Question

Using the continuous compounding equation, if someone invested $5,000 at an interest rate of 3.5%, and someone else invested $5,250 at an interest rate of 3.2%, how long would it take for both accounts to have the same value? What value would both accounts have?

Answer 1

Therefore after 16.26 unit of time, both accounts have same balance.

The both account have $8,834.43.

Step-by-step explanation:

Formula for continuous compounding :

`P(t)=P_0e^(rt)`

P(t)= value after t time

`P_0`= Initial principal

r= rate of interest annually

t=length of time.

Given that, someone invested $5,000 at an interest 3.5% and another one invested $5,250 at an interest 3.2% .

Let after t year the both accounts have same balance.

For the first case,

P= $5,000, r=3.5%=0.035

`P(t)=5000e^(0.035t)`

For the second case,

P= $5,250, r=3.5%=0.032

`P(t)=5250e^(0.032t)`

According to the problem,

`5000e^(0.035t)=5250e^(0.032t)`

`\Rightarrow (e^(0.035t))/(e^(0.032t))=(5250)/(5000)`

`\Rightarrow e^(0.035t-0.032t)=(21)/(20)`

`\Rightarrow e^(0.003t)=(21)/(20)`

Taking ln both sides

`\Rightarrow lne^(0.003t)=ln((21)/(20))`

`\Rightarrow 0.003t}=ln((21)/(20))`

`\Rightarrow t}=(ln((21)/(20)))/(0.003)`

`\Rightarrow t= 16.26`

Therefore after 16.26 unit of time, both accounts have same balance.

The account balance on that time is

`P(16.26)=5000e^(0.035* 16.26)`

=$8,834.43

The both account have $8,834.43.