Question 707572
A formula has been developed for this.  Terms being like a, ar, ar^2, ar^3,...
for n terms, the sum of those n terms is {{{(a-ar^n)/(1-r)}}}.


The problem description shows the sum (being 11.35), a=1.5, r=0.9, and the unknown variable is n for the number of years.  You might first want to use the formula {{{S[n]=(a-ar^n)/(1-r)}}}, and solve for n.  Once done that, THEN substitute the values.  Working that way, you should get to
{{{r^n=(a-S(1-r))/a}}}; I am dropping the subscript on S just for convenience.

Taking logarithms of both sides whatever base you choose,
{{{n*ln(r)=ln((a-S(1-r))/a)}}}
{{{n=ln((a-S(1-r))/a)/ln(r)}}}

Now just substitute the values.
log base ten of [(1.5-11.35(1-.9))/1.5]=-0.6138
log base ten of  0.9=-.045757
n=13.4 years