Find maximum of the function $f(x)=log_{2}^{4}x+12log_{2}^{2}x \cdot log_{2}(\frac{8}{x}) $ on interval $\left [ 1,64 \right ] $

Question

In solution it writes that we can write this function in this form???

$$f(x)=(log_{2}^{2}x-6log_{2}x)^{2} $$

And that quadratic function in brackets has minimum -9 for x=8

Anyway can someone explain me this?

Answer 1

Let $\ell:=\log_2x$ so $f-\ell^4+12\ell^2(3-\ell)=\ell^4-12\ell^3+36\ell^2=(6\ell-\ell^2)^2$, as claimed. On the given range, $6\ell-\ell^2\ge0$, so its maximum at $\ell=3$ gives $f=81$.