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    Hi guys, I'm stuck on a few logs questions on the Mixed Exercise in the Logs Chapter in the C2 Book. Please may someone give me some help?

    1) Find the possible values of x for which 2^2x+1

=3(2^x)-1

    and

    3) Given that p=log_q16, express in terms of p,

    a) log_q2,
    b) log_q(8_q).

    Thanks in advance!
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    2^(2x + 1) = 2^(2x) * 2^(1) by C1 rules, right?

    = 2 * 2^(2x)

    Can you see how to do the question now? If not, let u = 2^x and sub it in.

    Post some working for question 3.
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    Thanks for the help, but I am still a bit confused on question 1 >_< can you explain a bit more? O_o

    and on number 3, how do i start it? Do I sub log16 into log2?
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    (Original post by xsweetwhispers)
    Thanks for the help, but I am still a bit confused on question 1 >_< can you explain a bit more? O_o
    Well, what exactly is confusing you? Post your working and explanation of where + why you get stuck.

    and on number 3, how do i start it? Do I sub log16 into log2?
    Note that 16 = 2^4...
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    What are the inequality signs for?

    Baiscally you can turn the first one into a quadratic equation. y=2^x can be used to put that into a quad as 2^2x becomes y^2 and then you end up with a y term to the single power later in the equation...solve for y and then sub back in for 2^x
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    On number one, this is how far I got...

    2^(2x+1) = 3(2^x)-1
    2^(2x) * 2^1 = 3(2^x)-1
    2^(2x) * 2 = 3(2^x)-1

    let y = 2^x

    2 * y = 2^(2x) * 2 = 3(y)-1
    2 * y = 2^(2x) * 2 = 3y-1

    & now I'm back to stuck again >_<

    ----------------------------------

    On number 3a), this is as far I've got to...

    16 = 2^4
    log16 = log (2^4)
    log 16 = 4 log 2
    x = 4log2/log16 = 1

    I think it is wrong though, can someone correct me?
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    (Original post by xsweetwhispers)
    On number one, this is how far I got...

    2^(2x+1) = 3(2^x)-1
    2^(2x) * 2^1 = 3(2^x)-1
    2^(2x) * 2 = 3(2^x)-1

    let y = 2^x

    2 * y = 2^(2x) * 2 = 3(y)-1
    2 * y = 2^(2x) * 2 = 3y-1

    & now I'm back to stuck again >_<

    -

    I don't think you have typed out the question correctly, but if you have then I think the answer is x=0

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    thanks for the help ^^ I got it now
    but here goes another one I'm stuck on now...

    quest 9.

    If xy = 64 and log_xy + log_yx = 5/2, find x and y
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    (Original post by xsweetwhispers)
    let y = 2^x

    2 * y = 2^(2x) * 2 = 3(y)-1
    You went wrong here.

    2 * y = 2^x * 2, not 2^(2x) * 2.

    So what is 2^(2x) * 2 in terms of y? Well, 2^(2x) = (2^x)^2 isn't it? So it's 2^(2x) * 2 = (2^x)^2 * 2 = y^2 * 2 = 2y^2.

    Your solution to 3a doesn't make sense because you come up with x from nowhere and x isn't defined in the question.
 
 
 
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