Find constants a and b in the function ax^b/ln(x) such that f(1/5)=1 and there is a local min at x=1/5

Answer 1

y = ax^b/lnx
y' = (ax^(b-1) (b*logx - 1))/log^2(x)

since x^b is never zero, we want
b logx = 1
b = 1/ln(1/5) = -0.62133

y = ax^b/lnx
y(1/5) = 1, so

a (1/5)^(-0.62133)/-ln5 = 1
a = -ln5/((1/5)^(-0.62133)) -0.59208

y = -0.59208 x^-0.62133 / lnx

check at

http://www.wolframalpha.com/input/?i=-0.59208+x^-0.62133+%2F+lnx+for+0.1+%3C%3D+x+%3C%3D+0.3