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# Can you solve this equation? watch

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1. hi everyone

i'm having difficulty in solving this equation. It's Q1b from C3 June 2007

e^x + 3e^-x = 4

The solutions are 0 and ln 3. How do you get this?

I can get 0 by logic by saying that if 1 + 3 = 4, so the power that e is raised to should equal 1, hence the solution being 0 (e^0 = 1).
But how would you solve this in a way to achieve ln 3 as another solution??

2. what would you multiply the equation by inorder to get a quadratic....
3. Multiply all terms by e^x, shift all terms over to one side of the equation, and you'll have yourself a lovely quadratic in disguise to solve for x.
4. Multiply both sides by e^x. (You're allowed to do this because e^x is never 0.) Then substitute u = e^x, and you'll get a quadratic in u.
5. thanks all of you so much , and thanks for all replying so quickly
6. actually, i read your answers and thought 'oh that's easy' but im trying to do it and im getting stcuk. can someone give me a step or two please?

my attempt:
(e^2x) + 3e^-2x - 4e^x = 0

let u = e^x

u^2 - 3u^2 - 4u = 0
-2u^2 - 4y = 0
-2u^2 = 4u
(divide by 2) -u^2 = 2u

7. (Original post by sheena18)
actually, i read your answers and thought 'oh that's easy' but im trying to do it and im getting stcuk. can someone give me a step or two please?

my attempt:
(e^2x) + 3e^-2x - 4e^x = 0

let u = e^x

u^2 + 3....?

mistake
8. X multiplied by e^-x doesn't give e^-2x it gives 1...
9. oh yeah samson89 e^-x multiplied by e^-x is e^-x+x
= e^0
= 1
gotcha (gosh im dumb)
10. thanks again (a final thanks! i've got the answers now yay)
have a good evening!

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