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Math Help - solving for x, involving logs

  1. #1
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    solving for x, involving logs

    y-9 = x^0.25 = (x-1.5)^0.2

    How can I solve this? I guess logs would come in at some point. I tried rearranging a bit but then got this

    0.25/0.2 = logx / log(x-1.5)
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by theowne View Post
    y-9 = x^0.25 = (x-1.5)^0.2

    How can I solve this? I guess logs would come in at some point. I tried rearranging a bit but then got this

    0.25/0.2 = logx / log(x-1.5)
    What is this?

    Should it read

    y-9=\sqrt[4]{x}\color{red}{{-}} \sqrt[5]{x-1.5}?
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  3. #3
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    No

    y=(x)^0.2+9
    y=(x-1.5)^0.25+9

    y-9=x^0.2=(x-1.5)^0.25
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by theowne View Post
    No

    y=(x)^0.2+9
    y=(x-1.5)^0.25+9

    y-9=x^0.2=(x-1.5)^0.25
    ....uhm....I think you mean that you have two simultaneous equations, namely

    y=\sqrt[5]{x}+9
    and y=\sqrt[4]{x-1.5}+9

    Now seeing that if we subtract nine from both we can set them equal to each other getting

    \sqrt[5]{x}=\sqrt[4]{x-1.5}


    Now how are you proposing we solve this?
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  5. #5
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    I do not know. I tried to do log of both sides and get

    0.25 log (x-1.5) = 0.2 log x

    0.25/0.2 = log x / log (x-1.5)

    Don't know what to do further or if that's even right.
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  6. #6
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    Can someone tell me where to go from there, or if that's right?
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  7. #7
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    If this is indeed your question, I don't think logarithms will help you much. There's no way in isolating x from log(x - 1.5) so you can't do all that much after what you've done.

    If you were to attempt to solve it:
    x^{0.2} = (x-1.5)^{0.25}
    x^{\frac{1}{5}} = (x - 1.5)^{\frac{1}{4}}
    x^{4} = (x - 1.5)^{5} (Raised both sides to the power of 20)
    x^{4} = x^{5} + 5x^{4}(-1.5) + 10x^{3}(-1.5)^{2} + 10x^{2}(-1.5)^{3} + 5x(-1.5)^{4} + (-1.5)^{5} (via binomial theorem)
    0 = x^{5} - 8.5x^{4} + 22.5x^{3} - 33.75x^{2} + 25.3125x - 7.59375

    You would somehow have to find the roots of that quintic equation, keeping in mind that x-1.5\geq 0

    There seems to be one real root: x \approx 5.2926
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