Remember that
The formula to calculate continuously compounded interest is equal to
[tex]A=P\left(e\right)^{\left\{rt\right\}}[/tex]
where
A is the Final Investment Value
P is the Principal amount of money to be invested
r is the rate of interest in decimal
t is the Number of Time Periods
e is the mathematical constant number
In this problem, we have that
Part 1
P=$13,000
A=$26,000
t=14 years
substitute given values
[tex]\begin{gathered} 26,000=13,000\left(e\right)^{\left\{14r\right\}} \\ solve\text{ for r} \\ 2=e^{14r} \\ apply\text{ ln on both sides} \\ ln(2)=lne^{14r} \\ ln(2)=14r \\ r=\frac{ln(2)}{14} \\ \\ r=0.0495 \\ r=4.95\% \\ r=5\% \end{gathered}[/tex]
Part 2
we have
P=$13,000
r=5%=0.05
t=26 years
substitute in the formula
[tex]\begin{gathered} A=13,000(e)^{\operatorname{\{}0.05*26\operatorname{\}}} \\ A=\$47,700.86 \end{gathered}[/tex]
