Question 122036
{{{P = 3000}}}
{{{r = .08}}}
{{{n = 2}}} ({{{n}}} is the number of times per year interest is compounded)
{{{A = 6000}}} ({{{A}}} is the amount after {{{t}}} years. The problem
wants to know what {{{t}}} is when {{{A = 2P}}})
{{{A = P(1 + r/n)^nt}}}
{{{6000 = 3000(1 + .08/2)^(2t)}}}
divide both sides by {{{3000}}}
{{{2 = (1 + .04)^(2t)}}}
{{{2 = 1.04^(2t)}}}
Take the log to base 10 of both sides
{{{log(2) = 2t*log(1.04)}}}
{{{.30103 = 2t*.01703}}}
{{{t = .30103 / .03406}}}
{{{t = 8.8382}}} years
That's approximately (1) 8.8
This is {{{8}}} years + {{{.8382*12 = 10.059}}} months
And this is {{{8}}} years, {{{10}}} months {{{.059*30 = 2}}} day
The answer is {{{8}}} years {{{10}}} months {{{2}}} day
Unless I made a mistake, that is.