Question

e^x + e^-x = 2

Answer 1

x = 0

Answer 2

\[e^x + \frac{1}{e^x} = 2\]

Answer 3

Solve this you will get quadratic equation..

Answer 4

yes obviously x=0 as 1+1=2 gives a perfect solution :)

Answer 5

how will i get a quadratic equation?

Answer 6

Multiply both sides by e^x

Answer 7

Or let y = e^x, solve for y.

Answer 8

\[e^{2x} + 1 = 2e^{x} \implies e^{2x} - 2e^x + 1 = 0\]

Answer 9

\[(e^x -1)^2 = 0\]

Answer 10

\[e^x = 1\]

Answer 11

\[e^x + \frac{1}{e^x} = 2\]\[e^x(e^x + \frac{1}{e^x}) = 2e^x\]\[e^{2x} + 1 = 2e^x\]\[(e^{x})^2 + 1 = 2e^x\]\[(e^{x})^2 -2e^x+ 1 = 0\]

Answer 12

Clerly x is 0..

Answer 13

thank you :)

Answer 14

Anything raised to the power 0 except 0 is 1..

Answer 15

have a nice day! see you after next chapter :)

Answer 16

interesting...i see a trig function =))

Answer 17

So for e^x =1 , x must be 0..

Answer 18

@lgbasallote I too..

Answer 19

You probably mean a hyperbolic trig function, lg: cosh x = 1.

Answer 20

trig function where?

Answer 21

If x is pure imaginary, your equation can be expressed with a cosine, e.g. if x = i y then you have cos(y) = 1. You still get the same answer, of course, y = x = 0.

Answer 22

Oh com' on @bronzegoddess and @Carl_Pham lgba was just joking and you took him seriously..

Answer 23

lol my bad

Answer 24

So?