Year 13 Maths Help Thread

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    (Original post by kiiten)
    Ohhh so ....
    e^lny = e^2
    y = e^2
    yea thats the answer but im 99% you would lose marks because  \displaystyle \displaystyle \ln(a + b + c) \neq \ln(a)  + \ln(b) + \ln(c) but yet youve used it in your working, and it only didnt matter because  \displaystyle y = e^2

    Think of it like having to find x when  \displaystyle x = 2^2 , if you do  \displaystyle x = 2 \times 2 you still get the right answer but you're working it wrong and you've just fluked it, that's kind of similar to whats happened here.

    My previous post includes 2 ways of doing it, http://www.thestudentroom.co.uk/show...&postcount=875.

    Another way is to get to  \displaystyle \ln(e^5 - e^2 + y) + \ln y = 7 then "e" both sides to get;

     \displaystyle e^{\ln(e^5 - e^2 + y) + \ln y} = e^7

    you'll then get a quadratic to solve
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    (Original post by DylanJ42)
    yea thats the answer but im 99% you would lose marks because  \displaystyle \displaystyle \ln(a + b + c) \neq \ln(a)  + \ln(b) + \ln(c) but yet youve used it in your working, and it only didnt matter because  \displaystyle y = e^2

    Think of it like having to find x when  \displaystyle x = 2^2 , if you do  \displaystyle x = 2 \times 2 you still get the right answer but you're working it wrong and you've just fluked it, that's kind of similar to whats happened here.

    My previous post includes 2 ways of doing it, http://www.thestudentroom.co.uk/show...&postcount=875.

    Another way is to get to  \displaystyle \ln(e^5 - e^2 + y) + \ln y = 7 then "e" both sides to get;

     \displaystyle e^{\ln(e^5 - e^2 + y) + \ln y} = e^7

    you'll then get a quadratic to solve
    Thanks

    Like this?

    If this is right then im not sure how you would factorise after this.

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    (Original post by kiiten)
    Thanks

    Like this?

    If this is right then im not sure how you would factorise after this.

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    Anyone?

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    Your first half of the LHS is right but you have brought e^...+ lny down to give lny, which isn't quite right

    Remember that x^{a + b} = x^a \times x^b .
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    (Original post by SeanFM)
    Your first half of the LHS is right but you have brought e^...+ lny down to give lny, which isn't quite right

    Remember that x^{a + b} = x^a \times x^b .
    What do you mean i havent brought down lny??
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    (Original post by kiiten)
    What do you mean i havent brought down lny??
    Okay, how did you get from the line with e^ln(...) to the line below? how did you get the +lny in the equation below it?
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    (Original post by SeanFM)
    Okay, how did you get from the line with e^ln(...) to the line below? how did you get the +lny in the equation below it?
    I moved everything after e^ln in front of e^ln (e^5 - e^2 etc.). Then e^ln cancels out which leaves e^5 - e^2 etc.
    lny is in the equation because that was also part of the index notation?
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    (Original post by kiiten)
    Different ques

    ???? I got y = 2 / ln

    Where did i go wrong?

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    I'm not quite sure what you are trying to do but there is a whole lot simpler way to solve it if you get rid off the natural logarithm.

    Just use the fact that \ln(x)+\ln(y)=\ln(xy)=7

    So xy=e^7

    Then rearrange for, say, x and we get x=\frac{e^7}{y}

    Subbing it into the first equation we get: \frac{e^7}{y}-y=e^5-e^2

    Which leads to e^7-y^2=y(e^5-e^2) and just solve for y and then x.
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    (Original post by kiiten)
    I moved everything after e^ln in front of e^ln (e^5 - e^2 etc.). Then e^ln cancels out which leaves e^5 - e^2 etc.
    lny is in the equation because that was also part of the index notation?
    the e^5 - e^2 - y can be taken down but notice that it adds lny...

    What you've done isn't completely correct is what I'm trying to say.

     e^{a+b} = e^a \times e^b where a and b are whatever you want.
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    (Original post by RDKGames)
    I'm not quite sure what you are trying to do but there is a whole lot simpler way to solve it if you get rid off the natural logarithm.

    Just use the fact that \ln(x)+\ln(y)=\ln(xy)=7

    So xy=e^7

    Then rearrange for, say, x and we get x=\frac{e^7}{y}

    Subbing it into the first equation we get: \frac{e^7}{y}-y=e^5-e^2

    Which leads to e^7-y^2=y(e^5-e^2) and just solve for y and then x.
    Thanks - i understand what youre saying here but im not too sure how you would solve to find y.
    EDIT: sorry its in pencil :3
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    (Original post by SeanFM)
    the e^5 - e^2 - y can be taken down but notice that it adds lny...

