# Help with inequality with one unknown

Please could you help how to solve the inequality $$(\sqrt{x-9})(2^{x-8}+3^{x-9}-9)\geq 0$$

• It's not clear what the argument of the radical is. – lulu May 17 at 15:34
• Is it $$\sqrt{x-9}(2^{x-8}+3^{x-9}-9)\geq 0$$? – Dr. Sonnhard Graubner May 17 at 15:34
• What did you try? Did you try $x=9,10,11$ for example? – Dietrich Burde May 17 at 15:35
• @Dr. Sonnhard Graubner yes, this is it – ramhat lubumba May 17 at 15:37
• $$x=9$$ is one solution, to solve $$2^{x-8}+3^{x-9}-9\geq 0$$ you will Need a numerical method – Dr. Sonnhard Graubner May 17 at 15:39

Let $$x = t+9$$, then the radical needs $$t \geqslant 0$$. $$t=0$$ is a solution obviously, for others, we are left to solve $$2\cdot 2^t+3^t\geqslant 9$$. Note LHS is increasing, and starting from $$3<9$$, so there is a unique $$a>0$$ s.t. $$t \in [a, \infty)$$ are all solutions, i.e. the full solution set is $$x\in \{9\}\cup[a+9, \infty)$$, where $$2\cdot 2^a+3^a=9$$.

To find $$a$$, you will need to use numerical methods, it’s about $$1.288...$$.