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Interesting problem... degree level maybe?! c3 homework :S e^2x = 10 watch

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    e^(2x) = 10x

    how do you solve

    homework says, find exact solutions..
    i have found one solution using iteration but can't find the other (there are 2..)

    how do you find the exact solutions?
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    you are going with this the wrong way, by exact, they mean as a surd, or a fraction, you use ln function in this
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    (Original post by desertwarrior)
    e^(2x) = 10x

    how do you solve

    homework says, find exact solutions..
    i have found one solution using iteration but can't find the other (there are 2..)

    how do you find the exact solutions?
    The title says e^{2x}=10 and your post says e^{2x}=10x Which is correct?
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    (Original post by steve2005)
    The title says e^{2x}=10 and your post says e^{2x}=10x Which is correct?
    sorry!!

    it's

    e^(2x) = 10x
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    I don't think there are any exact solutions.

    If you're really interested, look up the Lambert W-function.
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    (Original post by DFranklin)
    I don't think there are any exact solutions.

    If you're really interested, look up the Lambert W-function.
    I think that's what wolfram alpha used.. thanks

    is there any way to use iteration to find the 2nd solution?
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    (Original post by desertwarrior)
    e^(2x) = 10x

    how do you solve

    homework says, find exact solutions..
    i have found one solution using iteration but can't find the other (there are 2..)

    how do you find the exact solutions?
    Since you are asked for EXACT solutions I think there is an error in the question.
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    if it = 10 it would be easy, just use natural logs.

    But 10x seems much harder than C3?
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    (Original post by mfc20)
    if it = 10 it would be easy, just use natural logs.

    But 10x seems much harder than C3?
    yea, weird o.0 use ln on both sides....
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    Ok, you seem to have a misconception that an answer that is accurate to a large number of decimal places is exact. It isn't.

    x_{n+1}=\frac{e^{2x_n}}{10}

    x_{n+1}=\frac{\ln {10 x_n}}{2}

    Try both with x_1 = 1
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    (Original post by Mr M)
    Ok, you seem to have a misconception that an answer that is accurate to a large number of decimal places is exact. It isn't.

    x_{n+1}=\frac{e^{2x_n}}{10}

    x_{n+1}=\frac{\ln {10 x_n}}{2}

    Try both with x_1 = 1
    Ah thanks Mr M!!! That's what i've been working on, but I get one solution converging to 1.2713.........
    From the shape of the y=e^2x and y=10x graphs I would assume that there surely should be 2 solutions.. but iteration can only find one from my understanding..

    am I forgetting something,
    P.s. repped you thanks
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    (Original post by desertwarrior)
    Ah thanks Mr M!!! That's what i've been working on, but I get one solution converging to 1.2713.........
    From the shape of the y=e^2x and y=10x graphs I would assume that there surely should be 2 solutions.. but iteration can only find one from my understanding..

    am I forgetting something,
    P.s. repped you thanks
    You said the question required EXACT solutions. So iteration will not give the required solution.
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    (Original post by steve2005)
    You said the question required EXACT solutions. So iteration will not give the required solution.
    yeah but still better than nothing right?
    do you know how to find the other solution?
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    How about substitution?
    let  u=2x
    so  e^u = 5
    therefore  u={\ln {5}}
    use that to find  x
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    I agree with \frac{1}{2}ln(5)

    I did it by differentiating both sides w.r.t x and it becomes simple to solve. However, the graphical idea of there being two solutions seems quite completing.

    I might have to get MatLab out.

    EDIT: How does the substitution work?
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    (Original post by ollimollimoll)
    How about substitution?
    let  u=2x
    so  e^u = 5
    therefore  u={\ln {5}}
    use that to find  x
    it needs a u after the 5, e^u = 5u
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    (Original post by desertwarrior)
    yeah but still better than nothing right?
    do you know how to find the other solution?
    This is not exact, obviously

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    (Original post by Henerz)

    EDIT: How does the substitution work?
    Good old substitution:
    replace 2x with u, and you can work it out easily. Then, afterwards when solved with u, put 2x back in. Simple

    EDIT:that doesn't work, sorry
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    (Original post by Monochrome)
    it needs a u after the 5, e^u = 5u
    does it?

    Ah! you are right. oh well, I tried...
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    (Original post by ollimollimoll)
    Good old substitution:
    replace 2x with u, and you can work it out easily. Then, afterwards when solved with u, put 2x back in. Simple
    i think that is wrong
 
 
 
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