Answer:
5.7 years
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=\$9,000\\A=\$11,800\\r=4.75\%=4.75/100=0.0475\\n=4[/tex]
substitute in the formula above
[tex]11,800=9,000(1+\frac{0.0475}{4})^{4t}[/tex]
Solve for t
[tex]\frac{11.8}{9} =(1.011875)^{4t}[/tex]
Applying property of exponents
[tex]\frac{11.8}{9} =(1.011875^{4})^{t}[/tex]
Applying log both sides
[tex]log(\frac{11.8}{9})=log[(1.011875^{4})^{t}][/tex]
[tex]log(\frac{11.8}{9})=(t)log[(1.011875^{4})][/tex]
[tex]t=log(\frac{11.8}{9})/log[(1.011875^{4})][/tex]
[tex]t=5.7\ years[/tex]
