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Given this problem $$8\cdot 3^{\sqrt{x}+\sqrt[4]{x}}+9^{\sqrt[4]{x}+1}\geq 9^{\sqrt{x}}$$

$$8\cdot 3^{\sqrt{x}+\sqrt[4]{x}}+3^{2\sqrt[4]{x}+2}\geq 3^{2\sqrt{x}}\\8\cdot 3^{\sqrt{x}+\sqrt[4]{x}-2\sqrt{x}}+3^{2\sqrt[4]{x}+2-2\sqrt{x}}\geq 1\\8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+3^{2\sqrt[4]{x}-2\sqrt{x}+2}\geq 1\\8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+9\cdot 3^{2\sqrt[4]{x}-2\sqrt{x}}\geq 1$$

After simplifying I get $8\cdot 3^{\sqrt[4]{x}-\sqrt{x}}+9\cdot 3^{2\sqrt[4]{x}-2\sqrt{x}}\geq 1$ now putting $t=3^{\sqrt[4]{x}-\sqrt{x}}$ we get $8t+9t^2\geq 1$ and solving we get $t\in (-\infty,-1)\cup(\frac{1}{9},+\infty)$ since $t$ is exponential function with positive base then $t>0$ hence we're only looking at the interval $(\frac{1}{9},\infty)$.Now I have no idea how to find the interval for $x$,I've tried substituting $\sqrt[4]{x}-\sqrt{x}=-2$ and I get $x=16$ and substituting $\sqrt[4]{x}-\sqrt{x}\to \infty$ which is impossible so I have no idea what to do now.

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you already have

$t=3^{\sqrt[4]{x}-\sqrt{x}} \ge \dfrac{1}{9} \iff \sqrt[4]{x}-\sqrt{x} \ge -2 $

$p= \sqrt[4]{x} \ge 0, \implies p-p^2\ge -2 \iff -1\le p \le2 \cap p\ge0 \implies 0 \le p \le 2 \implies 0\le x \le 16 $

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Your work seems correct: the only place equality takes place is at $x=16$. Substituting that into your original equation checks, with $6561$ on both sides.

Since both sides of your inequality are continuous for $x\ge 0$, the solution set for your inequality is either $[0,16]$ or $[16,\infty)$. A quick calculator check shows that the inequality is true for $x=15$ and false for $x=17$. Therefore the solution is

$x\in[0,16]$

This is confirmed with a graph:

enter image description here

I do not understand your last sentence: $\sqrt[4] x-\sqrt x\to-\infty$ as $x\to\infty$, but what does that have to do with the solution to this problem?

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  • $\begingroup$ Doing substitution $x=t^2$ and getting $t\in[3,4]$ you get $x\in [3^2,4^2]$ or differently if $t\in [3,\infty)$ you get $x\in [9,\infty)$ and you find that by replacing $3$ as $t$ and seeing that $x\to \infty$ as $t \to \infty$ $\endgroup$ – kingW3 Mar 5 '15 at 22:14

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