$\log_2\sqrt{x^2 - \frac{x}{2}+1} + 3\log_3(x^2-\frac{x}{2}) \leq 4$

Question

$$\log_2\sqrt{x^2 - \frac{x}{2}+1} + 3\log_3(x^2-\frac{x}{2}) \leq 4$$

I wrote $x^2 - \frac{x}{2} = a$

$\log_2(a+1) + 3\log_3(a) \leq 4$

$\log_2(a+1) + 3\log_3(a) \leq \log_2 16$

$\log_2\frac{(a+1)}{16} + 3\log_3(a) \leq 0$

$$\frac{\ln(\frac{a+1}{16})}{\ln2} + \frac{3\ln a}{\ln 3} \leq 0$$

After that I got stuck. Any help very much appreciated

Answer 1

You have some mistakes after your substitution. It is equivalent with:

$$\log_2\sqrt{a+1}+3\log_3 a\leq 4$$

or

$$\frac{\ln(a+1)}{2\ln 2}+\frac{3\ln a}{\ln 3} \leq 4$$

Let $f: (0,\infty) \to \mathbb{R}, f(x)=\dfrac{\ln(x+1)}{2\ln 2}+\dfrac{3\ln x}{\ln 3}$. Now, it is easy to see that $f'(x) \geq 0$ so $f$ is increasing. The inequality can be written as:

$$f(a) \leq f(3)$$

and thus $a \leq 3$. It remains to solve a quadratic in $x$:

$$x^2-\frac{x}{2}\leq 3$$

and combined with the restriction $x^2-\dfrac{x}{2}>0$ (for the logarithms to exist) gives the final result:

$$x \in \left[-\frac{3}{2},0\right)\cup \left(\frac{1}{2},2\right]$$