Inequality with prime

Condition: $p\ne3$ .

If $p=2$ , we have $\left|2+\log_{\frac{3}{p}}(3p^2)\right|=\left|2+\log_{\frac{3}{2}}12\right|>2>\dfrac{3}{2}$ , satisfied.

If $p>3$ we have: $\left|2+\log_{\frac{3}{p}}(3p^2)\right|>\dfrac{3}{2}$

$\Leftrightarrow \left|2+\dfrac{\ln(3p^2)}{\ln\left(\dfrac{3}{p}\right)}\right|>\dfrac{3}{2}$

$\Leftrightarrow \left|2+\dfrac{\ln3+2\ln p}{\ln3-\ln p}\right|>\dfrac{3}{2}$

$\Leftrightarrow \left|\dfrac{3\ln3}{\ln3-\ln p}\right|>\dfrac{3}{2}$

$\Leftrightarrow \dfrac{\ln 3}{\ln p-\ln 3}>\dfrac{1}{2}$

$\Leftrightarrow 2\ln3>\ln p-\ln3$

$\Leftrightarrow \ln p <3\ln3 \Leftrightarrow p<27$

Since $p$ is a prime and $p>3$ , we get $p\in\{5; 7; 11; 13; 17; 19; 23\}$

So, there are $\boxed{8}$ primes satisfy the inequality.

Cool solution Khang Nguyen Thanh ! I was going to try base changes like you but feared it would be too messy. Instead, I noted that $2=\log_{\frac{3}{p}}{\frac{9}{p^2}}$ , which simplifies the inside of the absolute value to $\log_{\frac{3}{p}}{27}$ . The rest of my analysis was largely the same as yours. Cool problem!