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    Log base 3 (4x + 7)/x = 2

    I used the log law ...Log base a X = y , a^y = X

    and got 3^2 = (4x + 7)/2

    However in the mark scheme there is a mark for stating or implying 2 = log base 3 9... I did not state this but still got the correct answer using my above method would I still get full marks?
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    it seems unfair for penalising you. Your method works fine. It is true that 2 can be written as log39 but that should be asked as a separate question really.
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    (Original post by the bear)
    it seems unfair for penalising you. Your method works fine. It is true that 2 can be written as log39 but that should be asked as a separate question really.

    Ok thank you... I guess it would depend on the mood of the examiner.
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    (Original post by IShouldBeRevising_)
    Log base 3 (4x + 7)/x = 2

    I used the log law ...Log base a X = y , a^y = X

    and got 3^2 = (4x + 7)/2

    However in the mark scheme there is a mark for stating or implying 2 = log base 3 9... I did not state this but still got the correct answer using my above method would I still get full marks?
    surely at some point you changed that 3^2 into 9 (or 18)

    thus, implied
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    Sorry for this stupid question but why did x become 2?
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    (Original post by JerzyDudek)
    Sorry for this stupid question but why did x become 2?
    There's a 2 on the RHS of the equation and you can think of this as the log to base 3 of 9.
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    (Original post by davros)
    There's a 2 on the RHS of the equation and you can think of this as the log to base 3 of 9.
    But  \[\log_3 \frac{4x + 7}{x} = 2 \Leftrightarrow 3^2 = \frac{4x + 7}{x}\], isn't it?

    Edit: there is no 3 outside the brackets, sorry.
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    (Original post by JerzyDudek)
    But  \[\log_3 \frac{3(4x + 7)}{x} = 2 \Leftrightarrow 3^2 = \frac{3(4x + 7)}{x}\], isn't it?
    You're correct.

    Either there's a typo in the OP's question or there's a typo/mistake in his working.
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    (Original post by JerzyDudek)
    But  \[\log_3 \frac{3(4x + 7)}{x} = 2 \Leftrightarrow 3^2 = \frac{3(4x + 7)}{x}\], isn't it?
    You're quite right...I hadn't spotted the OP's subtle replacement of x with 2 on the LHS!
 
 
 
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