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# 2003 paper II question 11

### #1

Posted 15 May 2005 - 09:17 PM

### #2

Posted 19 May 2005 - 07:56 AM

(It is 114kb so may take a few moments to download depending on your connection)

H tends 2 infinity

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Never argue with an idiot. They drag you down to their level then beat you with experience.

### #3

Posted 19 May 2005 - 10:08 AM

how do you get a.a(^x) = a(^x) + 1 from a(^x+1) = a(^x) + 1

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### #4

Posted 19 May 2005 - 10:20 AM

### #5

Posted 19 May 2005 - 10:22 AM

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

Posted 19 May 2005 - 04:46 PM

WWE RAW IS WAR

### #7

Posted 19 May 2005 - 04:56 PM

^{x+1}= a.a

^{x}

well lets use numbers

3

^{2}= 9

what is that the same as.....well 3x3 or 3

^{1}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 x

^{n+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

Posted 19 May 2005 - 06:50 PM

### #9

Posted 19 May 2005 - 06:54 PM

its using the defination of taking a number to the power of something

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### #10

Posted 19 May 2005 - 07:21 PM

WWE RAW IS WAR

### #11

Posted 19 May 2005 - 08:34 PM

### #13

Posted 19 May 2005 - 08:50 PM

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

Posted 19 May 2005 - 08:53 PM

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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