    What you've done isn't completely correct is what I'm trying to say.

     e^{a+b} = e^a \times e^b where a and b are whatever you want.
    So what happens to the lny? Do you mean it doesnt work because lny isnt inside the bracket?
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    (Original post by kiiten)
    So what happens to the lny? Do you mean it doesnt work because lny isnt inside the bracket?
    That is what I am trying to show you (what happens to the lny)

    Like I said from the previous post,

    In this case a is ln(e^5 - e^2 + y) and b is ln(y).

    So e^a+b is e^(ln(e^5 - e^2 + y) * e^(lny) = ....

    Now you should be able to see the difference between what you did in post 882 and what has just been done.
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    (Original post by SeanFM)
    That is what I am trying to show you (what happens to the lny)

    Like I said from the previous post,

    In this case a is ln(e^5 - e^2 + y) and b is ln(y).

    So e^a+b is e^(ln(e^5 - e^2 + y) * e^(lny) = ....

    Now you should be able to see the difference between what you did in post 882 and what has just been done.
    Ahh i see. Thank youu - sorry i just wasnt getting it at all but i understand now. Just to check does this lead to:

    y(e^5 - e^2 + y) = e^7
    y = e^7 / (e^5 - e^2 + y)
    y = e^2 - e^5 + y/(e^5 - e^2 + y)

    ??
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    (Original post by kiiten)
    Ahh i see. Thank youu - sorry i just wasnt getting it at all but i understand now. Just to check does this lead to:

    y(e^5 - e^2 + y) = e^7
    y = e^7 / (e^5 - e^2 + y)
    y = e^2 - e^5 + y/(e^5 - e^2 + y)

    ??
    After your first line of the quoted post it is easier to turn it into a quadratic, as you can see you've reached a dead end.. I think.
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    (Original post by SeanFM)
    After your first line of the quoted post it is easier to turn it into a quadratic, as you can see you've reached a dead end.. I think.
    Yep . Ohh So:

    ye^5 - ye^2 + y^2 = e^7
    y^2 - ye^2 + ye^5 - e^7 = 0
    y(y - e^2) + e^5 (y - e^2) = 0
    (y+ e^5)(y - e^2) = 0
    y = e^2

    Finally, yay - is this right? - question only asks for +ve solutions so the other bracket is invalid.
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    (Original post by kiiten)
    Yep . Ohh So:

    ye^5 - ye^2 + y^2 = e^7
    y^2 - ye^2 + ye^5 - e^7 = 0
    y(y - e^2) + e^5 (y - e^2) = 0
    (y+ e^5)(y - e^2) = 0
    y = e^2

    Finally, yay - is this right? - question only asks for +ve solutions so the other bracket is invalid.
    You can always check the solutions of an equation by putting them back into the equation
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    (Original post by SeanFM)
    You can always check the solutions of an equation by putting them back into the equation
    Yea I think i figured it out at last. The solutions work. Thanks so much for your help

    One more question - dy/dx of (x+1)^3(x+2)^6 do you use the chain rule?
    I ended up with 3(x+1)^2 6(x+2)^5 but the question wants it in the form (x+1)^2 (x+2)^5 (ax+b) ??
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    (Original post by kiiten)
    Yea I think i figured it out at last. The solutions work. Thanks so much for your help

    One more question - dy/dx of (x+1)^3(x+2)^6 do you use the chain rule?
    I ended up with 3(x+1)^2 6(x+2)^5 but the question wants it in the form (x+1)^2 (x+2)^5 (ax+b) ??
    Think about the product rule too
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    (Original post by SeanFM)
    Think about the product rule too
    Ah but im not sure about using the product rule with brackets. I tried it with (x+1)^3 and got:

    (x+1)3(x+1)^2 + (x+1)^3 ?
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    (Original post by kiiten)
    Ah but im not sure about using the product rule with brackets. I tried it with (x+1)^3 and got:

    (x+1)3(x+1)^2 + (x+1)^3 ?
    The product rule is when you have two functions multiplying eachother - you're mixing it up with the chain rule a bit, though you do have to (spoiler alert) use the chain rule here.

    If you say f(x) = (x+1)^3 and g(x) = (x+2)^6.. you do the rest
 
 
 
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