Answer:
Step-by-step explanation:
I have no idea what formula that is you're using but the one I teach in both algebra 2 and in precalculus for continuous compounding is
[tex]A(t)=Pe^{rt}[/tex]
where A(t) is the amount after the compounding, P is the initial investment, ee is Euler's number, r is the interest rate in decimal form, and t is the time in years. If our money doubles, we just have to come up with a number which will be P and then double it to get A(t). It doesn't matter what number we pick to double, the answer will come out the same regardless. I started with 2 and then doubled it to 4 and filled in the rest of the info given with time as my unknown:
[tex]4=2e^{(.062)(t)}[/tex]
Begin by dividing both sides by 2 to get
[tex]2=e^{.062t}[/tex]
The only way we can get that t out of its current position is to take the natural log of both sides. Natural logs have a base of e, so
[tex]ln_e(e)=1[/tex] This is because they are inverses of one another. Taking the natural log of both sides:
[tex]ln2=.062t[/tex] Now divide by .062 to get
t = 11.2 years
