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Math Help - Need help making X the subject

  1. #1
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    Need help making X the subject

    Hi all, the attached document shows 7 questions to "rearrange to make x the subject". I have answers for some of them but other I'm completely stumped on where to even begin. Here are my answers so far

    1)i) y= mx+c
    y-c= mx
    y-c/m = x

    ii) y=mx+cx+c
    y-c = mx+cx
    I'm not sure where to go after this, I would come out with a final answer of y-c/mc = 2x?

    iii) y=x+c/a/b - once again i'm not even sure where to go with this to start with

    iv) y=x^2+c
    y-c=x^2
    sqroot(y-c)=x

    v) y=e^Lx -1
    y+1 = e^Lx
    not sure where to go with this, I imagine you use logs somehow

    vi) y=xcos^2 + xsin^2
    Completely stumped

    vii) y=log(e^x)
    Not sure where to begin

    All help is apprecaited! many thanks
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  2. #2
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    Re: Need help making X the subject

    Quote Originally Posted by funbuoj View Post
    Hi all, the attached document shows 7 questions to "rearrange to make x the subject". I have answers for some of them but other I'm completely stumped on where to even begin. Here are my answers so far

    1)i) y= mx+c
    y-c= mx
    y-c/m = x
    Be careful how you write this! You are dividing y- c by m so that would be written (y- c)/m= x. What you wrote would normally be interpreted as y- (c/m)= x which is not correct.

    ii) y=mx+cx+c
    y-c = mx+cx
    I'm not sure where to go after this, I would come out with a final answer of y-c/mc = 2x?
    No, mx+ cx= (m+c)x. To solve (m+ c)x= y- c, you get x= (y- c)/(m+ c)

    iii) y=x+c/a/b - once again i'm not even sure where to go with this to start with
    I don't know what you mean by this. Is that supposed go be y= ((x+ c)/a)/b or y= x+ (c/a)/b?

    iv) y=x^2+c
    y-c=x^2
    sqroot(y-c)=x
    The equation x^2= y- c has two solutions, x= sqrt(y-c), which you give and x= -sqrt(y- c).

    [quote]v) y=e^Lx -1
    y+1 = e^Lx
    not sure where to go with this, I imagine you use logs somehow
    Assuming you mean e^(Lx), yes. The definition of the (natural) logarithm is that it is the inverse function to e^x: log(e^x)= x. So if you take the logarithm of both sides, log(y+ 1)= log(e^(Lx))= Lx. Now divide both sides by L.

    vi) y=xcos^2 + xsin^2
    Completely stumped
    So am I! I have no idea what "cos" and "sin", could possibly mean without an argument! Is it possible you meant cos^2(x)+ sin^2(x) or xsin^2(x)+ xcos^2(x)? If it is either of those, use the fact that sin^2(x)+ cos^2(x)= 1.

    vii) y=log(e^x)
    Not sure where to begin

    All help is apprecaited! many thanks
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