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Math Help - Please solve the equation for "Y"?

  1. #1
    ADH
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    Cool Please solve the equation for "Y"?

    Hello,

    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.

    So let's start.

    1. 3y^e = x-5
    I got y = e root((x-5)/3)

    2. lny^3 = 3x
    I got y = 3rd root(e^3x)

    3. 3 + ln(abs y) = -x
    I got e^(-x-3) = y

    4. ln(y-3) = x-2
    I WAS HAVING TROUBLE WITH THIS ONE! PLEASE HELP!

    5. 3e^y = x-5
    I got y = ln(x-5)/3

    Thank you!
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  2. #2
    Senior Member Educated's Avatar
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    The first 3 seem correct to me.

    4. ln(y-3) = x-2

    \ln{(y-3)} = x-2

    e^{\ln{(y-3)}} = e^{x-2}

    y-3 = e^{x-2}

    y = e^{x-2} + 3

    5 seems incorrect. Here's what got:

    3e^y = x-5

    e^y = \frac{x-5}{3}

    \ln(e^y) = \ln(\frac{x-5}{3})

    y= \ln(\frac{x-5}{3})
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  3. #3
    Member Traveller's Avatar
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    1) seems okay.

    2) y= e^(3rd root ( 3x) )

    3) y = +- e^(-x-3)

    4) y = e^(x-2) + 3

    5) y = ln((x-5)/3) which I think you meant anyway.

    EDIT : I see now that by lny^3 you meant ln(y^3), in which case your answer is correct.
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  4. #4
    ADH
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    Wow. Thank you so much. I made one mistake, for number 4, the actual question was ln(abs(y-3)) = x-2 . So would the answer be plus/minus? Thank you!
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  5. #5
    Senior Member Educated's Avatar
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    Actually the second one can be simplified down to just:

    y = e^x
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  6. #6
    Member Traveller's Avatar
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    Quote Originally Posted by ADH View Post
    Wow. Thank you so much. I made one mistake, for number 4, the actual question was ln(abs(y-3)) = x-2 . So would the answer be plus/minus? Thank you!
    y = 3 +- e^(x-2)
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