MHB Prove Inequality: 1 < √3 < 2 ⇒ 6 < 3^√3 < 7

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The inequality 1 < √3 < 2 leads to the conclusion that 6 < 3^√3 < 7 through a series of mathematical deductions. By manipulating the initial inequality, it is shown that 0 < (√3 - 3/2)^2 < 1/4, which further refines the bounds of √3. This results in the inequalities 3^(5/3) < 3^√3 < 3^(7/4). The values 6 and 7 are confirmed to be less than 3^(5/3) and 3^(7/4) respectively, solidifying the overall inequality. The discussion highlights the logical steps needed to prove the inequality effectively.
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Deduce from the simple estimate that if $$1<\sqrt{3}<2$$, then $$6<3^{\sqrt{3}}<7$$.

Hi members of the forum,

This problem says the resulting inequality may be deduced from the simple estimate, but I was unable to do so; could anyone shed some light on how to deduce the intended result?

Thanks in advance.
 
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Re: Proving an inequality

$$1\,<\,\sqrt3\,<\,2$$​

$$\Rightarrow\ -\frac12\,<\,\sqrt3-\frac32\,<\,\frac12$$

$$\Rightarrow\ 0<\,\left(\sqrt3-\frac32\right)^2\,<\,\frac14$$

$$\Rightarrow\ 0<\,\frac{21}4-3\sqrt3\,<\,\frac14$$

$$\Rightarrow\ \frac53<\,\sqrt3\,<\,\frac74$$

$$\Rightarrow\ 3^{5/3}<\,3^{\sqrt3}\,<\,3^{7/4}$$

Note that $$6=216^{1/3}<243^{1/3}=3^{5/3}$$ and $$3^{7/4}=2187^{1/4}<2401^{1/4}=7$$.
 
Re: Proving an inequality

Nehushtan said:
$$1\,<\,\sqrt3\,<\,2$$​

$$\Rightarrow\ -\frac12\,<\,\sqrt3-\frac32\,<\,\frac12$$

$$\Rightarrow\ 0<\,\left(\sqrt3-\frac32\right)^2\,<\,\frac14$$

$$\Rightarrow\ 0<\,\frac{21}4-3\sqrt3\,<\,\frac14$$

$$\Rightarrow\ \frac53<\,\sqrt3\,<\,\frac74$$

$$\Rightarrow\ 3^{5/3}<\,3^{\sqrt3}\,<\,3^{7/4}$$

Note that $$6=216^{1/3}<243^{1/3}=3^{5/3}$$ and $$3^{7/4}=2187^{1/4}<2401^{1/4}=7$$.
Hi Nehushtan, thanks to your simple explanation because it is now very clear to me! I appreciate it! :)
 
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