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Edexcel C2 June 2011 Question 6D help!!

Can someone tell me how n = 14 in Q 6d June 2011 C2 Paper: http://www.edexcel.com/migrationdocuments/QP%20GCE%20Curriculum%202000/June%202011%20-%20QP/6664_01_que_20110526.pdf?
I worked out a = 256, r=256 which was correct.

For d i did:

used the Sn formula and got: 256(1−0.75n)1−0.75>1000\frac{256(1-0.75^n)}{1-0.75}>1000
256(1−0.75n)>1000(0.25)256(1-0.75^n)>1000(0.25)
256−192n>250256-192^n>250
−192n>−6-192^n>-6
and i end up with n>−log6−log192n>\frac{-log6}{-log192} Clearly not right :frown:
I don't know what i'm doing wrong, please help!!
Original post by jadecross

256(1−0.75n)>1000(0.25)256(1-0.75^n)>1000(0.25)


This line is fine.


256−192n>250256-192^n>250


This line is not.

256×0.75n≠192n256\times 0.75^n \not=192^n

I presume/hope that was just a silly mistake.
Reply 2
Avatar for jadecross
jadecross
OP
Original post by ghostwalker
This line is fine.



This line is not.

256×0.75n≠192n256\times 0.75^n \not=192^n

I presume/hope that was just a silly mistake.
Yep silly mistake, I got the right answer in the end thank you so much for your help!! :smile:
(edited 10 years ago)

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