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Math Help - logarithmic problem

  1. #1
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    logarithmic problem

    5e^.2x=7
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  2. #2
    Super Member bigwave's Avatar
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    Cool

    first just need to clarify do you mean

    5 e^{.2}x=7 or 5e^{.2x}=7
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  3. #3
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    Quote Originally Posted by math321 View Post
    5e^.2x=7

    Hint: if e^b=y , then taking natural logarithm on both sides we get  b=\ln y , provided, of course, y>0

    Tonio
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  4. #4
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    the second sum


    and i want to find out...instead of I using the up arrow how do i get someting raised to a power
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  5. #5
    Super Member bigwave's Avatar
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    ln(5e^.2x) = ln7

    is

     <br />
\frac{x}{5} + ln5 = ln7<br />

    so x = 5(ln7 - ln5) or 5ln\frac{7}{5}

    use latex to get the power

    x^2 then highlight the expression with \Sigma icon and you get x^2 after it is compiled in preview or after saving and posting
    Last edited by bigwave; December 21st 2009 at 11:04 AM. Reason: spelling
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  6. #6
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    thanks alot

    i got an exam tomoro so i am just practicing
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