# Which radical expression is larger?

#### Archie

Archie, (in effect) you supported that 2 < $$\displaystyle \ \dfrac{log(10)}{log(3)}, \ \$$ and we know that 2 = 16/8 < 22/8.

But I do not see support as to where you show the relative size of $$\displaystyle \ \dfrac{log(10)}{log(3)} \$$ versus 22/8.
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 $\frac{\log{10}}{\log{3}} < \frac{\log{15}}{\log{3}} < 2.5$

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#### Archie

Or, perhaps better:
$$\frac{\log{10}}{\log{3}} < \frac{\log{\sqrt{243}}}{\log{3}} = \frac{\log{243}}{2\log{3}} = \frac{\log{3^5}}{2\log{3}} = \frac{5\log{3}}{2\log{2}} = \frac52 < \frac{22}{8}$$

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#### greg1313

Archie, your fifth denominator has a typo. It should be 2log3. You may have noticed
that I gave that post a "Thanks."

#### greg1313

$$\displaystyle 10^{1/22} \ \ vs. \ \ 30^{1/30}$$

$$\displaystyle (10^{1/22})^{(2*11*15)} \ \ vs. \ \ (30^{1/30})^{(2*11*15)}$$

$$\displaystyle 10^{15} \ \ vs. \ \ 30^{11}$$

$$\displaystyle \dfrac{10^{15}}{10^{11}} \ \ vs. \ \ \dfrac{30^{11}}{10^{11}}$$

$$\displaystyle 10^4 \ \ vs. \ \ \bigg(\dfrac{30}{10}\bigg)^{11}$$

$$\displaystyle 10^4 \ \ vs. \ \ 3^{11}$$

$$\displaystyle \dfrac{10^4}{3^{12}} \ \ vs. \ \ \dfrac{3^{11}}{3^{12}}$$

$$\displaystyle \dfrac{10^4}{(3^3)^4} \ \ vs. \ \ \dfrac{1}{3}$$

$$\displaystyle \bigg(\dfrac{10}{27}\bigg)^4 \ \ vs. \ \ \dfrac{1}{3}$$

$$\displaystyle \bigg(\dfrac{10}{27}\bigg)^4 \ < \bigg(\dfrac{10}{20}\bigg)^4 \ = \ \bigg(\dfrac{1}{2}\bigg)^4 \ = \ \dfrac{1}{16} \ < \ \dfrac{1}{3}$$

Therefore, $$\displaystyle \ \ 30^{1/30} \ > \ 10^{1/22}$$

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