Question 1100973
Let n be number of compounding periods  ( one fourth of a year each ).


{{{28000*(1.015)^n=29718}}}


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{{{log((28000))+n*log((1.015))=log((29718))}}}

{{{n*log((1.015))=log((29718))-log((28000))}}}

{{{n*(0.006466)=4.4730196-4.447158}}}

{{{n=(0.0258616)/(0.006466)}}}

{{{n=3.9996}}}

{{{close}}}{{{enough}}}{{{to}}}{{{n=4}}}----------four periods, same as 1 year