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    (Original post by dongonaeatu)
    dont understand the second step
    I have used the change of base law in order to make both of the bases the same so I can then combine the logs into one single log.

    The change of base law is as follows:

    logax = logbx/logba

    ^ Learn that formula.

    In this case we had:

    log2x + log4x = 2

    a = 4
    b = 2

    So it turned into:

    log2x + log2x/log24 = 2
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    (Original post by Joshmeid)
    log2x + log4x = 2

    log2x + log2x/log24 = 2

    log2x + log2x/2 = 2

    2log2x + log2x = 4

    3log2x = 4

    log2x^3 = 4

    x^3 = 16

    x = 2.52(3sf)

    Ask if you are confused by anything.
    Please don't post full solutions, they are considered a last resort.
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    (Original post by Joshmeid)
    I have used the change of base law in order to make both of the bases the same so I can then combine the logs into one single log.

    The change of base law is as follows:

    logax = logbx/logba

    ^ Learn that formula.

    In this case we had:

    log2x + log4x = 2

    a = 4
    b = 2

    So it turned into:

    log2x + log2x/log24 = 2
    thanks i go it
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    (Original post by avtar95)
    could u put that in letters
    According to the change of base rule  \displaystyle log_ab can be written as  \displaystyle \frac{log_cb}{log_ca}

    In the question it was  \displaystyle log_4x
    So here  \displaystyle a=4 \text{ and } b=x

    Lets use the change of base rule, we want the base of logs to be 2 hence let  \displaystyle c=2
     \displaystyle log_4x = \frac{log_2x}{log_24}

    Do you get it?
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    (Original post by raheem94)
    According to the change of base rule  \displaystyle log_ab can be written as  \displaystyle \frac{log_cb}{log_ca}

    In the question it was  \displaystyle log_4x
    So here  \displaystyle a=4 \text{ and } b=x

    Lets use the change of base rule, we want the base of logs to be 2 hence let  \displaystyle c=2
     \displaystyle log_4x = \frac{log_2x}{log_24}

    Do you get it?
    kinda its quite complicated though
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    (Original post by dongonaeatu)
    is that rule just for log2x and log4x i've never been taught that
    Remember  \displaystyle log_ab + log_ac = log_a(bc)

    So  \displaystyle log_2x + log_4x cannot be written in that way as both have different bases.
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    (Original post by raheem94)
    Please don't post full solutions, they are considered a last resort.
    thanks i go it
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    (Original post by dongonaeatu)
    kinda its quite complicated though
    It isn't complicated with a bit of practice you will be fine with it.
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    (Original post by raheem94)
    Remember  \displaystyle log_ab + log_ac = log_a(bc)

    So  \displaystyle log_2x + log_4x cannot be written in that way as both have different bases.
    are the bases the a's so the number in front of the log
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    (Original post by dongonaeatu)
    are the bases the a's so the number in front of the log
    The a's are the bases.
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    (Original post by raheem94)
    It isn't complicated with a bit of practice you will be fine with it.
    so loga 5 +loga 4= loga20 as the base is the same (a)

    but loga5+ logb4= not log20 as different bases what would i do in this instance
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    (Original post by dongonaeatu)
    so loga 5 +loga 4= loga20 as the base is the same (a)

    but loga5+ logb4= not log20 as different bases what would i do in this instance
    Yes,  \displaystyle log_a5 + log_a4 = log_a(5 \times 4) = log_a(20)


    And  \displaystyle log_a5+log_b4 \not= log20

    For solving such questions you need to put both on the same base. Lets change the base of  \displaystyle log_b4 to  \displaystyle a

     \displaystyle log_b4 = \frac{log_a4}{log_ab}

    So now the equation becomes,  \displaystyle log_a5 + \frac{log_a4}{log_ab}
    You can see that now all the logarithms have the same base.

    Do i make any sense?
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    (Original post by raheem94)
    Yes,  \displaystyle log_a5 + log_a4 = log_a(5 \times 4) = log_a(20)


    And  \displaystyle log_a5+log_b4 \not= log20

    For solving such questions you need to put both on the same base. Lets change the base of  \displaystyle log_b4 to  \displaystyle a

     \displaystyle log_b4 = \frac{log_a4}{log_ab}

    So now the equation becomes,  \displaystyle log_a5 + \frac{log_a4}{log_ab}
    You can see that now all the logarithms have the same base.

    Do i make any sense?
    did u use the change of base loga^b=logc to the b/logc to the a

    and u make C the base ur changing it to
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    (Original post by dongonaeatu)
    did u use the change of base loga^b=logc to the b/logc to the a

    and u make C the base ur changing it to
    I am not understanding your question.

    I used the change of base rule to convert  \displaystyle log_b4 to  \displaystyle \frac{log_a4}{log_ab}
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    (Original post by raheem94)
    I am not understanding your question.

    I used the change of base rule to convert  \displaystyle log_b4 to  \displaystyle \frac{log_a4}{log_ab}
    do i have to learn the change of base formula

    and the C in the equation is the log of what u are converting it to so if i want it in base of a i have the C as a
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    (Original post by dongonaeatu)
    do i have to learn the change of base formula

    and the C in the equation is the log of what u are converting it to so if i want it in base of a i have the C as a
    Yes, you have to learn the change of base formula.

    And yes the C in my equation is 'a', and your concept about this is correct. I wanted the log to have the base 'a' hence i set 'C' to 'a'.
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    (Original post by raheem94)
    Yes, you have to learn the change of base formula.

    And yes the C in my equation is 'a', and your concept about this is correct. I wanted the log to have the base 'a' hence i set 'C' to 'a'.
    do i get given the formula in the exam and how many hours of maths do u do a day u are amazing
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    (Original post by dongonaeatu)
    do i get given the formula in the exam and how many hours of maths do u do a day u are amazing
    I don't know if the formula will be given or not, but this is a simple formula. Just understand it, don't try to mug it up.

    I have already finished A-Level maths(C1-4, M1-2), with an A* , now i am doing further maths, so C2 is simple for me. I don't work too much, A-Level maths isn't hard. Just go through the chapters in the book and you will be fine with it. Try to understand things, do you use a book?
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    (Original post by raheem94)
    I don't know if the formula will be given or not, but this is a simple formula. Just understand it, don't try to mug it up.

    I have already finished A-Level maths(C1-4, M1-2), with an A* , now i am doing further maths, so C2 is simple for me. I don't work too much, A-Level maths isn't hard. Just go through the chapters in the book and you will be fine with it. Try to understand things, do you use a book?
    wow A* thats great well done. i have a green edxcel textbook and a c2 resource pack and access to past papers and marked solutions
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    (Original post by dongonaeatu)
    wow A* thats great well done. i have a green edxcel textbook and a c2 resource pack and access to past papers and marked solutions
    I also used the green edexcel text books, they are great. I self-studied all units using them, they explain really well. Have you read the chapters properly?

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