exponentials problem (1 Viewer)

[ezzy85]

How would you solve this?

e^2x = 3e^-x

Thanks

 

[Huy]

this can't be right,
x=0?

e^2x = 3e^-x
e^2x = e^-3x, recall logaB^c = clogaB

loge.e 2x = loge.e -3x; n.b. loge.e = 1

2x = -3x
2x + 3x = 0
5x = 0
x = 0?

what the!

 

[BlackJack]

Originally posted by ezzy85
How would you solve this?

e^2x = 3e^-x

Thanks

e^2x = 3e^-x

A = e^{ln A} = ln (e^A)

e^2x = e^{ln 3} . e^-x

e^2x = e^{ln 3 - x}

Equating superscripts...

2x = ln3 - x

x = ln3 / 3

 

[ND]

Originally posted by Huy


e^2x = 3e^-x
e^2x = e^-3x

That doesn't work because e^-x + e^-x + e^-x doesn't equal e^-x*e^-x*e^-x

 

[ezzy85]

thanks guys

 

[TimTheTutor]

Mmm.. rather than use "tricks" such as 3 = e^(ln3) I would go for the straightforward option using indice laws then solving equation

e^(2x) = 3e^(-x)
e^(2x) = 3 / e^x
e^(2x + x) = 3
e^(3x) = 3
3x = ln 3
x = (ln3)/3