Question 1103291
<pre>
{{{A=P(1+r/n)^(nt)}}}

We solve for t before substituting, because it's
easier than substituting first:

{{{ln(A)=ln(P(1+r/n)^(nt))}}}

{{{ln(A)=ln(P)+ln(1+r/n)^(nt)}}}

{{{ln(A)=ln(P)+(nt)ln(1+r/n)}}}

Divide both sides by {{{(n)ln(1+r/n)}}}

{{{(ln(A)-ln(P))/((n)ln(1+r/n))=t}}}

A = final amount = $18221.60
P = beginning amount =$17000
n = number of compoundings per year = 4 (since it's quarterly)
r = 7% as a decimal, 0.07
t = number of years = ???

Plug in:

{{{t=(ln(18221.60)-ln(17000))/((4)ln(1+0.07/4))}}}

{{{t="0.9999972068..."}}}

Rounds to 1 year.

Edwin</pre>