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Math Help - Please solve the equation for "Y"? (a few more problems!)

  1. #1
    ADH
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    Cool Please solve the equation for "Y"? (a few more problems!)



    Hello,

    I wasn't completly sure on the ones I have solved.

    Most of these I have already solved, and have included the answer I got! Please tell me if I have gotten the answer right. Thank you. Some of these I do need a bit of help with.
    If you don't have time to do all of them and check them, you can just do a couple! Anything will help! Thank you.

    1. (e^-y) + (e^y) = 2
    I WASN'T SURE ON THIS ONE. PLEASE HELP!

    2. (x^2)(ln y) - xlny = 3
    I got y = e^(3/(x^2)-x)

    3. (ln y )^3 = 5^x
    I got e^third root(5x)

    4. ln(1+ln y ) = 3
    I WASN'T SURE ON THIS ONE. PLEASE HELP! (I didn't know if I was supposed to distribute the "ln" or what. Thank you.

    5. (e^(-2y)) + 1 = 3e^x
    I got y = (ln3(e^x) - 1)/(-2lne)

    6. y^x = lnx + 2
    I got y = x root (lnx + 2)

    Thank you everyone!
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  2. #2
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    e^{-y} + e^y = 2

    \frac{1}{e^y} + e^y = 2

    1 + e^{2y} = 2e^y after multiplying both sides by e^{y}

    e^{2y} - 2e^y + 1 = 0

    Y^2 - 2Y + 1 = 0 if we let Y = e^y

    (Y-1)^2 = 0

    Y-1 = 0

    Y = 1

    e^y = 1

    y = \ln{1}

    y= 0.
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  3. #3
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    Quote Originally Posted by ADH View Post

    Hello,

    I wasn't completly sure on the ones I have solved.

    Most of these I have already solved, and have included the answer I got! Please tell me if I have gotten the answer right. Thank you. Some of these I do need a bit of help with.
    If you don't have time to do all of them and check them, you can just do a couple! Anything will help! Thank you.

    1. (e^-y) + (e^y) = 2
    I WASN'T SURE ON THIS ONE. PLEASE HELP!

    2. (x^2)(ln y) - xlny = 3
    I got y = e^(3/(x^2)-x)

    3. (ln y )^3 = 5^x
    I got e^third root(5x)

    4. ln(1+ln y ) = 3
    I WASN'T SURE ON THIS ONE. PLEASE HELP! (I didn't know if I was supposed to distribute the "ln" or what. Thank you.

    5. (e^(-2y)) + 1 = 3e^x
    I got y = (ln3(e^x) - 1)/(-2lne)

    6. y^x = lnx + 2
    I got y = x root (lnx + 2)

    Thank you everyone!
    \ln{(1 + \ln{y})} = 3

    1 + \ln{y} = e^3

    \ln{y} = e^3 - 1

    y = e^{e^3 - 1}.
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  4. #4
    Senior Member Educated's Avatar
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    The rest (2,3,5 and 6) seem correct to me.

    For number 5:

    y = (ln3(e^x) - 1)/(-2lne) can be simplified to: y = (ln3(e^x) - 1)/(-2)

    Remember ln(e^1) = 1
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  5. #5
    ADH
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    Wow thanks everyone! Hey "Educated", are you from Hastings, Michigan?
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  6. #6
    Senior Member Educated's Avatar
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    That's off the topic, but no.

    I'm from Hastings, New Zealand. If you go into my profile, you will see.
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