Question

solve for x? (e^x+e^-x)/(e^x-e^-x)=3

Answer 1

nice question!!

e^2x/e^x + 1/e^x =3

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

e^2x + 1 = 3e^x

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

e^x^2 -3e^x + 1 = 0 <---note this is a quadratic equation

e^x = (3 +/- √(9-4))/2 <---so use the quadratic formula

e^x = (3+√(5))/2 or (3-√(5))/2

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

x = 0.96242365 or -0.96242365


hope this helps!!
:)

Answer 2

i guess it will be easier to apply componendo-dividendo

Answer 3

@Rohangrr it seems you just covered the numerator..there's still a denominator

Answer 4

i tried several times but i could not solve it.

Answer 5

@Igbasallote ya!! but @MHAMMED u can solve for deno.. also !!

Answer 6

Using the same way ? give me few second i am trying to understand what you did.

Answer 7

oh wait..forgot to distribute...lemme repeat that...

Answer 8

Guys really thankful for your help all of you

Answer 9

rohan's first steps were right..

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

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

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

e^2x + 1 = 3(e^2x - 1)
e^2x + 1 = 3e^2x - 3
1+3 = 3e^2x - e^2x
4 = 2e^2x
2 = e^2x

sqrt both sides...

sqrt (2) = e^x

ln both sides...

x = ln (sqrt 2)

Answer 10

well, i think my method is easier :|

Answer 11

i think i got it wrong lol..wolfram agrees with @Arnab09

Answer 12

all of you did well. thanks so much .

Answer 13

cot hx = 3
rest is here
http://en.wikipedia.org/wiki/Inverse_hyperbolic_function
x = 1/2 (ln (4/2)) = 1/2 ln 2

Answer 14

@lgbasallote I don't think you were necessarily wrong, your answer just wasn't simplified. Since sqrt(2) = 2^(1/2) is in the logarithm, it can be taken outside using the power rule. So you'd still have ln(2)/2. :P

Answer 15

oh oh yeah..glad i was right :DDD thanks @AccessDenied (Y) boost to the morale hehe