Exponential Equation?

Question

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

Find the values of y!

Please show your workings so I can actually understand it!

Thanks Ben and Koalahash that really helped.

DLM - There is a minus before the y on the 5 bit, so you can combine the terms...

That should say CANT above!!

Answer 1

so, if you make x = e^y then you get

3x + 5x^-1 = 16
3x +5/x = 16 make the left side have x as a denominator

(3x^2 + 5)/x = 16 multiply both sides by x

3x^2 +5 = 16x
3x^2 - 16x + 5 = 0 use quadratic equation or factor

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

3x -1 = 0 and x - 5 = 0

x = 1/3 and 5 but x = e^y

so e^y = 1/3, 5 take the natural log (ln) of both sides

ln e^y = ln (1/3) and ln 5 and ln e^y = y

so,

y = ln (1/3) and ln 5

hope that helps

Answer 2

Let x=e^y, then the equation becomes a quadratic, which you can factor or use the quadratic formula for.

More details:

Plugging in x=e^y gives:

3x + 5/x = 16

multiplying through by x gives:

3x^2 - 16x + 5 = 0
(3x - 1)(x - 5) = 0
x = 1/3, 5

so e^y = 1/3, 5

so y = ln(1/3), ln(5)

Answer 3

3e^y + 5e^-y = 16
3e^2y + 5 = 16e^y
3e^2y - 16e^y + 5 = 0
Let X = e^y so X² = e^2y
3x² - 16x + 5 = 0
(3x - 1)(x - 5) = 0
x = 1/3 , x = 5
e^y = 1/3 , 5
yloge(e) = log1 - log3
y = 0 - log3
y = - log3 and the other answer is given by:-
e^y = 5
yloge(e) = log5
y = log 5

Answer 4

3e^y + 5e^-y = 16
multiply through by e^y
3e^2y + 5 = 16e^y
3e^2y-16e^y+5=0
factorise this quadratic
in e^y-
(e^y-5)(3e^y-1)=0
>>>e^y=5 or e^y=1/3

when e^y=5, y=ln5
when e^y=1/3, y =ln(1/3)
= -ln3

hence,y= ln5 or -ln3

i hope that this helps

Answer 5

3 (e^y + e^-y ) + 2 e^-y = 16

3 ( e ^ y-y ) + 2 e ^-y = 16 e ^ y-y = e^0 = 1

3 ( 1 ) + 2 e ^-y = 16

2 e ^ -y = 13

e^-y = 13/2

ln ( e ^-y) = ln ( 13/2)

-y = ln (13/2)

y = -ln (13/2)

you can find the value of ln (13/2) by your calculator.
good luck :-)

Answer 6

Combine like terms
8e^y = 16

Divide both sides by 8
e^y = 2

Take the natural logarithm of both sides (ln "cancels" e^)
y = ln 2