Answer:
"To the nearest year, it would be about 9 years"
Explanation:
11c)
This is compound growth problem. It goes by the formula:
![F=P(1+r)^t](https://img.qammunity.org/2020/formulas/mathematics/middle-school/2l5txdvbnn6otiepxf4q78p59bxjtd8j5b.png)
Where
F is the future amount
P is the present (initial) amount
r is the rate of growth, in decimal
t is the time in years
Given,
P = 20,000
r = 8% = 8/100 = 0.08
F = double of initial amount = 2 * 20,000 = 40,000
We need to find t:
![F=P(1+r)^t\\40,000=20,000(1+0.08)^t\\2=(1.08)^t](https://img.qammunity.org/2020/formulas/mathematics/middle-school/dxijm8gxqs26efvupw7oqfbkzesshomkdh.png)
To solve exponentials, we can take Natural Log (Ln) of both sides:
![2=(1.08)^t\\Ln(2)=Ln((1.08)^t)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/d2g4uzg16ipq18zsy6rp0ccdxcjy7tmp4i.png)
Using the rule shown below we can simplify and solve:
![Ln(a^b)=bLn(a)](https://img.qammunity.org/2020/formulas/mathematics/high-school/7wsoqgrzb9lt4bwduqjatxijle8yk2ax4h.png)
We can write:
![Ln(2)=Ln((1.08)^t)\\Ln(2)=tLn(1.08)\\t=(Ln(2))/(Ln(1.08))\\t=9.0064](https://img.qammunity.org/2020/formulas/mathematics/middle-school/peomvxtweunjbh4hphdq9qdt9ftslv8ba3.png)
To the nearest year, that would be about 9 years