An exercise about rationalizing denominators

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To express the fraction $$\frac{5^{\frac{1}{3}}}{5^{\frac{5}{3}}$$ with a rational denominator, participants discussed various methods, including simplifying the expression before rationalizing. One effective approach involves rewriting the expression as $$\frac{5^{1/3}}{5 \cdot 5^{2/3}}$$ and then multiplying both the numerator and denominator by $$5^{1/3}$$. This results in a final expression of $$\frac{5^{\frac{2}{3}}}{25}$$. The discussion also addressed minor language corrections and formatting suggestions for clarity. The problem was ultimately resolved, and participants expressed gratitude for the assistance.
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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?
 
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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))
 
Nekomimi said:
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
 
Mark44 said:
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?

Mark44 said:
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.
 
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
 
symbolipoint said:
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?
 
I agree with post #3 and #4, since I just now worked the problem on paper. I did the work two different ways.
 
symbolipoint said:
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?
 
Nekomimi said:
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.
symbolipoint said:
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/
Nekomimi said:
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}##.
 
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  • #10
Mark44 said:
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?
 
  • #11
Nekomimi said:
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.
Nekomimi said:
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.
 
  • #12
Mark44 said:
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!
 
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