E^x + e^-x = 3 (functions and logs)?

Question

please explain the steps to reach the answer

Answer 1

e^x=t

obtain

t+1/t=3 or

t^2-3t+1=0

solve for t (second degree equation) and find t1,t2

then solve e^x=t1 or e^x=t2

t=(3+/-sqrt(5))/2

then

x=ln((3+sqr(5))/2) or

x=ln((3-sqr(5))/2)

Answer 2

okay.... add a log onto each term, but because it uses an e add a ln to each term,

ln e^x + ln e^-x = ln 3

take the exponents from the logs and put them infront of their logs. like so

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

and sorry, i dont really remember the rest, (and i dont have a calculator to figure out the logs of each...) but pretty much, just find out what all the logs are, and isolate x. (but honestly i might have done something wrong, because it seems that x is going to disappear once your done witht he problem)

hope it helps..... but it probably didnt :P

Answer 3

e^x + e^-x = 3

Subtract 3 from each side:

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

Multiply both sides by e^x:

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

Replace e^x with t temporarily:

t^2 - 3t + 1 = 0

Solve using the quadratic formula:

t = (3 + √5) / 2, t = (3 - √5) / 2

Put e^x back in for t:

e^x = (3 + √5) / 2, e^x = (3 - √5) / 2

Take natural logarithms to solve for x:

x = ln ((3 + √5) / 2), x = ln ((3 - √5) / 2).

Answer 4

e^x+(1/e^x)=3

[e^(x^2)+1]/(e^x)=3

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

x^2+1=ln(3e^x)=ln3+lne^x

x^2+1=ln3+x

x^2-x+1-ln3=0 (use calculator to find zeros)