Can you please solve this for me? e^x-2e^-x=1?
can you please give the complete solution..?
explanation: OPTIONAL
Pr0bLem:
e^x-2e^-x=1
(read as “e raised to x minus two e raised to negative x equals one”)
thank you so much..! 🙂
Favorite Answer
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
