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Solving equations with e - C3 Watch

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    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
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    (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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    (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.
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    (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?
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    (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.
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    (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 ?
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    (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.
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    (Original post by NotNotBatman)
    yes.
    Thanks
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    (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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