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Logarithms question

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?
Reply 1
Avatar for the bear
the bear
20
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.
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.
Reply 3
Avatar for TenOfThem
TenOfThem
18
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
Sorry for this stupid question but why did x become 2?
Reply 5
Avatar for davros
davros
16
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.
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
Unparseable latex formula:

\[\log_3 \frac{4x + 7}{x} = 2 \Leftrightarrow 3^2 = \frac{4x + 7}{x}\]

, isn't it?

Edit: there is no 3 outside the brackets, sorry.
(edited 11 years ago)
Reply 7
Avatar for Notnek
Notnek
21
Original post by JerzyDudek
But
Unparseable latex formula:

\[\log_3 \frac{3(4x + 7)}{x} = 2 \Leftrightarrow 3^2 = \frac{3(4x + 7)}{x}\]

, 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.
Reply 8
Avatar for davros
davros
16
Original post by JerzyDudek
But
Unparseable latex formula:

\[\log_3 \frac{3(4x + 7)}{x} = 2 \Leftrightarrow 3^2 = \frac{3(4x + 7)}{x}\]

, 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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