Year 13 Maths Help Thread

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    (Original post by tiny hobbit)
    The end bit should be dx + e

    You can then equate coefficients of the x terms to find d.

    The 24 is e
    (Original post by NotNotBatman)
    on the first line you just have +d, you should have +dx+e
    I should have noticed that earlier! I've got the answer now, thank you!
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    Have i done something wrong ....

    e^2x - e^x = 0
    e^2x = e^x
    lne^x = lne^2x
    x=2x

    ??

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    (Original post by kiiten)
    Have i done something wrong ....

    e^2x - e^x = 0
    e^2x = e^x
    lne^x = lne^2x
    x=2x

    ??

    Posted from TSR Mobile
    looks fine;

    now just find the value of x for which  \displaystyle x = 2x
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    (Original post by kiiten)
    Have i done something wrong ....

    e^2x - e^x = 0
    e^2x = e^x
    lne^x = lne^2x
    x=2x

    ??

    Posted from TSR Mobile
    I think you should probably form a quadratic equation in e^x and go from there.
    Doing it this way you may struggle to solve  e^{2x}-e^x-1=0 .
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    (Original post by B_9710)
    I think you should probably form a quadratic equation in e^x and go from there.
    Doing it this way you may struggle to solve  e^{2x}-e^x-1=0 .
    But there is no 1 in the original equation?
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    (Original post by DylanJ42)
    looks fine;

    now just find the value of x for which  \displaystyle x = 2x

    So zero? :dontknow:
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    (Original post by kiiten)
    So zero? :dontknow:
    perfect

    as a matter of fact, going on what B_9710 said, how would you solve  \displaystyle e^{2x}-e^x-1=0

 (just to make sure you know the method as you could be tested on it)
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    (Original post by kiiten)
    So zero? :dontknow :




    :dots::dontknow::dontknow::dontknow::dontknow::dontknow::dontknow:
    Yes, it is 0

    What B_9710 was trying to say was you'd struggle to solve a similar equation involving a number, such as the one in his example, with the method you used. So if you knew how to solve it as a quadratic equation in ex it would be to your benefit.
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    (Original post by DylanJ42)
    perfect
    Thanks - i got confused because i did

    x=2x
    0 = 2x /x
    0 = 2 ????

    Is that wrong??

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    (Original post by kiiten)
    Thanks - i got confused because i did

    x=2x
    0 = 2x /x
    0 = 2 ????

    Is that wrong??

    Posted from TSR Mobile
    yea because x = 0 and so dividing by x creates a whole world of problems

    from x = 2x just bring everything to the right of the equals sign(or the left, it doesnt matter) and get 0 = x
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    (Original post by DylanJ42)
    yea because x = 0 and so dividing by x creates a whole world of problems

    from x = 2x just bring everything to the right of the equals sign(or the left, it doesnt matter) and get 0 = x
    2x / x isnt x though? (Bringing everything to the right of the equation)

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    Different ques

    ???? I got y = 2 / ln

    Where did i go wrong?

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    (Original post by kiiten)
    2x / x isnt x though? (Bringing everything to the right of the equation)

    Posted from TSR Mobile
    sorry I worded that awfully, i mean "-x" from both sides of the equation to get 0 = x
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    (Original post by kiiten)
    Different ques

    ???? I got y = 2 / ln

    Where did i go wrong?

    Posted from TSR Mobile
    log(a+b+c) \neq loga + logb + logc

    For your other question, your equation is right - just solve it. If you had gone on to solve it and then put it back into the original question you would have been right - and you could have validated that for yourself rather than asking. Do not be afraid to try
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    (Original post by kiiten)
    Different ques

    ???? I got y = 2 / ln

    Where did i go wrong?

    Posted from TSR Mobile
    first off,  \displaystyle \ln(a + b + c) \neq \ln(a)  + \ln(b) + \ln(c)

    also you cant say  \displaystyle ln(y) = 2 \implies y = \frac{2}{ln}

    that is similar to saying  \displaystyle  f(x) = 4 \implies x = \frac{4}{f} , it just doesnt make any sense.

    I think the easiest way to do this question is to realise x and y must take the form  \displaystyle e^n (where n is just a number). So write  \displaystyle x = e^a ,  \displaystyle  y = e^b and go from there.

    If not take equation 2 and "e" both sides (ive no idea what this step is actually called) to get  \displaystyle e^{(\text{Left hand side})} = e^{(\text{Right hand side})} then get in terms in x and sub into equation 1, then work out the quadratic which will follow

    Edit: whoops you had the answer already, i didnt even notice :getmecoat:
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    (Original post by SeanFM)
    log(a+b+c) \neq loga + logb + logc

    For your other question, your equation is right - just solve it. If you had gone on to solve it and then put it back into the original question you would have been right - and you could have validated that for yourself rather than asking. Do not be afraid to try
    (Original post by DylanJ42)
    first off,  \displaystyle \ln(a + b + c) \neq \ln(a)  + \ln(b) + \ln(c)

    also you cant say  \displaystyle ln(y) = 2 \implies y = \frac{2}{ln}

    that is similar to saying  \displaystyle  f(x) = 4 \implies x = \frac{4}{f} , it just doesnt make any sense.

    I think the easiest way to do this question is to realise x and y must take the form  \displaystyle e^n (where n is just a number). So write  \displaystyle x = e^a ,  \displaystyle  y = e^b and go from there.

    If not take equation 2 and "e" both sides (ive no idea what this step is actually called) to get  \displaystyle e^{(\text{Left hand side})} = e^{(\text{Right hand side})} then get in terms in x and sub into equation 1, then work out the quadratic which will follow

    Edit: whoops you had the answer already, i didnt even notice :getmecoat:
    I dont already have the answer? unless you mean some of my working is correct.

    Im confused - i rearranged equation 1 to get x. Then i subbed that into the 2nd equation to get

    ln (e^5 - e^2 + y) + lny = 7
    Now ? - do you bring lny to the other side then divide by ln?
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    (Original post by kiiten)
    I dont already have the answer? unless you mean some of my working is correct.

    Im confused - i rearranged equation 1 to get x. Then i subbed that into the 2nd equation to get

    ln (e^5 - e^2 + y) + lny = 7
    Now ? - do you bring lny to the other side then divide by ln?
    You can't can't can't divide by ln on its own - as Dylan has already said in his post. lne^3 = 3 does not imply that e^3 = 3 / ln(....) because ln(...) without an argument is meaningless. Just like how you wouldn't divide sinx + cosx by sin(nothing) - it would be meaningless.

    As Dylan has taken time to give you hints I will not proceed any further as it may be confusing to have input from multiple people. Good luck!
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    (Original post by SeanFM)
    As Dylan has taken time to give you hints I will not proceed any further as it may be confusing to have input from multiple people. Good luck!
    I was on the reply screen as you replied but you were first, it was my bad. :sorry:
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    (Original post by kiiten)
    I dont already have the answer? unless you mean some of my working is correct.

    Im confused - i rearranged equation 1 to get x. Then i subbed that into the 2nd equation to get

    ln (e^5 - e^2 + y) + lny = 7
    Now ? - do you bring lny to the other side then divide by ln?
    look on the right hand side of your page, you have  \displaystyle \ln y = 2 . How do you get y on its own here?

    Edit: wait i think you fluked getting this right :laugh:
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    (Original post by DylanJ42)
    look on the right hand side of your page, you have  \displaystyle \ln y = 2 . How do you get y on its own here?
    Ohhh so ....
    e^lny = e^2
    y = e^2
 
 
 
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