Compound interest formula:
[tex]A=P(1+\frac{r}{n})\placeholder{⬚}^{nt}[/tex]A is the amount after t years
P is the principal
r is the interest rate in decimals
n is the number of times interest is compound
t is the time in years
For the given situation:
A=65,000
P=36,000
r=0.0775
n=1
t=t
[tex]65,000=36,000(1+\frac{0.0775}{1})\placeholder{⬚}^{1*t}[/tex]Solve the equation for t:
[tex]\begin{gathered} 65,000=36,000(1+0.0775)\placeholder{⬚}^t \\ 65,000=36,000(1.0775)\placeholder{⬚}^t \\ \frac{65,000}{36,000}=1.0775^t \\ \\ \frac{65}{36}=1.0775^t \\ \\ log(\frac{65}{36})=log(1.0775)\placeholder{⬚}^t \\ \\ log(\frac{65}{36})=t*log(1.0775) \\ \\ t=\frac{log(\frac{65}{36})}{log(1.0775)} \\ \\ t=7.91 \end{gathered}[/tex]Then, after approximately 8 years the amount due will reach $65,000 or more
