Exponents, division, Algebra 2

[peaches]

This is a challenged question. My teacher said the answer is 3[sup:21po5swm]sqrt2[/sup:21po5swm] but the book has been know to be wrong and he said he would work them tonight. I'm confused.
Question:
1/(9[sup:21po5swm](sqrt 2)[/sup:21po5swm])

This is what I've done, but I think it's wrong:
1/(9[sup:21po5swm](sqrt 2)[/sup:21po5swm])
1/((3[sup:21po5swm]2[/sup:21po5swm])[sup:21po5swm]sqrt 2[/sup:21po5swm])

Help!!!!

 

[chrisr]

\(\displaystyle Since\ \sqrt{2}>1,\ 3^{\sqrt{2}}>3.\)

\(\displaystyle 9^{\sqrt{2}}>9,\ so\ \frac{1}{9^{\sqrt{2}}}<\frac{1}{9}\)

I think you've left out some information.

 

[soroban]

Hello, peaches!

This is a challenge question.
\(\displaystyle \text{My teacher said the answer is: }\: 3^{\sqrt{12}}\) . . . . This is incredibly wrong! **

\(\displaystyle \text{Question: }\;\frac{1}{9^{\sqrt{2}}}\)

This is what I've done, but I think it's wrong:

. . \(\displaystyle \frac{1}{9^{\sqrt{2}}} \;=\;\frac{1}{(3^2)^{\sqrt{2}}}\)

Your work looks good . . .

We can go a few more steps . . .

. . \(\displaystyle \frac{1}{(3^2)^{\sqrt{2}}} \;=\;\frac{1}{3^{2\sqrt{2}}} \;=\;3^{-2\sqrt{2}}\)


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**

[1] We would never leave it as \(\displaystyle \sqrt{12}\)
. . .We would simplify it: .\(\displaystyle 2\sqrt{3}\)

[2] It is not \(\displaystyle \sqrt{12} \,=\,2\sqrt{3}\)
. . .Instead, it is \(\displaystyle 2\sqrt{2}\)

[3] Since the "action" is in the denomiator,
. . .his answer should have had a negative exponent.


Other than that, he's correct.
Is your teacher's first name "Forrest"?

 

[peaches]

thank you sooooooo much!!!! Oh, my teacher's name is Mr. Coen (I think his first name might be Kendall, but I'm not sure) Once, again, Thanks!!!!

 

[chrisr]

\(\displaystyle \frac{1}{9^{\sqrt{2}}}=\frac{1}{3^{2\sqrt{2}}}=\frac{1}{3^{\sqrt{2}\sqrt{2}\sqrt{2}}}=\frac{1}{3^{\sqrt{2}^3}}=3^{-\sqrt{2}^3}\)

is as close as i can get to the original \(\displaystyle 3^{\sqrt{2}\)

i dunno where the 12 is coming from, what am i missing?