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

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  • $\begingroup$ It's not clear what the argument of the radical is. $\endgroup$ – lulu May 17 at 15:34
  • $\begingroup$ Is it $$\sqrt{x-9}(2^{x-8}+3^{x-9}-9)\geq 0$$? $\endgroup$ – Dr. Sonnhard Graubner May 17 at 15:34
  • $\begingroup$ What did you try? Did you try $x=9,10,11$ for example? $\endgroup$ – Dietrich Burde May 17 at 15:35
  • $\begingroup$ @Dr. Sonnhard Graubner yes, this is it $\endgroup$ – ramhat lubumba May 17 at 15:37
  • $\begingroup$ $$x=9$$ is one solution, to solve $$2^{x-8}+3^{x-9}-9\geq 0$$ you will Need a numerical method $\endgroup$ – Dr. Sonnhard Graubner May 17 at 15:39
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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 $1.288...$.

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