Solve #10 9 Discuss the function h log2 10 Recall that an annual interest rate o

Question

Solve #10

9 Discuss the function h log2 10 Recall that an annual interest rate of percent compounded n times annually means that, if the calendar year is divided into n equal periods (thus each has 3 days), an account of P dollars earns an amount of 100 dollars at the end of each period. (a) Derive a formula which gives the amount of money in a savings account at the end of k years, where k is a positive integer, if the initial deposit in the account is P dollars, the annual interest rate is x t compounded n times annually, and no money is ever withdrawn from the account. (b) Let In be the logarithm with base e as usual, and let D be number greater than P. How many years (the number of years understood

Explanation / Answer

Let interest compounded =I

Using the Simple Interest formula I = P×x×t let t= 1/n where t is the no. Of times interest in compounded n times in 1 year and we also assume that x is in the decimal form, that x/100= .0x is written as x throughout the derivation.

I1 (interest amount at the end of first year)= P×x×t

A1(amount after 1 year in the account )= P+ I1

= P+ P (xt)=P+ P (x/n)= P (1+x/n)

I2=A1(x/n)= P (1+x/n)(x/n)

A2= A1+I2= P (1+x/n)+ P (1+x/n)(x/n)

= P (1+x/n)[1+x/n]= P (1+x/n)2

Similarly, A3=A2+I3 = P (1+x/n)3 thus by induction we can prove that for k years

A= P (1+x/(100n))1k on substitution of x in fraction form.

Since but if the interest is compounded n times then

A= P (1+x/(100n))nk

since the amount without any withdrawl is D

D= P (1+x/(100n))nk

To find the formula for time when P will become D

D/P= (1+x /(100n)nk taking log on both sides to the base e

ln (D/P)= nk ln (1+x/(100n))

ln (D/P)/ ln (1+x/(100n))= nk

(I/n)ln (D/P)/ ln (1+x/100n)= k

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