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 ?
a=1, and r=2
so tn = 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.
To find out how many days it will take Simon to earn £1,000, we can work backwards from the desired amount. Let's break down the problem step by step:
1. Simon earns 1p on the first day.
2. On the second day, he earns 2p.
3. On the third day, his pay doubles again to 4p.
4. This doubling pattern continues each day, so on the fourth day, he earns 8p, and so on.
From this pattern, we can see that Simon's daily earnings can be expressed as 2^(n-1), where n is the number of days worked.
To find the number of days it takes to earn £1,000, we need to solve the equation:
2^(n-1) = 1,000
To do this, we can use logarithms. Taking the logarithm (base 2) of both sides will help us solve for n.
log2(2^(n-1)) = log2(1,000)
Since log2(2^(n-1)) simplifies to (n-1), we have:
(n-1) = log2(1,000)
Now, we can solve for n by rearranging the equation:
n = log2(1,000) + 1
Using a calculator, the value of log2(1,000) is approximately 9.966. Add 1 to this to get:
n ≈ 9.966 + 1
n ≈ 10.966
Therefore, it will take approximately 10.966 days for Simon to earn £1,000 in a day. Since we cannot have a fraction of a day, we round up to the nearest whole number.
Therefore, it will take Simon approximately 11 days to earn £1,000 in a day.