Question about changing roots to powers

[e.spyck]

Amateur here..

I am trying to change this fraction to a number with powers. According to the book, solution should be [7][4/5]. I always arrive at [7][/3/5]. What am I doing wrong?
Thanks for your help!

 

[Cubist]

It seems that the question is asking you to show the following...

[math] \frac{7}{\sqrt[5]{7}} = 7^{\frac{4}{5}} [/math]

But on your first 2 lines you imply...

[math] \frac{7}{\sqrt[5]{7}}=\sqrt[5]{\frac{\color{red}7\color{black}}{7}}[/math] but this isn't correct.


Think about the numerator only, what value of "x" would satisfy...

[math] 7=\sqrt[5]{7^x}[/math]

EDIT: When you find x, then the following progression would be correct...

[math] \frac{7}{\sqrt[5]{7}}= \frac{\sqrt[5]{7^x}}{\sqrt[5]{7}} = \sqrt[5]{\frac{7^x}{7}}[/math]... and continue

 

[HallsofIvy]

I would do it the other way around:
\(\displaystyle \frac{7}{\sqrt[5]{7}}= \frac{7^1}{7^{1/5}}= 7^{1-1/5}= 7^{5/5- 1/5}\).

\(\displaystyle \sqrt[n]{a}= a^{1/n}\) and \(\displaystyle \frac{1}{a^n}= a^{-n}\).

 

[JeffM]

Or you could it do your way but correctly

[MATH]\dfrac{7}{\sqrt[5]{7}} = \dfrac{\sqrt[5]{7^5}}{\sqrt[5]{7}} =[/MATH]
[MATH]\sqrt[5]{\dfrac{7^5}{7}} = \sqrt[5]{7^4} = 7^{4/5}.[/MATH]