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

[anemone]

Deduce from the simple estimate that if \(\displaystyle 1<\sqrt{3}<2\), then \(\displaystyle 6<3^{\sqrt{3}}<7\).

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.

 

[Nehushtan]

Re: Proving an inequality

\(\displaystyle 1\,<\,\sqrt3\,<\,2\)​

\(\displaystyle \Rightarrow\ -\frac12\,<\,\sqrt3-\frac32\,<\,\frac12\)

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

\(\displaystyle \Rightarrow\ 0<\,\frac{21}4-3\sqrt3\,<\,\frac14\)

\(\displaystyle \Rightarrow\ \frac53<\,\sqrt3\,<\,\frac74\)

\(\displaystyle \Rightarrow\ 3^{5/3}<\,3^{\sqrt3}\,<\,3^{7/4}\)

Note that \(\displaystyle 6=216^{1/3}<243^{1/3}=3^{5/3}\) and \(\displaystyle 3^{7/4}=2187^{1/4}<2401^{1/4}=7\).

 

[anemone]

Re: Proving an inequality

\(\displaystyle 1\,<\,\sqrt3\,<\,2\)​

\(\displaystyle \Rightarrow\ -\frac12\,<\,\sqrt3-\frac32\,<\,\frac12\)

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

\(\displaystyle \Rightarrow\ 0<\,\frac{21}4-3\sqrt3\,<\,\frac14\)

\(\displaystyle \Rightarrow\ \frac53<\,\sqrt3\,<\,\frac74\)

\(\displaystyle \Rightarrow\ 3^{5/3}<\,3^{\sqrt3}\,<\,3^{7/4}\)

Note that \(\displaystyle 6=216^{1/3}<243^{1/3}=3^{5/3}\) and \(\displaystyle 3^{7/4}=2187^{1/4}<2401^{1/4}=7\).

Hi Nehushtan, thanks to your simple explanation because it is now very clear to me! I appreciate it! :)