Respuesta :
Answer:
20 years.
Step-by-step explanation:
We have been given a formula [tex]A(t)=A_0\cdot e^{kt}[/tex], which represents the value of an investment A (in dollars) after t years.
Substitute the given values:
[tex]\$3,000=\$1,000\cdot e^{k*10}[/tex]
Let us solve for k.
[tex]\frac{\$3,000}{\$1,000}=\frac{\$1,000\cdot e^{k*10}}{\$1,000}[/tex]
[tex]3=e^{k*10}[/tex]
Take natural log of both sides:
[tex]\text{ln}(3)=\text{ln}(e^{k*10})[/tex]
Using property [tex]\text{ln}(a^b)=b\cdot \text{ln}(a)[/tex], we will get:
[tex]\text{ln}(3)=10k\cdot\text{ln}(e)[/tex]
We know that [tex]\text{ln}(e)=1[/tex], so
[tex]\text{ln}(3)=10k\cdot 1[/tex]
[tex]\text{ln}(3)=10k[/tex]
[tex]\frac{\text{ln}(3)}{10}=\frac{10k}{10}[/tex]
[tex]\frac{\text{ln}(3)}{10}=k[/tex]
[tex]\$9,000=\$1,000\cdot e^{\frac{\text{ln}(3)}{10}*t}[/tex]
Dividing both sides by 1000, we will get:
[tex]9=e^{\frac{\text{ln}(3)}{10}*t}[/tex]
Take natural log of both sides:
[tex]\text{ln}(9)=\text{ln}(e^{\frac{\text{ln}(3)}{10}*t)[/tex]
[tex]\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot\text{ln}(e)[/tex]
[tex]\text{ln}(9)=\frac{\text{ln}(3)}{10}*t\cdot1[/tex]
[tex]10*\text{ln}(9)=10*\frac{\text{ln}(3)}{10}*t[/tex]
[tex]10\text{ln}(9)=\text{ln}(3)*t[/tex]
[tex]10\text{ln}(3^2)=\text{ln}(3)*t[/tex]
[tex]2\cdot 10\text{ln}(3)=\text{ln}(3)*t[/tex]
[tex]20\text{ln}(3)=\text{ln}(3)*t[/tex]
Divide both sides by [tex]\text{ln}(3)[/tex]:
[tex]\frac{20\text{ln}(3)}{\text{ln}(3)}=\frac{\text{ln}(3)*t}{\text{ln}(3)}[/tex]
[tex]20=t[/tex]
Therefore, it will take 20 years for the investment to be $9,000.
