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    Hey.
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    Is this a suitible method for 7(i)?
    https://b940eb267c31463d4d23ccfd3105...20C2%20OCR.pdf
    Thanks.
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    No, that is wrong, for your third step it should be  y- 1 = (2^1)(2^{log_2(x)}) and not  y - 1 = 2^1 + 2^{log_2x} due to the laws of exponentials  (x^a)(x^b) = x^{a+b} . Then go on to simplify to required result.
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    log2^(y-1)=1 + log2^x
    log2^(y-1) - log2^x = 1

    log2 ^ y-1/x = 1

    y-1/x = 2^1

    y-1/x = 2

    Simplify and you get the answer
 
 
 
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