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    Given that 64^y = 2^3y-1, find the value of y. I don't even know where to begin.
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    Use logs I think
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    Look up "logarithms".
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    so Logy64=Log(3y-1)2
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    64 = 8^2
    8 = 2^3

    therefore 2^6 = 64

    2^6y = 2^3y-1

    6y = 3y-1 (base the same)

    rearrange to get y (I presume -1 is part of the power)
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    (Original post by Texxers)
    so Logy64=Log(3y-1)2
    logs is in C2, not C1
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    (Original post by Chef289)
    64 = 8^2
    8 = 2^3

    therefore 2^6 = 64

    2^6y = 2^3y-1

    6y = 3y-1 (base the same)

    rearrange to get y (I presume -1 is part of the power)
    I wish I could find more similar problems in my C1 textbook. How did you go about doing it? I understand what you did, but I don't get why.
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    Firstly change them all to the same base Once you have that you can equal the powers
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    Laws of indices: (2^3)^y is the same as 2^3y

    basically keep changing them until its the same base, and then you'll have something where both sides are equivalent and you can equal to powers
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    (Original post by Student1914)
    I wish I could find more similar problems in my C1 textbook. How did you go about doing it? I understand what you did, but I don't get why.
    Here's a website that tells you the basics. Law of Indices.

    http://mathematics.laerd.com/maths/indices-intro.php
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    (Original post by nisha.sri)
    Firstly change them all to the same base Once you have that you can equal the powers
    (Original post by potassiumnitrate)
    Laws of indices: (2^3)^y is the same as 2^3y

    basically keep changing them until its the same base, and then you'll have something where both sides are equivalent and you can equal to powers
    (Original post by Chef289)
    Here's a website that tells you the basics. Law of Indices.

    http://mathematics.laerd.com/maths/indices-intro.php
    Thanks guys, I get it now.
 
 
 
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