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    Now, I did actually arrive to the right answer for this question, but my question is in regards to the solution given in the mark scheme, as I don't understand the solution given back there.

    The question is:

    6^r^+^s/ 8^r \times 9^r^+^2^s \times 12^r^-^s / 8^r \times 9^r^+^2^s

    The solution in the mark scheme converts this into powers of 2 and 3, but I'm not sure how...

    This is question Q1A from the 2007 paper.
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    6^{r+s} = (3 \times 2)^{r+s} = 3^{r+s}2^{r+s}
    Similarly for the other terms.
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    (Original post by DFranklin)
    6^{r+s} = (3 \times 2)^{r+s} = 3^{r+s}2^{r+s}
    Similarly for the other terms.
    Just another quick question in regards to this: I've noticed that if I say s = -0.3, or most negative decimals, then this does not actually produce an integer, but a decimal, even though s <= 0.

    For instance, 2^0.3 x 3^1.2 produces 4.601...
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    (Original post by UCASLord)
    Just another quick question in regards to this: I've noticed that if I say s = -0.3, or most negative decimals, then this does not actually produce an integer, but a decimal, even though s <= 0.

    For instance, 2^0.3 x 3^1.2 produces 4.601...
    Missing your point?
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    (Original post by HelpMe??)
    Missing your point?
    You're right, there is no point.

    I missed the "let r and s b integers" part of the question, and then thought why using a non-integer s doesn't produce an integer solution.
 
 
 
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