An exercise about rationalizing denominators

[Nekomimi]

Homework Statement

Described below.

Relevant Equations

None.

Express the following as a fraction with rational denominator: $$\frac{5^{\frac{1}{3}}}{5^{\frac{5}{3}}}$$

If I try to start by multiplicating both the numerator and denominator by $5^{-\frac{2}{3}}$, I get:

$$\begin{align}
\nonumber \frac{5^{\frac{1}{3}}}{5^{\frac{5}{3}}} & = \frac{5^{\frac{1}{3}}}{5^{\frac{5}{3}}} \times \frac{5^{-\frac{2}{3}}}{5^{-\frac{2}{3}}} = \frac{5^{-\frac{1}{3}}}{5} \\
\end{align}$$

Which apparently makes me stuck because that negative exponent will make the radical go back to the denominator anyway, and I can't seem to get it out of there. Any help?

 

[symbolipoint]

I may be missing the point about rationalizing denominator, but given expression is also 5^(1/3-5/3)
= 5^(-4/3)
= 1/(5^(4/3))

Rationalizing that might be easier, involving cube root of 5. Your step here may be to use (5^(1/3)*5^(1/3))/(5^(1/3)*5^(1/3))

 

[Mark44]

Problem Statement: Described below.
Relevant Equations: None.

If I try to start by multiplicating both the numerator and denominator by $5^{-\frac{2}{3}}$, I get:

$$\begin{align}
\nonumber \frac{5^{\frac{1}{3}}}{5^{\frac{5}{3}}} & = \frac{5^{\frac{1}{3}}}{5^{\frac{5}{3}}} \times \frac{5^{-\frac{2}{3}}}{5^{-\frac{2}{3}}} = \frac{5^{-\frac{1}{3}}}{5} \\
\end{align}$$

Here's another thought -- do some simplification first before attempting to make the denominator rational.

$$\frac{5^{1/3}}{5^{5/3}} = \frac{5^{1/3}}{5 \cdot 5^{2/3}}$$
Now multiply numerator and denominator by $5^{1/3}$ over itself.

BTW, the word is "multiplying," not multiplicating, which isn't a word in English

 

[Nekomimi]

Here's another thought -- do some simplification first before attempting to make the denominator rational.

$$\frac{5^{1/3}}{5^{5/3}} = \frac{5^{1/3}}{5 \cdot 5^{2/3}}$$
Now multiply numerator and denominator by $5^{1/3}$ over itself.

Right, so

$$\frac{5^{1/3}}{5 \cdot 5^{2/3}} \cdot \frac{5^{\frac{1}{3}}}{5^{\frac{1}{3}}} = \frac{5^{\frac{2}{3}}}{5 \cdot 5} = \frac{5^{\frac{2}{3}}}{25}$$

Is that right?

BTW, the word is "multiplying," not multiplicating, which isn't a word in English.

I'm sorry for the mistake, English is not my first language.

 

[symbolipoint]

Watch carefully (also because my typesetting skill is not very good) that:

5^(1/3)*5^(1/3)*5^(1/3)=5^(1/3+1/3+1/3)=5^1=5

 

[Nekomimi]

Watch carefully (also because my typesetting skill is not very good) that:

5^(1/3)*5^(1/3)*5^(1/3)=5^(1/3+1/3+1/3)=5^1=5

I don't understand where I should do that.
Was my last try incorrect?

 

[symbolipoint]

I agree with post #3 and #4, since I just now worked the problem on paper. I did the work two different ways.

 

[Nekomimi]

I agree with post #3 and #4, since I just now worked the problem on paper. I did the work two different ways.

Oh, I think I get it now. Your tip was for solving it through the way I initially started with? (That is, by not simplifying $5^{\frac{5}{3}}$.)

Well then, let's see:

$$ \frac{5^{-\frac{1}{3}}}{5} = \frac{1}{5 \cdot 5^{\frac{1}{3}}} = \frac{1}{5 \cdot 5^{\frac{1}{3}}} \cdot \frac{5^{\frac{2}{3}}}{5^{\frac{2}{3}}} = \frac{5^{\frac{2}{3}}}{25}$$

Is that right?

 

[Mark44]

I'm sorry for the mistake, English is not my first language.

Not a problem -- your English is very good. I was just letting you know for future reference.

Watch carefully (also because my typesetting skill is not very good) that:

5^(1/3)*5^(1/3)*5^(1/3)=5^(1/3+1/3+1/3)=5^1=5

This would be much more readable in LaTeX. Here's the same work in a more readable form:
$5^{1/3}*5^{1/3}*5^{1/3}=5^{1/3+1/3+1/3}=5^1=5$
All I did was add a pair of # symbols at each end and replace each parenthesis with a brace --{ or }. Take a look at our LaTeX primer -- https://www.physicsforums.com/help/latexhelp/

Well then, let's see:
$$ \frac{5^{-\frac{1}{3}}}{5} = \frac{1}{5 \cdot 5^{\frac{1}{3}}} = \frac{1}{5 \cdot 5^{\frac{1}{3}}} \cdot \frac{5^{\frac{2}{3}}}{5^{\frac{2}{3}}} = \frac{5^{\frac{2}{3}}}{25}$$

Yes, that's right. My only quibble is that you should start with the first fraction of the given problem, $\frac {5^{1/3}}{5^{5/3}}$. The step you don't show above is multiplying top and bottom by $5^{-1/3}$.

 

[Nekomimi]

Not a problem -- your English is very good. I was just letting you know for future reference.
This would be much more readable in LaTeX. Here's the same work in a more readable form:
$5^{1/3}*5^{1/3}*5^{1/3}=5^{1/3+1/3+1/3}=5^1=5$
All I did was add a pair of # symbols at each end and replace each parenthesis with a brace --{ or }. Take a look at our LaTeX primer -- https://www.physicsforums.com/help/latexhelp/
Yes, that's right. My only quibble is that you should start with the first fraction of the given problem, $\frac {5^{1/3}}{5^{5/3}}$. The step you don't show above is multiplying top and bottom by $5^{-1/3}$.

Don't you mean multiplying by $5^{-\frac{2}{3}}$? In any case, yes, I skipped that step because it was previously mentioned. I'll keep that in mind for the next time, though.

Since it is solved, is there a way to close this thread?

 

[Mark44]

Don't you mean multiplying by $5^{-\frac{2}{3}}$? In any case, yes, I skipped that step because it was previously mentioned. I'll keep that in mind for the next time, though.

Maybe what I wrote wasn't exactly what I meant.
Probably the easiest approach is what I suggested in post #3:
$\frac{5^{1/3}}{5^{5/3}} = \frac {5^{1/3}}{5 \cdot 5^{2/3}}$
Now, multiply top and bottom by $5^{1/3}$ to get a denominator of 25.

Since it is solved, is there a way to close this thread?

Not that I know of. There used to be, but we've updated to a new system that doesn't have that feature, I don't believe.

 

[Nekomimi]

Not that I know of. There used to be, but we've updated to a new system that doesn't have that feature, I don't believe.

OK. Thank you very much, anyway!