- Thread starter greg1313
- Start date

If you do that, then you will have 10^(15/11) versus 30. What would be your nextHint: try raising both expressions to the 30th power and see which is larger.

step? I have it as a challenge problem, which means I'm challenging others.

My strategy involves getting rid of fractional exponents for both sides at the outset.

\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}\(\displaystyle \sqrt[22]{10} \ \ \ or \ \ \ \sqrt[30]{30}\)

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.

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It looks like you're missing a step or two between these two.\(\displaystyle \frac{\sqrt[30]{10}\sqrt[30]{3}}{\sqrt[22]{10}} = \frac{\sqrt[150]{243}}{\sqrt[165]{100}} \)

No, I don't see how you have justified this part of it.\(\displaystyle \frac{\sqrt[150]{243}}{\sqrt[165]{100}} > 1 \)

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

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

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\(\displaystyle \frac{\sqrt[30]{10}}{\sqrt[22]{10}} = 10^{\frac{1}{30}-\frac{1}{22}} = 10^{-\frac{2}{165}} = 100^{-\frac{1}{165}}\)

\(\displaystyle \sqrt[30]{3} = \left(243^{\frac{1}{5}}\right)^{\frac{1}{30}} = 243^{\frac{1}{150}}\)

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).

\(\displaystyle \sqrt[30]{3} = \left(243^{\frac{1}{5}}\right)^{\frac{1}{30}} = 243^{\frac{1}{150}}\)

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).

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