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Math Help - Solving equations and inequalities (logs and e)

  1. #1
    Junior Member
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    Solving equations and inequalities (logs and e)

    I am not sure what to do with the following problems.
    We are suppose to solve using algebraic method and give an exact answer in order to receive the most credit.

    These are just some examples (there is alot more) but i feel that once i understand how to do these the rest will be some wat easier to figure out.


    a)x^log(x)= x^3/100

    b)3< 3e^(1/x^2+1)<3e

    c)|10^x^2-10000|<9000

    Once again thank you very much for any help. I really appreciate it.

    -Gus
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  2. #2
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    Hello, Danktoker!

    a) x^log() = x^3/100

    Is that: . x^{\log x} \:=\:\frac{x^3}{100}\;\;\text{ or }\;\;x^{\log x} \:=\:x^{\frac{3}{100}}



    b) 3 < 3e^(1/x^2+1) < 3e

    Is that: . 3\:<\:3e^{\frac{1}{x^2} + 1} \:<\:3e\;\;\text{ or }\;\;3\:<\:3e^{\frac{1}{x^2+1}} \:<\:3e



    c) |10^x^2-10000| < 9000

    We have: . \left|10^{x^2} - 10,\!000\right| \;<\;9,\!000

    which equals: . -9,\!000 \;<\;10^{x^2} - 10,\!000\;<\;9,\!000

    Add 10,000: . 1,\!000 \;<\; 10^{x^2} \;<\;19,\!000

    Take logs: . \log(1000) \;<\;\log\left(10^{x^2}\right) \;<\;\log(19,\!000)

    . . Note: . \log(19,\!000) \:=\:\log(19\cdot10^3) \:=\:\log(19) + \log(10^3)


    \text{So we have: }\;\underbrace{\log\left(10^3\right)}_{\text{This is 3}} \;<\;x^2\underbrace{\log(10)}_{\text{This is 1}} \;<\;\log(19) + \underbrace{\log\left(10^3\right)}_{\text{This is 3}}

    . . . . . . . . . . . . . . 3 \quad<\quad x^2 \quad<\quad \log(19) + 3

    Therefore:. . . . . . . \sqrt{3}\;\;<\;\;x\;\;<\;\;\sqrt{\log(19) + 3}

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  3. #3
    Junior Member
    Joined
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    for the first one it is the first one.

    for the second one it is the second one.

    Sory i dnot know how to copy what u did..

    I hope that is clear.
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