Kristen invests $5,745 in a bank. The bank pays 6.5% interest compounded monthly. How long must she leave the money in the bank for it to double? Round to nearest tenth of a year.

To solve this we are going to use the compounded interest formula: [tex]A=P(1+ \frac{r}{n})^{nt} [/tex]
where
[tex]A[/tex] is the final amount 
[tex]P[/tex] is the initial investment
[tex]r[/tex] is the interest rate in decimal form 
[tex]t[/tex] is the time in years 

We know that initial investment is $5,745, so [tex]P=5745[/tex]. We also know that Kristen wants his investment to double, so [tex]A=2*5745=11490[/tex]. Now, to convert the interest rate to decimal form, we are going to divide the rate by 100%: [tex]r= \frac{6.5}{100}=0.065 [/tex]. Since the interest is compounded monthly, it is compounded 12 times per year: therefore, [tex]n=12[/tex]. Now that we have all the information we need, lets replace the values in our formula and solve for [tex]t[/tex]:
[tex]A=P(1+ \frac{r}{n})^{nt} [/tex]
[tex]11490=5745(1+ \frac{0.065}{12})^{12t} [/tex]
[tex] \frac{11490}{5745} =(1+ \frac{0.065}{12})^{12t}[/tex]
[tex]2=( \frac{2413}{2400} )^{12t}[/tex]
[tex]ln(2)=ln(\frac{2413}{2400} )^{12t}[/tex]
[tex]ln(2)=12tln(\frac{2413}{2400} )[/tex]
[tex] \frac{ln(2)}{12ln(\frac{2413}{2400} )} =t[/tex]
[tex]t=\frac{ln(2)}{12ln(\frac{2413}{2400} )}[/tex]
[tex]t=10.7[/tex]

We can conclude that she must leave the money for 10.7 years in the bank for it to double.