Maximize $8\cdot 27^{\log_{6}x}+27\cdot 8^{\log_{6}x}-x^3$

Question

Find the maximum possible value of $$8 \cdot 27^{\log_{6}x}+27 \cdot 8^{\log_{6}x}-x^3$$ $x>0$

How do I deal with the powers which involve logarithm of $x$?

I'm not being able to simplify.

Any suggestion?

Edit:Solved it.It was simple.Just substitute $x^3=6^t$.That makes the expression very simple.Put derivative to 0.x is $6$ simply by observation!

Answer 1

The problem being obviously (according to the input to Wolfran Alpha) $$f(x)=8\times 27^{\log_{6}x}+27\times 8^{\log_{6}x}-x^3$$ start changing variable $y=\log_6(x)$ that is to say $x=6 ^y$. Replacing, $$f(y)=8\times 27^y+27 \times 8^y-6^{3y}$$ Computing derivatives $$f'(y)=8\times 27^y \log (27)+27\times 8^y \log (8)-3\times 6^{3 y} \log (6)$$ Now, inspection $$f'(0)=78 \log (2)+21 \log (3)$$ $$f'(1)=0$$ So, $y=1$ corresponds to an extremum. Now, the second derivative $$f''(y)=8\times 27^y \log ^2(27)+27\times 8^y \log ^2(8)-9\times 6^{3 y} \log ^2(6)$$ $$f''(1)=-3888 \log (2) \log (3) <0$$ So $y=1\implies f(1)=216$ is the maximum.

Answer 2

Hint: If you think the dot is multiplication, I would also guess they intend the reading to be $(8\cdot 27)^{\log_{6}x}+(27\cdot 8)^{\log_{6}x}-x^3$ Now remember that $6^{\log_6 x}=x$. What can you do with the $8 \cdot 27$? Unfortunately, in this approach I do not find a maximum.

Another reading would be for the first term to be $8 \cdot 27^{\log_6 x}$. You can do $27^{\log_6 x}=\left(6^{\log_6 27}\right)^{\log_6 x}=6^{\log_6 27\cdot\log_6 x}=\left(6^{\log_6 x}\right)^{\log_6 27}=x^{\log_6 27}\approx x^{1.83944}$