solve for y,, 3*e^(y) + 5*e^(-y) = 16?

Question

how do i solve this,, to find y

been tryin this for awhile now

thanks

Answer 1

3 e^y + 5 e^-y = 16

let e^y = u
3u + 5u^-1 = 16
multiply u at both sides
3u^2 + 5 = 16u
3u^2 -16u + 5 = 0
(3u-1)(u-5) = 0
3u-1 = 0 or u-5 = 0
u = 1/3 or u = 5
e^y = 1/3 or e^y = 5
y = ln 1/3 or ln 5
y = -ln 3 or ln 5

Answer 2

Let e^y=x
then e^(-y)= 1/x

3x+5/x=16
3x^2+5=16x
3x^2-16x+5=0

x1=(8+7)/3=5
x2=(8-7)/3=1/3

e^y=5

y=ln5

e^y=1/3

y=-ln3

Answer 3

5e^(-y) = 5/e^y

Multiply through by e^y

3e^(y²) + 5 = 16e^y

Now you can substitute x=e^y and solve as a quadratic...

I would have you do this on your own, but feel compelled to check the work of the person who submitted before me. (No one had answered when I began).

3x² - 16x + 5 = 0

(3x - 1)(x - 5) = 0

3e^y = 1

e^y = 1/3

y = ln(1/3)

e^y = 5

y=ln(5)

Answer 4

=> 3*e^y + 5/e^y = 16
=> 3*e^y*e^y + 5 = 16*e^y
=> 3*e^2y + 5 = 16*e^y
apply log on both sides
=> log3 + 2y + log5 = log16 + y
=> 2y - y + log15 = log16
=> y = log16 - log15
=> y = log (16/15)