Quick algebra Problem help?!?

6=[(84)/(1+7e^(-x))]. six is equal to 84 over 1 + 7e, the e raised to the -x. Thanks to whoever tries!

Comments

  • multiply both sides by 1+7e^-x to get

    6*(1+7e^-x) = 84

    divide by 6

    1+7e^-x = 84/6 = 14

    subtract 1

    7e^-x = 13

    divide by 7

    e^-x = 13/7

    natural log of both sides

    ln(e^-x) = ln(13/7)

    remember ln (e^x) = x

    so, -x = ln(13/7)

    x = -ln13/7

    also remember -ln y = ln (1/y)

    so x = ln (7/13)

  • 6=[(84)/(1+7e^(-x))]

    OK, first thing is first. You have to get rid of the denominator.

    Multiply both sides by 1+7e^-x

    6(1+7e^-x) = 84

    Distribute the 6

    6+42e^-x = 84

    Subtract the 6

    42e^-x = 72

    Divide by 42

    e^-x = 72/42 = 13/7

    Use the reciprocal of e to eliminate the negative exponent

    1/e^x = 13/7

    Multiply both sides by e^x

    1 = 13/7e^x

    Divide both sides by 13/7

    7/13 = e^x

    Time to go logging

    ln(7/13) = ln(e^x)

    Bring the x down front

    ln(7/13) = x(ln)e

    ln(e) cancels

    ln(7/13) = x

    -----------------

    >.<

    That was a doozie. =D

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