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

Answer:

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

The both account have $8,834.43.

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,

$5000e^{0.035t}=5250e^{0.032t}$

$\Rightarrow \frac{e^{0.035t}}{e^{0.032t}}=\frac{5250}{5000}$

$\Rightarrow e^{0.035t-0.032t}=\frac{21}{20}$

$\Rightarrow e^{0.003t}=\frac{21}{20}$

Taking ln both sides

$\Rightarrow lne^{0.003t}=ln(\frac{21}{20})$

$\Rightarrow 0.003t}=ln(\frac{21}{20})$

$\Rightarrow t}=\frac{ln(\frac{21}{20})}{0.003}$

$\Rightarrow t= 16.26$

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

The account balance on that time is

$P(16.26)=5000e^{0.035\times 16.26}$

              =$8,834.43

The both account have $8,834.43.