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Old 03-22-2007, 01:54 PM   #1
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Default maths help needed asap!

i have some tricky work to do. i dont understand it at all. i need to do it asap. heres an example of a qu i need help with. (see question b)

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Old 03-22-2007, 03:11 PM   #2
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Default Re: maths help needed asap!

Eh, sorry, we haven't done logs yet.
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Old 03-22-2007, 05:59 PM   #3
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Default Re: maths help needed asap!

ok. i guess ill have to spend an hour tomorrow in the "support session" lol. i still wont finish it. nothing i can do about it though. maths is v hard and she storms through the work. thats why im failing
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Old 03-22-2007, 06:03 PM   #4
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Default Re: maths help needed asap!

Is that GCSE higher?
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Old 03-22-2007, 06:09 PM   #5
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Default Re: maths help needed asap!

google it lol....
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Old 03-22-2007, 06:52 PM   #6
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Default Re: maths help needed asap!

Quote:
Originally Posted by rudster816
google it lol....
i tried lol. failed.... its first year a level. i used to get a's easily in gsce. now in alevel i struggle to get c's. its fcuked up
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Old 03-22-2007, 08:59 PM   #7
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Default Re: maths help needed asap!

all logarithms are is a fancy way of writing a power. what happens is the that base of the log is 10, which would be 10 raised to a power. the number the follows the log is the answer, so when you do the log of 4 on your calculator, you are finding 10^what ever power=4.

to help with those two problems, what you need to do is remember the fact that logs are exponents. so if you have 2log4=x, it is also equalivent to log16=x. that can also be reversed, and log100 would be 2log10.

the first one, what you would do is take the log of both sides, so you get:

log(a^x)=log(10^[2x+1])
which with what I stated above can be simplified to:
x*log(a)=2x+1*log(10)
being that logarithms have a given base of 10 (except for natural logs- ln, and logs with subscript after them that specifies the base) you now have:
x*log(a)=)2x+1)*1
which with some basic math, you finally get
log(a)=x+.5
and again using the rule of logs from the second paragraph you get:
a=10^(x+.5)

now b is a little more complex looking, but it is the same concept. You have to remember that logs can be added, subtracted, and multiplied just like exponents.
2*log(2x)=1+log(a)
subtract 1 from each side
2*log(2x)-1=log(a)
now just use the fundamental rule of logs to get:
a=10^(2*log(2x)-1)
and there you go.
hope this helps.
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Old 03-22-2007, 09:13 PM   #8
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Default Re: maths help needed asap!

Hmm, we just did logs, but Im in high school advanced algebra 2 and didnt do stuff quite this complex.
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Old 03-23-2007, 01:07 PM   #9
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Default Re: maths help needed asap!

Quote:
Originally Posted by wood_workur
all logarithms are is a fancy way of writing a power. what happens is the that base of the log is 10, which would be 10 raised to a power. the number the follows the log is the answer, so when you do the log of 4 on your calculator, you are finding 10^what ever power=4.

to help with those two problems, what you need to do is remember the fact that logs are exponents. so if you have 2log4=x, it is also equalivent to log16=x. that can also be reversed, and log100 would be 2log10.

the first one, what you would do is take the log of both sides, so you get:

log(a^x)=log(10^[2x+1])
which with what I stated above can be simplified to:
x*log(a)=2x+1*log(10)
being that logarithms have a given base of 10 (except for natural logs- ln, and logs with subscript after them that specifies the base) you now have:
x*log(a)=)2x+1)*1
which with some basic math, you finally get
log(a)=x+.5
and again using the rule of logs from the second paragraph you get:
a=10^(x+.5)

now b is a little more complex looking, but it is the same concept. You have to remember that logs can be added, subtracted, and multiplied just like exponents.
2*log(2x)=1+log(a)
subtract 1 from each side
2*log(2x)-1=log(a)
now just use the fundamental rule of logs to get:
a=10^(2*log(2x)-1)
and there you go.
hope this helps.
cheers dude. i think i might save this for future reference :P
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Old 03-23-2007, 03:44 PM   #10
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Default Re: maths help needed asap!

no problem.
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