Maximum of a weird function

First, note that $27^{\log_6 x} = 3^{3\cdot \log_6 x} = 3^{\log_6(x^3)}$ and similarly, $8^{\log_6x} = 2^{\log_6 (x^3)}$ . As $x$ ranges over positive reals, $x^3$ also ranges over positive reals. Thus, we can substitute $x^3=6^y$ where $y$ ranges over all reals.

It suffices to find the maximum value of $g(y)=8\cdot 3^y + 27\cdot 2^y - 6^y=8\cdot 3^y + 27\cdot 2^y - 2^y\cdot 3^y$ . By Simon's Favorite Factoring Theorem, $g(y)=216-(3^y-27)(2^y-8)$ .

Finally, notice that for $y>3$ , $3^y > 27$ and $2^y > 8$ implying $(3^y-27)(2^y-8) > 0$ . Similarly, when $y<3$ , $3^y < 27$ and $2^y < 8$ implying $(3^y-27)(2^y-8) > 0$ . Either way, if $y\ne 3$ then $g(y) < 216$ . Since $g(3)=216$ , this must be our maximum value. Final answer: $\boxed{216}$ .