Find the time necessary for $1000 to double when it is invested at rate r =7% and compounded (a) annually, (b) monthly, (c) daily, and (d) continuously

College Algebra
1st Edition
ISBN:9781938168383
Author:Jay Abramson
Publisher:Jay Abramson
Chapter6: Exponential And Logarithmic Functions
Section6.1: Exponential Functions
Problem 68SE: An investment account with an annual interest rateof 7 was opened with an initial deposit of 4,000...
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Find the time necessary for $1000 to double when it is invested at rate r =7% and compounded (a) annually, (b) monthly, (c) daily, and (d) continuously

Expert Solution
Step 1

(As per bartleby guidelines, for questions containing multiple subparts, only three subparts will be solved. Upload the other parts separately)

a)

Given data:

Principal, P=1000Rate, R=7%Amount, A=2000

The amount for compound interest is given by,

A=P1+R100n

Substitute the values,

2000=10001+7100n2=1.07n

Step 2

Take logarithm both sides,

ln2=n·ln1.07n=ln2ln1.07n10.245

Thus, $1000 at a rate of 7% compounded annually will take approximately 10.245 years to double itself.

b)

Rate per month,

 r=712=0.5833%

Apply the formula,

A=P1+R100n

Step 3

Substitute the values,

2000=10001+712100n2=1.00583n

Take logarithm both sides,

ln2=n·ln1.005833n=ln2ln1.005833n119.239

The number of years will be,

T=119.23912=9.9366

Thus, $1000 at an annual rate of 7% compounded monthly will take approximately 9.9366 years to double itself.

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