Question 868846
You want the formula this way:  A=P(1+r/n)^(nt)
and as rendered, is {{{A=P(1+r/n)^(nt)}}}.


n is the number of compounding for a year.
t is time in years.
r is the interest rate as a decimal number.
P is "principle", starting amount.



You want to find t for n=52, P=1800, and r=0.032.  A=2*P=3600.
Whether Natural log or Common log is your choice.  You <i>SHOULD</i> solve for t <b>symbolically, first</b>; and then substitute the values.


My choice is log base 10.
{{{log((A))=log((P))+log(((1+r/n)^(nt)))}}}
{{{log((A))=log((P))+nt*log((1+r/n))}}}
{{{nt*log((1+r/n))=log((A))-log((P))}}}
{{{nt*log((1+r/n))=log((A/P))}}}, and you know that {{{A/P}}} will be 2, so make that change now.
{{{nt*log((1+r/n))=log((2))}}}
{{{t=(1/n)(1/log((1+r/n)))log((2))}}}
{{{highlight(t=(1/n(log((1+r/n))))log((2)))}}}-----<i>these steps and result can look much better on paper or display wall.</i>


Now substitute the values.
{{{t=(1/52log((1+0.032/52)))*log((2))}}}
should be {{{21&2/3}}} year or 21 years 8 months