C2 Log Question - Please help me and explain

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blobbybill
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Given that Log2x = a, find in terms of a, the simplest form of:

a) Log2(16x )

So far I have this:
= Log2(16) + Log2(X)

And then I don't know how to continue from there, can you please explain why to me, referencing log laws so I can understand why you have done what you have done.

b) Log2(x^4 / 2)

So far I have this:
= Log2(x^4) - Log2(2)

And then I don't know how to continue from there, can you please explain why to me, referencing log laws so I can understand why you have done what you have done.

Thanks a lot! I really struggle with logs.
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takurino
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Do you have the answers for the questions? I think the first one might be 4a and would the second one be (a+1)/4? If I'm right I'll gladly explain
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blobbybill
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(Original post by takurino)
Do you have the answers for the questions? I think the first one might be 4a and would the second one be (a+1)/4? If I'm right I'll gladly explain
The answer for part a is: 4 + a and the answer for part 2 is: 4a - 1
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takurino
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(Original post by blobbybill)
The answer for part a is: 4 + a and the answer for part 2 is: 4a - 1
okay I see what it was looking for now aha. so for the first one you broke it into log2(16) + log2(x) and we know what log2(16) = 4. we also know log2(x) = a because we are given that information so it would be 4+a.

For part 2 log2(x^4) - log2(2) is correct. We know log2(2) = 1 and you can re-write log2(x^4) as 4log2(x) (by moving the power infront) so that would equal 4a. So you would have 4a - 1.
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theguywhosaidhi
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you started it correctly!

basically... Log2(16) = Log2(2^4) = 4Log2(2) = 4

this is because we can take powers of the number inside to outside the Log.

the reason 4Log2(2) = 4 is because Log2(2) = 1

hence.. 4 * 1 = 4

hence... 4Log2(2) + Log2(X) = 4 + a

thanks this is good revision, i'm resitting maths core 2

edit: you can cheat and put Log2(16) in your calculator to get = 4 though understanding is better just in case the actual exam contains algebra or something

edit2: using the knowledge of the power law, you can do part (b) now, try it yourself. good luck
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blobbybill
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(Original post by takurino)
okay I see what it was looking for now aha. so for the first one you broke it into log2(16) + log2(x) and we know what log2(16) = 4. we also know log2(x) = a because we are given that information so it would be 4+a.

For part 2 log2(x^4) - log2(2) is correct. We know log2(2) = 1 and you can re-write log2(x^4) as 4log2(x) (by moving the power infront) so that would equal 4a. So you would have 4a - 1.
How do you "know log2(x) = a because we are given that information"? I don't see that information in the question. And also, how does that make it 4+a?
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takurino
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(Original post by blobbybill)
How do you "know log2(x) = a because we are given that information"? I don't see that information in the question. And also, how does that make it 4+a?
So in the first bit of the question its says given log2(x) = a so thats where the 'a' came from. The question was log2(16x) and you broke it up correctly into log2(16) + log2(x). If you put log2(16) into your calc you will get 4 so thats where the 4 comes from. Then we have to figure out log2(x) but at the beginning of the question they tell us log2(x) = a. So you know log2(16) = 4 and you know log2(x) = a. Put these together and you have 4+a.

Hope this helps!
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K-Man_PhysCheM
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(Original post by blobbybill)
How do you "know log2(x) = a because we are given that information"? I don't see that information in the question. And also, how does that make it 4+a?
Re-read your OP. In the very first line, you wrote "Let \log_2 x = a".

So, you've split \log_2 16x into \log_2 16 + \log_2 x

By the definition of the log, \log_2 16= y \Leftrightarrow 2^y = 16

Therefore, \log_2 16 = 4, because 2^4 = 16

So \log_2 16x \equiv \log_2 16 + \log_2 x \equiv 4 + a
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takurino
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(Original post by K-Man_PhysCheM)
Re-read your OP. In the very first line, you wrote "Let \log_2 x = a".

So, you've split \log_2 16x into \log_2 16 + \log_2 x

By the definition of the log, \log_2 16= y \Leftrightarrow 2^y = 16

Therefore, \log_2 16 = 4, because 2^4 = 16

So \log_2 16x \equiv \log_2 16 + \log_2 x \equiv 4 + a
Damn how do you get all those fancy effects
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K-Man_PhysCheM
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(Original post by takurino)
Damn how do you get all those fancy effects
I have used LaTeX, a piece of mathematics text software that has been integrated into TSR.

Here is a link to the information page: https://www.thestudentroom.co.uk/wiki/LaTex
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takurino
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(Original post by K-Man_PhysCheM)
I have used LaTeX, a piece of mathematics text software that has been integrated into TSR.

Here is a link to the information page: https://www.thestudentroom.co.uk/wiki/LaTex
Thank you so much! Always wondered how people got their text in that format
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