As you mentioned C++ in the tag, try one of this method: double pow (double base , double exponent); float pow (float base , float exponent); long double pow (long double base, long double exponent); double pow (Type1 base , Type2 exponent); // additional overloads Returns base raised to the power exponent. See reference here. … Read more
D[case] = (Dref[case]*((50/ t)^ m )) * ( math.exp((20/6)*((1/295) – (1/Temp))) It seems to be extra parentheses Try D[case] = (Dref[case]*((50/ t)^ m )) * math.exp((20/6)*((1/295) – (1/Temp))) 1 solved Python math.exp function Syntax
You can try k >= 2 and limit N with sqrt(w), start i with 2 in the loop. If solution is not found then k = 1, n = w. Also you do not need pow, you could multiply intermediate product by N on each iteration. solved Given an integer, w, find n and k … Read more