Question

$3550 $ ⁢ 3550 is invested at 10.0% 10.0 % compounded continuously. How long will it take for the balance to reach $7100 $ ⁢ 7100 ? Round your answer to two decimal places, if necessary.

Answer 1

Answer:

6.93 years.

Step-by-step explanation:

We have been given that $3550 is invested at 10.0% compounded continuously.

To solve our given problem, we will use continuous compounding formula.

$A=P\cdot e^{rt}$, where,

A = Final amount,

P = Principal amount,

e = Mathematical constant,

r = Interest rate in decimal form,

t = Time

$10\%=\frac{10}{100}=0.10$

Substitute the given values:

$7100=3550\cdot e^{0.10t}$

$\frac{7100}{3550}=\frac{3550\cdot e^{0.10t}}{3550}$

$2=e^{0.10t}$

Take natural log of both sides:

$\text{ln}(2)=\text{ln}(e^{0.10t})$

Using property $\text{ln}(a^b)=b\cdot \text{ln}(a)$, we will get:

$\text{ln}(2)=0.10t\cdot \text{ln}(e)$

$0.6931471805599453=0.10t\cdot 1$

$0.6931471805599453=0.10t$

Switch sides:

$0.10t=0.6931471805599453$

$\frac{0.10t}{0.10}=\frac{0.6931471805599453}{0.10}$

$t=6.9314718$

$t\approx 6.93$

Therefore, it will take approximately 6.93 years for the balance to reach $7100.