Can you please solve this for me? e^x-2e^-x=1?

1.) Factor out e^x.

This becomes: e^x (1 – 2e^-2x) = 1

2.) Divide both sides by e^x.

This becomes: 1 – 2e^-2x = e^-x

3.) Transpose the left side to the right side of the equation.

This becomes: 2e^-2x + e^-x -1 = 0

4.) Factor out.

This becomes: ( 2e^-x – 1 ) ( e^-x + 1 ) = 0

5.) Therefore,

(equation 1) 2e^-x – 1 = 0

(equation 2) e^-x + 1 = 0

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We solve equation 1:

2e^-x = 1

e^-x = 1/2

Get ln of both sides:

ln(e^-x) = ln(1/2)

-x = -ln2

So, x = ln 2

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Solving equation 2:

e^-x = 1

Get ln of both sides:

ln(e^-x) = ln 1

-x = 0

So, x = 0

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The answers are x = ln 2 and x = 0

e^x – (2/e^x) = 1

e^2x – 2 = e^x

e^2x – e^x – 2 = 0

(e^x – 2) (e^x + 1) = 0

e^x – 2 = 0 or e^x + 1 = 0

e^x = 2 or e^x = -1

But, e^x cannot be negative, because e is a positive real number.

Therefore, e^x = 2

Applying ln to both sides:

ln e^x = ln 2

x (ln e) = ln 2

x = ln 2 = 0.693