The answer in the book is $\ln(3 + \sqrt{10})$

Could someone help me?

 

[Rectifier]

I made a mistake at the second step $e^{-x} \neq e^{-1} \cdot e^x$

 

[member 587159]

I made a mistake at the second step $e^{-x} \neq e^{-1} \cdot e^x$

Make the substitution y = e^x.

 

[Ray Vickson]

I made a mistake at the second step $e^{-x} \neq e^{-1} \cdot e^x$

Right: $e^{-x} = \frac{1}{e^x}$.