1. The problem statement, all variables and given/known data Using the series expansion of e^KX and the fact that a^X=e^Xlna, evaluate 2^-3.4 accurate to 3 dp? 2. Relevant equations 3. The attempt at a solution So a^X=e^Xlna. Basically we need to expand e^-3.4ln(2). e^x=1+X+(X^2)/2!+(X^3)/3!... e^-3.4ln(2)= 1-3.4ln(2)+(-3.4ln(2))^2/2+(-3.4ln(2))^3/6… See any errors in the expansion? The answer is supposedly 0.095 Can't seem to get it right. Thank you.