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    I know.. I know.. this is probably really basic,,, I just can't seem to do it

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    (Original post by Sulpha)
    The question would help.
    Sorry my attachments weren't working for a moment, but it's there now!
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    (Original post by creativebuzz)
    Sorry my attachments weren't working for a moment, but it's there now!
    If it helps, treat the second one as a quadratic where x=7^x


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    In the first try rewriting 9^x as (3^2)^x .... then can you change the second term into a power of 3 as well?

    Sulpha - your solution is incorrect - you can't take logs like that.
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    (Original post by creativebuzz)
    I know.. I know.. this is probably really basic,,, I just can't seem to do it

    Both of these questions are quadratics

    For the first one 9^x = 3^{2x} = (3^x)^2 and 3^{x+2} = 3^x \times 3^2 = 9(3^x)
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    (Original post by TenOfThem)
    Both of these questions are quadratics

    For the first one 9^x = 3^{2x} = (3^x)^2 and 3^{x+2} = 3^x \times 3^2 = 9(3^x)
    I don't understand why we're are multiply them if the question is showing that you must add them? :/
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    (Original post by creativebuzz)
    I don't understand why we're are multiply them if the question is showing that you must add them? :/
    if you do not understand what TenOfThem is explaining it is best to take it to you teacher at the earliest opportunity.
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    (Original post by TeeEm)
    if you do not understand what TenOfThem is explaining it is best to take it to you teacher at the earliest opportunity.
    Ah I managed to see what TenOfThem was doing once I started writing everything down! It turned out to be a quadratic so I just let y = 3^x and then factorised etc and then I managed to get the right answer!
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    (Original post by creativebuzz)
    Ah I managed to see what TenOfThem was doing once I started writing everything down! It turned out to be a quadratic so I just let y = 3^x and then factorised etc and then I managed to get the right answer!
    correct!
 
 
 
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