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    (Original post by Damask-)
    That's another expression. Do you mean 2^{2x} - 2^x - 6=0?

    You can factorise it like a quadratic.
    Thanks
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    Anyone help on how ro solve this one
    log base 3 (log base 4 4 cubed)
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    (Original post by Niki5)
    Anyone help on how ro solve this one
    log base 3 (log base 4 4 cubed)
    Take it a piece at a time

    What's log_{4} 4^3 ?
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    (Original post by Niki5)
    Anyone help on how ro solve this one
    log base 3 (log base 4 4 cubed)
    1. What is the question, is that supposed to equal something or do you need to simplify.

    2. http://www.thestudentroom.co.uk/wiki/latex please use this to write your question, it makes thing easier to read and avoids ambiguity.
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    (Original post by joostan)
    Correct
    Just rules of indices . . .
    So \ (2a)^2 = 2^2 \times a^2 = 4a^2
    i think this is wrong because you can also have a negative answer \ (-2a)^2 = (-2)^2 \times a^2 = 4a^2
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    (Original post by suba93)
    i think his is wrong because \ (-2a)^2 = -2^2 \times a^2 = 4a^2
    what seems to be the problem?
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    (Original post by jsmithy11)
    1. What is the question, is that supposed to equal something or do you need to simplify.

    2. http://www.thestudentroom.co.uk/wiki/latex please use this to write your question, it makes thing easier to read and avoids ambiguity.
    I think it's meant to be log_{3} (log_{4} 4^3), which should simplify nicely
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    (Original post by Niki5)
    Solve log base 2a 4a squared
    log_2a ^4a^2

    log_2a ^4 + log_2a ^a^2=0

    log_2a^4 =- log_2a^a^2
 
 
 
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