Question

A) Find the interest rate needed for an investment of $8,000 to double in 14 yr if interest is compounded monthly.

B)How long will it take for an investment of $10,000 to grow to $12,500 if the investment earns interest at the rate of 8%/year compounded annually?

C)How long will it take for an investment of $15,000 to double if the investment earns interest at the rate of 7%/year compounded continuously?

Answer #1

a.We use the formula:

A=P(1+r/1200)^12n

where

A=future value

P=present value

r=rate of interest

n=time period.

(2*8000)=8000(1+r/12)^(12*14)

2^(1/168)=(1+r/12)

(1+r/12)=1.004134399

Hence r=(1.004134399-1)*12

=**4.96%(Approx).**

**2.**

We use the formula:

A=P(1+r/100)^n

where

A=future value

P=present value

r=rate of interest

n=time period.

12500=10000(1.08)^n

(1.08)^n=(12500/10000)

(1.08)^n=1.25

Taking log on both sides;

n*log (1.08)=log 1.25

Hence n=log 1.25/log 1.08

=**2.90 years(Approx)**

**3.**We use the formula:

A=P(e)^rn

where

A=future value

P=present value

r=rate of interest

n=time period.

e=2.71828

(2*15000)=15000(2.71828)^0.07n

2=(2.71828)^0.07n

Taking log on both sides;

log 2=0.07n*log 2.71828

Hence n=1/0.07[log 2/log 2.71828]

=**9.90 years(Approx).**

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monthly? (Round your answer to one decimal place.

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compounded continuously?

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compounded continuously and
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2
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(Discount Rate and Number of Time Periods)
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