If $x \geq 5$ then $3^{x-2} >x^2$ I have the following in some old course notes from UF and I am still curious about possible solutions. 

If $x \geq 5$ then $$3^{x-2}>x^2$$

I am asking for help proving that statement. To see an example of this just observe that $$3^{5-2}>5^2$$ I tried using logarithms to bring down the exponents but that went nowhere.
 A: I multiplied by 9 and then took logs to get $$x\ln 3 > \ln 9 +2 \ln x.$$  Taking the derivative of both sides gives $$\ln3 > \frac{2}{x}.$$  Since $\ln 3>1$, we know that this last inequality will be true if $x>2$.  This means that the left side of the original inequality is increasing faster than the right side when $x>2$.  So once we observe that the left side is greater than the right side at $x=5$, then the inequality is true for all larger $x$.
A: you will get by taking the logarithm $$(x-2)\ln(3)>2\ln(x)$$
now set $$f(x)=x-2-\frac{2}{\ln(3)}\ln(x)$$ show that for $$x\geq 5$$ $$f(x)$$ is strictly increasing and $$f(5)>0$$ holds
A: Letting $x=u+5$ with $u\ge0$, the expression to prove is $3^{u+3}\gt(u+5)^2$, which can be rewritten as
$$3^u\gt{25\over27}+{10\over27}u+{1\over27}u^2$$
But since $3\gt e$, we have $3^u\ge e^u$ for $u\ge0$, hence, using the Taylor series for $e^u$ (in particular, the fact that its terms are all positive when $u\ge0$), we have
$$3^u\ge e^u\ge1+u+{1\over2}u^2\gt{25\over27}+{10\over27}u+{1\over27}u^2$$
(The final inequality is strict even at $u=0$, since $1\gt{25\over27}$.)
