Solving recurance relation with master theorem

Question

I'm studying asympotic analysis and I encountered this problem:

Given a recurrence relation: $$T(n)= aT(n/b)+cn^a (n>0;a>=1;b>=1)$$ prove that
if $a>a^b$ then T(n)=$\mathfrak\theta(n^{log_b(a)})$
if $a=a^b$ then T(n)=$\mathfrak\theta(n^alog(n))$
if $a<a^b$ then T(n)=$\mathfrak\theta(n^a)$

This problem screams master theorem to me, but I can't seem to solve it. To be exact, I can't find the relation between $n^a$ and $n^{log_b(a)}$. I have also tried out substitution but it didn't work out.

Answer 1

Apparently, you can't prove it because it is incorrect. Let's assume that $a=1$ and $b>1$. So, we get: $T(n) = aT(n/b) + cn^a = T(n/b) + cn = \Theta(n)$

But you are trying to show that if $a=a^b$, then $T(n)=\Theta(n^a\log(n))$. Since $a=1$, this holds and we get $T(n)=\Theta(n\log(n))$, which is not $\Theta(n)$.