I get the graph for part (a) (i) but could someone explain part (ii) and (b)?
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2003 paper II question 11
Started by Larry, May 15 2005 09:17 PM
13 replies to this topic
#2
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Posted 19 May 2005 - 07:56 AM
Here's a solution: http://www.mathsroom.co.uk/hsnposts/2003_P2_Q11.pdf
(It is 114kb so may take a few moments to download depending on your connection)
(It is 114kb so may take a few moments to download depending on your connection)
H tends 2 infinity
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#3
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Posted 19 May 2005 - 10:08 AM
hmm that solution doesnt make any sense to me
how do you get a.a(^x) = a(^x) + 1 from a(^x+1) = a(^x) + 1
how do you get a.a(^x) = a(^x) + 1 from a(^x+1) = a(^x) + 1
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#4
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Posted 19 May 2005 - 10:20 AM
Yeah same here...hmmm... how?
#5
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Posted 19 May 2005 - 10:22 AM
Oh I think I got it! Laws of exponents
a ^ (X + 1) = a ^ x x a ^ 1 (cause if you multiply the two, add the powers right? So the reverse should also apply.)
So a^1 is simply a, which makes a ^ (x + 1) = a.a^x
a ^ (X + 1) = a ^ x x a ^ 1 (cause if you multiply the two, add the powers right? So the reverse should also apply.)
So a^1 is simply a, which makes a ^ (x + 1) = a.a^x
#6
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Posted 19 May 2005 - 04:46 PM
i still dont have a clue whats going on. Hopefuly this wont come up 2moro.
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#7
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Posted 19 May 2005 - 04:56 PM
why does ax+1 = a.ax
well lets use numbers
32 = 9
what is that the same as.....well 3x3 or 31 x 1
the exponent is a way of showing you the number of times the number has been multiplied by itself.
so if you had xn+1 that is saying x has been multiplied by itself n+1 times
which is the same as multiplying x "n" number of times then multiplying by "n" 1 one time
well lets use numbers
32 = 9
what is that the same as.....well 3x3 or 31 x 1
the exponent is a way of showing you the number of times the number has been multiplied by itself.
so if you had xn+1 that is saying x has been multiplied by itself n+1 times
which is the same as multiplying x "n" number of times then multiplying by "n" 1 one time
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#8
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Posted 19 May 2005 - 06:50 PM
I don't get this where does it all come from.
#9
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Posted 19 May 2005 - 06:54 PM
its just maths number work essentially
its using the defination of taking a number to the power of something
its using the defination of taking a number to the power of something
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#10
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Posted 19 May 2005 - 07:21 PM
i still dont understand it!
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#11
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Posted 19 May 2005 - 08:34 PM
I don't get where the a.a ^x comes from.
#12
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Posted 19 May 2005 - 08:44 PM
QUOTE(gary @ May 19 2005, 08:34 PM)
I don't get where the a.a ^x comes from.
yeah.....its totally confusing me.
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#13
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Posted 19 May 2005 - 08:50 PM
what is a.a^ x
its "a" times a ^x
yes??
so that x number of "a's" plus another "a"
yes??
so thats
a^x+1
hopefully yes???
its "a" times a ^x
yes??
so that x number of "a's" plus another "a"
yes??
so thats
a^x+1
hopefully yes???
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#14
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Posted 19 May 2005 - 08:53 PM
See..
a ^(x +1) can be expanded into a^x multiplied by a^1
Remember your laws of exponents? When you multiply two numbers with powers and the same base, the powers get added right? Like 2^4 * 2^5 = 2 ^ (4 + 5 ) = 2^9
SO, to get a.a^x, you just do the above procedure in REVERSE. Which means a ^ (x + 1) = a^ x * a ^1 < try multiplying this out, and you get a ^ ( x + 1 ), see?
SO, now that we know in expanded form its a ^ x * a ^ 1
BUT, a^1 is just a. so that makes a * a^x = a.a^x
a ^(x +1) can be expanded into a^x multiplied by a^1
Remember your laws of exponents? When you multiply two numbers with powers and the same base, the powers get added right? Like 2^4 * 2^5 = 2 ^ (4 + 5 ) = 2^9
SO, to get a.a^x, you just do the above procedure in REVERSE. Which means a ^ (x + 1) = a^ x * a ^1 < try multiplying this out, and you get a ^ ( x + 1 ), see?
SO, now that we know in expanded form its a ^ x * a ^ 1
BUT, a^1 is just a. so that makes a * a^x = a.a^x
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