Comparing the greatest values of two functions (Derivatives)

Question

I've tried doing this task, and for this kind of task I should be using derivatives. When I done all the calculus, everything I got were some weird result which I do not know how to compare.

Task says that if $M_1$ is the greatest value of function $f_1(x)= (\log_56)^{\sin{x}}$ and $M_2$ is the greatest value of function $f_2(x)= (\log_65)^{\sin{x}}$ than $M_1 \leq M_2$. Prove it.

I tried solving this and I got derivative of the first one being: $(\log_56)^{\sin{x}}\cos{x}\ln{(\log_56)}$ and for the second one I got $(\log_65)^{\cos{x}}\sin{x}(-\ln{(\log_56)})$ . From here If I say that derivatives are equal to zero, and then observe logarithms as constants and cancel them out i still get $\sin{x}$ and $\cos{x}$ which are not the same on $0$.

Answer 1

Hint: let $a>0$, $a \neq 1$, be a fixed number. Can you compute explicitly the maximum value of $x \mapsto a^{\sin x}$ as $x \in [0,2\pi)$? It is just a matter of monotonicity. Here is a sample graph.

Now compare the two vales corresponding to $a=\log_6 5$ and to $a=\log_5 6$.

Answer 2

I think you made a mistake in finding $f_2'(x)$.

Since $$f_1'(x)=(\log_56)^{\sin x}\cdot \cos x \cdot \ln(\log_56)$$and$$f_2'(x)=(\log_65)^{\sin x}\cdot \cos x \cdot \ln(\log_65),$$ both $f_1'(x)=0$ and $f_2'(x)=0$ lead $\cos x=0$. Hence, we have $$M_1=\max\{(\log_56)^1,(\log_56)^{-1}\}=\log_56$$and $$M_2=\max\{(\log_65)^1,(\log_65)^{-1}\}=\log_56.$$ Thus, we have $M_1=M_2$.