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find the value of k such that the equation e^(2x)=krootx has exactly one solution.

  • math -

    consider the two separate functions
    y = e^(2x) and y = k√x
    To have 1 real root, they must touch each other
    that is, they must have a common tangent at their point of contact
    so for the first : dy/dx = 2e^(2x)
    for the second: dy/dx = (1/2)(k)x^(-1/2)

    they must be equal
    4e^(2x) = k/√x ---> k = 4e^(2x) √x
    from the original: k = e^(2x)/√x

    so 4e^(2x) √x = e^(2x)/√x
    4√x = 1/√x
    x = 4

    then using one of the expressions for k
    k = e(2x)/√x = e^8/2

    check my arithmetic

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