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# Solving equations with e - C3

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1. e^2x + e^x = 6

I made e^x = y and e^2x = y^2
then i factorised to get y=-3 and y=2
replace y with e^x
NOTE: i used y instead if x because it makes the next part easier to understand.

Then i have e^x = -3 and e^x = 2
Is this wrong because when you add an ln in front to get x you end up with x = ln-3 and x = ln2
2. (Original post by kiiten)
e^2x + e^x = 6

I made e^x = y and e^2x = y^2
then i factorised to get y=-3 and y=2
replace y with e^x
NOTE: i used y instead if x because it makes the next part easier to understand.

Then i have e^x = -3 and e^x = 2
Is this wrong because when you add an ln in front to get x you end up with x = ln-3 and x = ln2
It is right, but you discard the negative solution as being invalid

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3. (Original post by kiiten)
e^2x + e^x = 6

I made e^x = y and e^2x = y^2
then i factorised to get y=-3 and y=2
replace y with e^x
NOTE: i used y instead if x because it makes the next part easier to understand.

Then i have e^x = -3 and e^x = 2
Is this wrong because when you add an ln in front to get x you end up with x = ln-3 and x = ln2
No, just reject x=ln(-3) as you can't take the natural log of a number less than or equal to 0.
4. (Original post by kingaaran)
It is right, but you discard the negative solution as being invalid

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(Original post by NotNotBatman)
No, just reject x=ln(-3) as you can't take the natural log of a number less than or equal to 0.
Ahh i see, so i was right. Does that mean that x = ln2 or am i supposed to write it in a different form?
5. (Original post by kiiten)
Ahh i see, so i was right. Does that mean that x = ln2 or am i supposed to write it in a different form?
Yes, normally you would leave it in exact form unless told otherwise.
6. (Original post by NotNotBatman)
Yes, normally you would leave it in exact form unless told otherwise.
Right and would a similar approach work for

e^4x - e^2x = 6 making e^4x = x^2 and e^2x = x ?
7. (Original post by kiiten)
Right and would a similar approach work for

e^4x - e^2x = 6 making e^4x = x^2 and e^2x = x ?
yes.
8. (Original post by NotNotBatman)
yes.
Thanks
9. (Original post by kiiten)
Right and would a similar approach work for

e^4x - e^2x = 6 making e^4x = x^2 and e^2x = x ?
However, instead of making that substitution, you could note that:

e^4x - e^2x - 6 = 0

(e^2x - 3 ) (e^2x + 2) = 0

Hence, e^2x = 3 or e^2x = -2 (the latter which doesn't have a solution!).

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