Jump to content

  • You cannot start a new topic
  • You cannot reply to this topic

2003 paper II question 11

Larry

Posted 15 May 2005 - 09:17 PM

I get the graph for part (a) (i) but could someone explain part (ii) and (b)?

Mr H

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)

tupacshakur

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

dfx

Posted 19 May 2005 - 10:20 AM

Yeah same here...hmmm... how?

dfx

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

tupacshakur

Posted 19 May 2005 - 04:46 PM

i still dont have a clue whats going on. Hopefuly this wont come up 2moro.

Dave

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

gary

Posted 19 May 2005 - 06:50 PM

I don't get this where does it all come from.

Dave

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

tupacshakur

Posted 19 May 2005 - 07:21 PM

i still dont understand it!

gary

Posted 19 May 2005 - 08:34 PM

I don't get where the a.a ^x comes from.

tupacshakur

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.

View Post



yeah.....its totally confusing me.

Dave

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???

dfx

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




  • You cannot start a new topic
  • You cannot reply to this topic