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kristen invests money in a bank and makes no additional deposits or withdrawals. The bank pays 1.4% interest compounded annually. To the nearest tenth of a year, how long must she leave the money in the bank for it to double? SOLVE ALGEBRAICALLY.

Respuesta :

Answer:approximately 50 years.

Step-by-step explanation:

Let $P represent the initial amount that she deposited. It means that principal,

P = $P

It was compounded annually. This means that it was compounded once in a year. So

n = 1

The rate at which the principal was compounded is 1.4%. So

r = 1.4/100 = 0.014

The formula for compound interest is

A = P(1+r/n)^nt

A = total amount in the account at the end of t years. For the initial amount to double, it means that

A = 2P

Therefore

2P = P (1+0.014/1)^1×t

2P/P = (1.014)^t

2 = (1.014)^t

Taking log to base 10 of both sides, it becomes

Log 2 = log 1.014^t

Log 2 = tlog 1.014

0.301 = 0.006t

t = 0.301/0.006 = 50.2 years

bayosanuade

Answer: 51.4 years

Step-by-step explanation:

There is what we call rule of 72 in Accounting to find the time  a sum of money will double if it is compounded annually at any given rate . The formula is given as :

t = [tex]\frac{72}{r}[/tex]

where t is the time and r is the rate.

t = ?

r = 1.4

Therefore :

t = 72/1.4

t = 51.4286

to the nearest tenth

t = 51. 4 years

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