Asked by azam
Simon starts a new job . On the first day he gets paid 1p . on the second day he gets paid 2p , and every day after that his pay doubles . how many days will it take him to earn £1,000 in a day ?
Answers
Answered by
Reiny
a=1, and r=2
so t<sub>n</sub> = a(r^(n-1))
100000 = 1(2)^(n-1)
log 100000 = log(2)^(n-1)
n-1 = log100000/log2
n-1 = 16.6
n = 17.6 or appr 18 days
check:
1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072
on the 17th he is under,
on the 18th day he is over.
so t<sub>n</sub> = a(r^(n-1))
100000 = 1(2)^(n-1)
log 100000 = log(2)^(n-1)
n-1 = log100000/log2
n-1 = 16.6
n = 17.6 or appr 18 days
check:
1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536,131072
on the 17th he is under,
on the 18th day he is over.
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