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Thread: Urgent help

  1. April 30th 2009, 10:28 AM #1
    manutd4life
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    Urgent help

    How to find the value of x in e^x - 8e^(-2x) = 0
    please help

    i used ln but cant do it please help
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  2. April 30th 2009, 12:13 PM #2
    TwistedOne151
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    e^x-8e^{-2x}=0, so
    e^x=8e^{-2x}. Multiplying both sides by e^{2x}, we get
    e^{3x}=8
    so then
    3x=\ln{8}
    x=\frac{1}{3}\ln{8}=\ln\left(\sqrt[3]{8}\right)=\ln2.

    --Kevin C.
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  3. April 30th 2009, 12:30 PM #3
    manutd4life
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    Thank you very much
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  4. April 30th 2009, 01:47 PM #4
    HallsofIvy
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    Quote Originally Posted by TwistedOne151 View Post
    e^x-8e^{-2x}=0, so
    e^x=8e^{-2x}. Multiplying both sides by e^{2x}, we get
    e^{3x}=8
    e^{3x}= (e^x)^3 so this is (e^x)^3= 8 so you can take the cube root at this point.
    e^x= 2

    so then
    3x=\ln{8}
    x=\frac{1}{3}\ln{8}=\ln\left(\sqrt[3]{8}\right)=\ln2.

    --Kevin C.
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  5. April 30th 2009, 02:10 PM #5
    pflo
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    e^x=8e^{-2x}
    \ln{e^x}=\ln(8e^{-2x})
    \ln{e^x}=\ln(8)+\ln(e^{-2x})
    x=\ln(8)+(-2x)
    3x=\ln(8)
    x=\frac{1}{3}\ln(8)
    x=\ln(8^\frac{1}{3})
    x=\ln(2)
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