pre calculous

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solve each equation algebraically and check it by substituing into the orignall equation.


Method of your choice by solving.






  • pre calculous -

    e^(.035x) = 4
    take ln of both sides
    ln(e^(.035x)) = ln4
    .035x lne = ln4, but lne = 1
    x = ln4 / .035 = appr. 39.6

    log x^2 = 6
    by definition:
    x^2 = 10^6
    x = 10^3 = 1000

    do logx^4 = 2 the same way

    2x-2^-x/2=4 : I have a feeling that is not what you meant, without brackets I cannot tell what the equation is. I think the first term is probably 2^x


    Again, I think you meant:
    e^x+e^(-x/2) = 4

    let e^(-x/2) = y , where y is a positive real number
    then e^(x/2) = 1/y
    and e^x = 1/y^2

    so 1/y2 + y = 4
    1 + y^3 = 4y^2
    y^3 - 4y^2 + 1 = 0
    I used a program to find
    y = .54 or y = 3.94 appr. , there is also a negative root which would not be allowed

    so e^(-x/2) = .54 or ..... use the other root
    -x/2 = ln.54
    x = 1.2324 or ......

    last question, way too ambiguous without brackets.

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