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# a log question from p2 watch

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1. here is the question: (the q in bold is the base)

given that p=log q16 express in terms of p,

(a) log q2
(b) log q(8q)

for (a) i got the answer as p/4 but i cant remember how to solve (b).
hope someone can help
cheers
2. think rules of logs for a
a) since 2^4=16 it's 1/4p
b)*all in base q) log8q=(LN8/LN16)p
3. i am sorry but i still do not understand how to solve (b). could u possibly expain a bit more. it is a 4 mark question
thanks
4. (Original post by theunforgiven)
i am sorry but i still do not understand how to solve (b). could u possibly expain a bit more. it is a 4 mark question
thanks
erm lets see even this is the right solution. as far as i recall it's much simpler than this but lets have a look.
p=logq of 16.
y=logq of 8q

I don't like logq so i make p=LN16/LNq
where q=16/p
=>y=logq of 128/p,
which means
logq of 128-logq of p
and p was from above log base q of 16, so substituting it back in

and since logq of 128 is 7/4p then you have
7/4p-logq of p, which i leave to you to solve.

It's such a pitty latex isn't working, or i would have written it out neatly. I'm hoping this is right.

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Updated: February 5, 2006
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