Answer:
[tex]t=15.6\ years[/tex]
Step-by-step explanation:
we know that
The compound interest formula is equal to
[tex]A=P(1+\frac{r}{n})^{nt}[/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 Number of Time Periods
n is the number of times interest is compounded per year
in this problem we have
[tex]t=?\ years\\ P=\$9,500\\A=\$19,300\\r=4.5\%=4.5/100=0.045\\n=365[/tex]
substitute in the formula above
[tex]19,300=9,500(1+\frac{0.045}{365})^{365t}[/tex]
solve for t
simplify
[tex](19,300/9,500)=(\frac{365.045}{365})^{365t}[/tex]
Apply log both sides
[tex]log(19,300/9,500)=log[(\frac{365.045}{365})^{365t}][/tex]
Apply property of logarithms in the right side
[tex]log(19,300/9,500)=(365t)log(\frac{365.045}{365})[/tex]
[tex]t=log(19,300/9,500)/[(365)log(\frac{365.045}{365})][/tex]
[tex]t=15.6\ years[/tex]
