Question

The value of an investment A (in dollars) after t years is given by the function A(t) = A0ekt. If it takes 10 years for an investment of $1,000 to triple, how many years will it take for the investment to be $9,000? Simplify your answer completely.

Answer 1

Answer:

20 years.

Step-by-step explanation:

We have been given a formula $A(t)=A_0\cdot e^{kt}$, which represents the  value of an investment A (in dollars) after t years.

Substitute the given values:

$\ 3,000=\ 1,000\cdot e^{k*10}$

Let us solve for k.

$\frac{\ 3,000}{\ 1,000}=\frac{\ 1,000\cdot e^{k*10}}{\ 1,000}$

$3=e^{k*10}$

Take natural log of both sides:

$\text{ln}(3)=\text{ln}(e^{k*10})$

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

$\text{ln}(3)=10k\cdot\text{ln}(e)$

We know that $\text{ln}(e)=1$, so

$\text{ln}(3)=10k\cdot 1$

$\text{ln}(3)=10k$

$\frac{\text{ln}(3)}{10}=\frac{10k}{10}$

$\frac{\text{ln}(3)}{10}=k$

$\ 9,000=\ 1,000\cdot e^{\frac{\text{ln}(3)}{10}*t}$

Dividing both sides by 1000, we will get:

$9=e^{\frac{\text{ln}(3)}{10}*t}$

Take natural log of both sides:

$\text{ln}(9)=\text{ln}(e^{\frac{\text{ln}(3)}{10}*t)$

$\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot\text{ln}(e)$

$\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot1$

$10*\text{ln}(9)=10*\frac{\text{ln}(3)}{10}*t$

$10\text{ln}(9)=\text{ln}(3)*t$

$10\text{ln}(3^2)=\text{ln}(3)*t$

$2\cdot 10\text{ln}(3)=\text{ln}(3)*t$

$20\text{ln}(3)=\text{ln}(3)*t$

Divide both sides by $\text{ln}(3)$:

$\frac{20\text{ln}(3)}{\text{ln}(3)}=\frac{\text{ln}(3)*t}{\text{ln}(3)}$

$20=t$

Therefore, it will take 20 years for the investment to be $9,000.