Question 980996
That is two compoundings per year for ten years, making twenty compounding periods.


{{{110000*(1+r/2)^20=250000}}}, the growth equation with the values filled; only interest rate r, is unknown; r is a decimal value, a fraction.  You could also treat the binomial as a variable, if this is easier.


{{{(1+r)^20=25/11}}}


{{{log(10,(1+r)^20)=log(10,(25/11))}}}


{{{20*log(10,(1+r))=log(10,(25/11))}}}


{{{log(10,(1+r))=(1/20)log(10,(25/11))}}}


Evaluate or compute the righthand member.


{{{log(10,(1+r))=0.0178273661756906284411161119654}}}, far more accuracy than meaningful.


Antilog of 0.0178273661756906284411161119654  will be  1+r.
...  {{{1+r=1.0419}}}
For ONE compounding period, rate is  4.19%, so for a whole year,  8.38%.