# Solve an exponential equation

Gold Member
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?

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

Math_QED
Homework Helper
2019 Award
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: $e^{-x} = \frac{1}{e^x}$.