Can somebody solve: e^(x) + e^(-x) = 3?

Question

i don't know why i can't figure this out, seems like i've forgotten basic exponent rules

Is there a rule where i can combine the two since they each have the same base

Answer 1

Let e^x = T so e^-x = 1/T.

So T +(1/T) = 3

Means T^2 - 3T +1 =0 now use qtc eqn and solve for T and x = lnT

Answer 2

e^(x) + e^(-x) = 3

e^(x) + e^(-x) -3 = 0

e^(2x) -3e^(x) + 1 = 0

let u = e^(x)

u² - 3u + 1 = 0

u² - 3u + 9/4 = 5/4

(u - 3/2)² = 5/4

(u - 3/2) = ± √5/2

u = ± √5/2 + 3/2 = (± √5 + 3)/2

e^(x) = (± √5 + 3)/2

x = ln((± √5 + 3)/2)

Answer: ln((± √5 + 3)/2) or  ÃƒÂ‚± 0.962

Answer 3

e^(x) + e^(-x) = 3

lne^x+lne^-x=ln3

1x-1x=ln3

0=ln3

ln3/0.......ln3 divided by zero=0

Answer 4

e^x + e^-x = 3

e^x - 3 + e^-x = 0

Multiply through by e^x (don't worry, it can't equal zero):

(e^x)^2 - 3e^x + 1 = 0

Using the substitution u = e^x:

u^2 - 3u + 1 = 0

Now we can use the quadratic formula to solve for u and then

x = ln u

will let us solve for x.

Answer 5

Not sure of the answer, but the first person was wrong.

e^(x) + e^(-x) = 3

ln[e^(x) + e^(-x)]  = ln(3)

Not sure where to go from here.

Note that:

ln[e^(x) + e^(-x)]  = ln(3)

is NOT the same as:

ln[e^(x)] + ln[e^(-x)]  = ln(3)