Sure, let's solve this step by step.
1. Finding [tex]\( f(0) \)[/tex]:
[tex]\( f(t) \)[/tex] represents the amount in the account at time [tex]\( t \)[/tex], and is given by the formula:
[tex]\[
A = f(t) = 8900 e^{0.057 t}
\][/tex]
To find [tex]\( f(0) \)[/tex], we simply substitute [tex]\( t = 0 \)[/tex] into the formula:
[tex]\[
f(0) = 8900 e^{0.057 \cdot 0} = 8900 e^0
\][/tex]
Since [tex]\( e^0 = 1 \)[/tex]:
[tex]\[
f(0) = 8900 \cdot 1 = 8900
\][/tex]
So, [tex]\( f(0) = \$8900.00 \)[/tex].
2. Finding the amount in the account after 8 years:
We need to find [tex]\( f(8) \)[/tex], so we substitute [tex]\( t = 8 \)[/tex] into the formula:
[tex]\[
f(8) = 8900 e^{0.057 \cdot 8}
\][/tex]
After performing the calculation, the amount in the account after 8 years is approximately:
[tex]\[
f(8) = \$14041.98
\][/tex]
3. Determining the time it will take for the investment to grow to \[tex]$13100:
We need to find \( t \), given that \( A = \$[/tex]13100 \):
[tex]\[
13100 = 8900 e^{0.057 t}
\][/tex]
First, we isolate the exponential term by dividing both sides by 8900:
[tex]\[
\frac{13100}{8900} = e^{0.057 t}
\][/tex]
Simplifying the left-hand side:
[tex]\[
\frac{13100}{8900} \approx 1.4719
\][/tex]
So, we have:
[tex]\[
1.4719 = e^{0.057 t}
\][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm (ln) of both sides:
[tex]\[
\ln(1.4719) = \ln(e^{0.057 t})
\][/tex]
Since [tex]\( \ln(e^x) = x \)[/tex]:
[tex]\[
\ln(1.4719) = 0.057 t
\][/tex]
Solving for [tex]\( t \)[/tex]:
[tex]\[
t = \frac{\ln(1.4719)}{0.057} \approx 6.8
\][/tex]
So, it will take approximately [tex]\( 6.8 \)[/tex] years for the investment to grow to \[tex]$13100.
To summarize the answers:
- \( f(0) = \$[/tex]8900.00 \)
- The amount in the account after 8 years is [tex]\( \$14041.98 \)[/tex]
- It will take approximately [tex]\( 6.8 \)[/tex] years for the investment to grow to [tex]\( \$13100 \)[/tex]
