C2 - how to solve this (geometric series)?

www.examsolutions.net/a-level-maths-papers/Edexcel/Core-Maths/Core-Maths-C2/2010-January/paper.php#Q6

It's question 6c.. To work it out I did: 18000 x 0.8^n-1 = 1000, and then worked on from there to get n=13, but the actual answer is 13, I don't understand why I'm wrong


Posted from TSR Mobile

0.8^n-1= 1/18
log0.8^n-1=log1/18

n-1 = (log1/18)/(log0.8) = 12.953
n= (log1/18)/(log0.8) + 1

n = 13.953

And then to the nearest penny n=14 but the markscheme says n=13



Posted from TSR Mobile

0.8^n-1= 1/18
log0.8^n-1=log1/18

n-1 = (log1/18)/(log0.8) = 12.953
n= (log1/18)/(log0.8) + 1

n = 13.953.

But now you have to sub in 13 into the original equation, this is the largest and first number that makes the left hand side less than 1000.

so n= 13 i.e. number of years after original purchase in which the price falls below £1000

You said the answer was 13 and you got 13, but this makes more sense for 6)b) than it does for 6)c)

Sorry made silly mistakes, it was question 6b and my answer was n=14 ;$



Posted from TSR Mobile