Find constants a and b in the function f(x)=(ax^b)ln(x) such that f(1/5)=1 and the function has a local minimum at x=1/5.

Answer 1

f = ax^b lnx
f' = ax^b/x + ax^b lnx b/x
= ax^blnx (1/xlnx + b/x)
= f(x)*(1/xlnx + b/x)
f'(1/5) = 1/(1/5 ln(1/5)) + 5b
= -5/ln5 + 5b
so, b = 1/ln5

f(x) = ax^(1/ln5) lnx
a(1/5)^(1/ln5)(-ln5) = 1
a = -5^(1/ln5) / ln5 = -e/ln5

f(x) = -e/ln5 x^(1/ln5) lnx

This has a minimum at x = 1/5, but f(1/5) = -1.

On the other hand, with

f(x) = e/ln5 x^(1/ln5) lnx

f(1/5) = 1, but it has a local max there.

Better check for typos, and double-check my math.

See

http://www.wolframalpha.com/input/?i=plot+y%3De%2Fln5+x%5E%281%2Fln5%29+lnx+for+x%3D0..1