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put this solution on YOUR website!THE AMOUNT OF A RADIOACTIVE TRACER REMAINING AFTER t DAYS IS GIVEN BY A=Ao e-0.058t, WHERE Ao IS THE STARTING AMOUNT AT THE BEGINNING OF THE TIME PERIOD. HOW MANY DAYS WILL IT TAKE FOR ONE HALF OF THE ORIGINAL AMOUNT TO DECAY?
AT HALF LIFE A =A0/2..HENCE
A0/2=A0
E^(-0.58T)=1/2
OR
E^0.58T=2
0.58T=LN(2)
T=LN(2)/0.58=1.195081=1.2 DAYS