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    How would you go about solving the following question?
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    Name:  logs.png
Views: 11
Size:  71.4 KB
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    Part a is basically subbing 1 wherever you have x and you should get a final answer of 0 which therefore proves it.

    Part b you replace u with 2^x and this should give you a final answer of the equation shown.

    Part c is a bit more complicated but is actually relatively straigtforward if you look at it. I don't think it is actually logs to start off with and you just need your basic c1 knowledge because I would use factor theorem. This is because from part a, I know that (x-1) is a factor of the equation therefore, (u-1) has to be a factor.

    Carry on from here and let me know what you get.
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    Thanks for the explanation, do you mind showing how you did part b. So far I have only had failed attempts.
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    (Original post by SmartFailure)
    Thanks for the explanation, do you mind showing how you did part b. So far I have only had failed attempts.
    2^3x + 4(2^x) + 2^x + 6 =0

    now to substitute u = 2^x here's what you would do

    2^3x = (2^x)^3

    now where you see 2^x make it U

    Hope that helped
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    Thanks, how do you find the solutions of a cubic?
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    (Original post by SmartFailure)
    Thanks, how do you find the solutions of a cubic?
    You've already been given that x=1 solves the equation in x. So when x=1 u=2 and so u=2 is a solution to the cubic now use the fact that (u-2) is a factor of the cubic.
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    can someone show the solution to part c please
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    (Original post by SmartFailure)
    can someone show the solution to part c please
    (U-2) is a factor of the cubic equation so use long division to leave a quadratic in u and solve it.
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    (Original post by SmartFailure)
    can someone show the solution to part c please
    Do you understand the logic of the theories for this work? You seem like you need to recap the topic because this is quiet logical if you understand index and solving quadratics/cubics.

    (Original post by B_9710)
    (U-2) is a factor of the cubic equation so use long division to leave a quadratic in u and solve it.

    That coud make it more complex than necessary.
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    (Original post by ckfeister)


    That coud make it more complex than necessary.
    How?
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    (Original post by ckfeister)
    That coud make it more complex than necessary.
    How...? That's what's expected.
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    (Original post by B_9710)
    How?
    (Original post by RDKGames)
    How...? That's what's expected.
    Well for me anyway, wouldn't (u-2)(au^2 + bu + c) be more simple? Maybe I do another way.
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    (Original post by ckfeister)
    Well for me anyway, wouldn't (u-2)(au^2 + bu + c) be more simple? Maybe I do another way.
    Practically the same thing. You want to find that quadaric; you can either compare coeffs or just do long division, whichever OP prefers, neither are complicated.
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    (Original post by SmartFailure)
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    u^3-4u^2+u+6=0

    We know that x=1 is a solution, so u=2^x=2 is also one. This has been open a while, so I'll give you one more step:

    \therefore (u-2)(u^2-2u-3)=0

    Now factorise u^2-2u-3 and calculate the values of x corresponding to those factors.
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    U = 2 or -1

    x = log2 (3)
    x= 1.58?
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    (Original post by SmartFailure)
    U = 2 or -1

    x = log2 (3)
    x= 1.58?
    Seems good.
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    (Original post by RogerOxon)
    u^3-4u^2+u+6=0

    \therefore (u-2)(u^2-2u-3)=0
    Some more on the step above:

    Start with:

    (u-2)(au^2+bu+c)=u^3-4u^2+u+6

    We know that the coefficient of u^3 is 1, so a=1.
    We know that the constant is +6, so c=-3.
    So:
    (u-2)(u^2+bu-3)=u^3-4u^2+u+6

    Looking at the coefficient of u^2:
    (LHS) -2+b=-4 (RHS)

    \therefore b=-2

    I do this in my head as I write the new equation - just see how many of the next power of u you already have and adjust with the next term in the factor. It takes practice, but you can always use the approach above.
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    (Original post by ckfeister)
    Well for me anyway, wouldn't (u-2)(au^2 + bu + c) be more simple? Maybe I do another way.
    It's essentially the same.
 
 
 
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