A person invests 6500 dollars in a bank. The bank pays 6.25% interest compounded semi-annually. To the nearest tenth of a year, how long must the person leave the money in the bank until it reaches 9000 dollars?

Answer:[tex]5.3\ 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]A=\$9,000\\ P=\$6,500\\ r=0.0625\\n=2[/tex]  substitute in the formula above  and solve for t[tex]9,000=6,500(1+\frac{0.0625}{2})^{2t}[/tex]  [tex]1.38462=(1.03125)^{2t}[/tex]  applying log both sides[tex]log(1.38462)=(2t)log(1.03125)[/tex]  [tex]t=log(1.38462)/2log(1.03125)=5.3\ years[/tex]