Exponential growth: money invested according to P=100*1.05^t

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Monkeyseat

Full Member
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Jul 3, 2005
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298
Hi,

Question

The amount of money, £P, in a special savings account at time t years after 1 January 2000 is given by P = 100 * 1.05[sup:20t2r0o1]t[/sup:20t2r0o1].

a) State the amount of money in the account on 1 January 2000.
b) Calculate, to the nearest penny, the amount of money in the account on 1 January 2004.

Working

a) When t = 0:

P = 100 so £100.

b) When t = 4:

P = 100 * 1.05[sup:20t2r0o1]4[/sup:20t2r0o1]
P = 121.55 so £121.55

The book says for part (b) that the answer is £115.76. I realised you could get this answer using t = 3, but shouldn't t = 4 (1 January 2004 is four years after 1 January 2000)?

Thanks.
 
Re: Exponential growth

Monkeyseat said:
Hi,

Question

The amount of money, £P, in a special savings account at time t years after 1 January 2000 is given by P = 100 * 1.05[sup:hzlpz6p1]t[/sup:hzlpz6p1].

Jan 1, 2004 would be 3 years 365 days - 1 day short of four years

a) State the amount of money in the account on 1 January 2000.
b) Calculate, to the nearest penny, the amount of money in the account on 1 January 2004.

Working

a) When t = 0:

P = 100 so £100.

b) When t = 4:

P = 100 * 1.05[sup:hzlpz6p1]4[/sup:hzlpz6p1]
P = 121.55 so £121.55

The book says for part (b) that the answer is £115.76. I realised you could get this answer using t = 3, but shouldn't t = 4 (1 January 2004 is four years after 1 January 2000)?

Thanks.
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