Question 38110
I do get 10 years
{{{A= P(1 + r/n)^nt}}}
I'm assuming that, if t is in years, n means number of times
per year that the interest is compounded.
10000 is what you end up with (double the 5000)
{{{10000 = 5000*(1 + .07/4)^4t}}}
divide both sides by 5000
{{{2 = (1 + .07/4)^(4t)}}}
{{{2 = ((4 + .07) / 4)^(4t)}}}
{{{2 = (4.07/4)^(4t)}}}
take the log of both sides. Remember that log(a^b) = b*log(a).
{{{log(2) = 4t*log(1.0175)}}}
{{{.301 = 4t* .00753}}}
{{{.301 = t * .0301}}}
{{{t = .301 / .0301}}}
{{{t = 10}}}