Please explain why this expression holds true? I thank you.
e^(ln(1/2)*1/6*10)=2^(-10/6)
+130516 e^(ln(1/2)*1/6*10)=2^(-10/6)
e^[ln(1/2) * ( 10/6) ] =
e^ [ (10/6) ln (1/2) ] =
e^ [ ln (1/2)(10/6) ] =
(1/2)(10/6) =
2^ (-10/6) [ this is true because (1/a)b = a-b ]
Thank you very much CPhill. Do you recognize this equality and where it comes from? It is this expression used to calculate the amount remaining of a radioactive substance, namely:Ao=Ao.e^kt, which I calculated to be equivalent to 2^(-t/h), where t=years and h=half-life of the radioactive substance. See here: http://web2.0calc.com/questions/m-me-kt-m-25-t-10-how-can-i-possibly-work-out-m-if-i-don-t-know-k-the-question-asks-this-a-particular-radioactive-substance-modelled-by-equation
