Answer :
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 :
[tex]P(t)=P_0e^{rt}[/tex]
P(t)= value after t time
[tex]P_0[/tex]= 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
[tex]P(t)=5000e^{0.035t}[/tex]
For the second case,
P= $5,250, r=3.5%=0.032
[tex]P(t)=5250e^{0.032t}[/tex]
According to the problem,
[tex]5000e^{0.035t}=5250e^{0.032t}[/tex]
[tex]\Rightarrow \frac{e^{0.035t}}{e^{0.032t}}=\frac{5250}{5000}[/tex]
[tex]\Rightarrow e^{0.035t-0.032t}=\frac{21}{20}[/tex]
[tex]\Rightarrow e^{0.003t}=\frac{21}{20}[/tex]
Taking ln both sides
[tex]\Rightarrow lne^{0.003t}=ln(\frac{21}{20})[/tex]
[tex]\Rightarrow 0.003t}=ln(\frac{21}{20})[/tex]
[tex]\Rightarrow t}=\frac{ln(\frac{21}{20})}{0.003}[/tex]
[tex]\Rightarrow t= 16.26[/tex]
Therefore after 16.26 unit of time, both accounts have same balance.
The account balance on that time is
[tex]P(16.26)=5000e^{0.035\times 16.26}[/tex]
=$8,834.43
The both account have $8,834.43.