$\displaystyle \sqrt[22]{10} \ \ \ or \ \ \ \sqrt[30]{30}$
The type is relatively small. That is the equivalent of 10^(1/22) versus 30^(1/30).
You may use a pencil and paper. You may not use a calculator, computer, or logarithmic tables.
$\displaystyle \sqrt[22]{10} \ \ \ or \ \ \ \sqrt[30]{30}$
The type is relatively small. That is the equivalent of 10^(1/22) versus 30^(1/30).
You may use a pencil and paper. You may not use a calculator, computer, or logarithmic tables.
\begin{align} \sqrt[22]{10} &\leftrightarrow \sqrt[30]{30} \\ \frac1{22}\log{10} &\leftrightarrow \frac1{30}\log{30} \\ 30\log{10} &\leftrightarrow 22(\log{10} + \log{3}) \\ 8\log{10} &\leftrightarrow 22\log{3} \\ \frac{\log{10}}{\log{3}} &\leftrightarrow \frac{22}{8} \\ 2 = \frac{2\log{3}}{\log{3}} = \frac{\log{9}}{\log{3}} \approx \frac{\log{10}}{\log{3}} &\lt \frac{22}8 \approx 3 \end{align}
$\displaystyle \frac{\sqrt[30]{30}}{\sqrt[22]{10}} = \frac{\sqrt[30]{10}\sqrt[30]{3}}{\sqrt[22]{10}} = \frac{\sqrt[150]{243}}{\sqrt[165]{100}} > 1 \longrightarrow \sqrt[30]{30} > \sqrt[22]{10}$
$\displaystyle \frac{\sqrt[30]{10}}{\sqrt[22]{10}} = 10^{\frac{1}{30}-\frac{1}{22}} = 10^{-\frac{2}{165}} = 100^{-\frac{1}{165}}$
$\displaystyle \sqrt[30]{3} = \left(243^{\frac{1}{5}}\right)^{\frac{1}{30}} = 243^{\frac{1}{150}}$
The first inequality justified by the shallower root of a larger number in the numerator compared with the denominator (with both radicands greater than 1).
Well $\frac{\log{27}}{\log{3}}=3$ and $10$ is clearly much closer to $9$ than it is to $27$, so it is reasonable to assume that we have a number close to $2$. $\frac{22}{8}=2.75$ which is closer to $3$ than it is to $2$. With a bit of thought I could probably find something a little more rigorous, but it didn't seem necessary.
E.g: The geometric mean of $9$ and $27$ is $\sqrt{243}$ which is a little larger than $15$. Since $10<15$ we know that $\frac{\log{10}}{\log{3}} < \frac{\log{15}}{\log{3}} < 2.5$
$\displaystyle 10^{1/22} \ \ vs. \ \ 30^{1/30}$
$\displaystyle (10^{1/22})^{(2*11*15)} \ \ vs. \ \ (30^{1/30})^{(2*11*15)}$
$\displaystyle 10^{15} \ \ vs. \ \ 30^{11}$
$\displaystyle \dfrac{10^{15}}{10^{11}} \ \ vs. \ \ \dfrac{30^{11}}{10^{11}}$
$\displaystyle 10^4 \ \ vs. \ \ \bigg(\dfrac{30}{10}\bigg)^{11}$
$\displaystyle 10^4 \ \ vs. \ \ 3^{11}$
$\displaystyle \dfrac{10^4}{3^{12}} \ \ vs. \ \ \dfrac{3^{11}}{3^{12}}$
$\displaystyle \dfrac{10^4}{(3^3)^4} \ \ vs. \ \ \dfrac{1}{3}$
$\displaystyle \bigg(\dfrac{10}{27}\bigg)^4 \ \ vs. \ \ \dfrac{1}{3}$
$\displaystyle \bigg(\dfrac{10}{27}\bigg)^4 \ < \bigg(\dfrac{10}{20}\bigg)^4 \ = \ \bigg(\dfrac{1}{2}\bigg)^4 \ = \ \dfrac{1}{16} \ < \ \dfrac{1}{3}$
Therefore, $\displaystyle \ \ 30^{1/30} \ > \ 10^{1/22} $