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# solving an equation watch

1. can anyone help me solve this equation:

e^x + e^-x = 2

Cheers
2. (Original post by spikey_aman)
can anyone help me solve this equation:

e^x + e^-x = 2

Cheers
surely just be inspection you can tell x=0 e^0+e^0=1+1=2
3. k if i'm correct, then log a + log b = log ab
thefore, e^x + e^(-x) = log (-x^2)
..
log (-x^2) = 2

then just solve for x. k not sure if thats correct...soz..

i guess is already been solved but what i've done above would that be the algebraic way 2 slove it??
4. (Original post by spikey_aman)
can anyone help me solve this equation:

e^x + e^-x = 2

Cheers
e^x+1/e^x=2

(e^2x+1)/e^x=2
e^2x+1=2e^x

e^2x-2e^x+1=0

sub in y=e^x

y^2-2y+1=0
(y-1)^2=0
y=1

e^x=1
x=0

There may well be a quicker way, but I don't think I know it. You could also cheat by using hyperbolic functions:

(e^x+e^-x)/2=cosh(x)
2cosh(x)=2
cosh(x)=1
x=0
5. If you're familiar with hyperbolic functions, you're looking to solve cosh(x) = 1. Therefore:
x = +/- arcosh(1) = 0
6. (Original post by dvs)
If you're familiar with hyperbolic functions, you're looking to solve cosh(x) = 1. Therefore:
x = +/- arcosh(1) = 0
I'm sure that's cheating, though
7. (Original post by BovineBeast)
I'm sure that's cheating, though
9. Yea, thats cheating - havent met the hyperbolics yet :P lol

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