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# OCR MEI AS core 2- June 2009 Watch

1. Hey guys if anyone out there can explain the answer to a question for me that would be great!

The question is Q9 ii. (June 2009 mei c2)

It is a logs question but ill have a go at typing it out:

(Loga^5+Loga^1/2)/Loga

So what i did was bring the powers to the front and did this

5Loga+0.5Loga-1Loga

then just did 5+0.5-1= 4.5

However looking at it I can see you just need to do (5+0.5)/1 :/

so my question is: why do we not need to use log rules to make the equation Loga5+Loga0.5-Loga? (that is what our teacher taught us!)

If i haven't explained very well please tell me
2. (Original post by Davelittle)
Hey guys if anyone out there can explain the answer to a question for me that would be great!

The question is Q9 ii. (June 2009 mei c2)

It is a logs question but ill have a go at typing it out:

(Loga^5+Loga^1/2)/Loga

So what i did was bring the powers to the front and did this

5Loga+0.5Loga-1Loga

then just did 5+0.5-1= 4.5

However looking at it I can see you just need to do (5+0.5)/1 :/

so my question is: why do we not need to use log rules to make the equation Loga5+Loga0.5-Loga? (that is what our teacher taught us!)

If i haven't explained very well please tell me
Because you have something that looks like (k log a) / log a which is just k (in this case k = 5.5)

You would use -log a if you had a term that said log(1/a) because 1/a is the same as a^-1, but you don't have that here - 1/loga is NOT the same as log(1/a).
3. did this question quickly, see if it helps?

sorry for my hand writing

bassically the log a's cancel, not take away

I guess you could imagine them going 5log(a-a) + 1/2log (a-a) which would be (5*1) + (1/2*1)
4. (Original post by davros)
Because you have something that looks like (k log a) / log a which is just k (in this case k = 5.5)

You would use -log a if you had a term that said log(1/a) because 1/a is the same as a^-1, but you don't have that here - 1/loga is NOT the same as log(1/a).
ok thanks i understand!

i swear my teacher told us that subtracting logs is the same as dividing logs and dividing is the same as subtracting like this. loga-logb=loga/b
5. I think that is true, but loga/logb /=/ log a - log b
i think it's only log(a/b)= loga - logb?
6. (Original post by Davelittle)
ok thanks i understand!

i swear my teacher told us that subtracting logs is the same as dividing logs and dividing is the same as subtracting like this. loga-logb=loga/b
That's true, but as I tried to point out (perhaps not clearly) you have one log divided by another log, NOT the log of (one thing divided by another).
7. (Original post by davros)
That's true, but as I tried to point out (perhaps not clearly) you have one log divided by another log, NOT the log of (one thing divided by another).
ah because its a^5 and a?

thanks!!!!!
8. no, not because it's A^5 divided by A, it's because it's log A/Log A instead of Log (A/A)

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