Why is $frac{15sqrt[4]{125}}{sqrt[4]{5}}$ $15sqrt{5}$ and not $15sqrt[4]{25}$?












1












$begingroup$


I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus







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  • 2




    $begingroup$
    The fourth root is the square root of the square root ...
    $endgroup$
    – Ethan Bolker
    Jan 3 at 16:54
















1












$begingroup$


I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    The fourth root is the square root of the square root ...
    $endgroup$
    – Ethan Bolker
    Jan 3 at 16:54














1












1








1





$begingroup$


I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus







share|cite|improve this question









$endgroup$




I have an expression I am to simplify:



$$frac{15sqrt[4]{125}}{sqrt[4]{5}}$$



I arrived at $15sqrt[4]{25}$. My textbook tells me that the answer is in fact $15sqrt{5}$. Here is my thought process:



$frac{15sqrt[4]{125}}{sqrt[4]{5}}$ = $frac{15*sqrt[4]{5}*sqrt[4]{25}}{sqrt[4]{5}}$ =



(cancel out $sqrt[4]{5}$ present in both numerator and denominator)
leaving:
$$15sqrt[4]{25}$$



Where did I go wrong and how can I arrive at $15sqrt{5}$?






algebra-precalculus




algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 3 at 16:52









Doug FirDoug Fir

4098




4098








  • 2




    $begingroup$
    The fourth root is the square root of the square root ...
    $endgroup$
    – Ethan Bolker
    Jan 3 at 16:54














  • 2




    $begingroup$
    The fourth root is the square root of the square root ...
    $endgroup$
    – Ethan Bolker
    Jan 3 at 16:54








2




2




$begingroup$
The fourth root is the square root of the square root ...
$endgroup$
– Ethan Bolker
Jan 3 at 16:54




$begingroup$
The fourth root is the square root of the square root ...
$endgroup$
– Ethan Bolker
Jan 3 at 16:54










3 Answers
3






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1












$begingroup$

$$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






share|cite|improve this answer









$endgroup$





















    2












    $begingroup$

    Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
        $endgroup$
        – Doug Fir
        Jan 3 at 17:09











      Your Answer





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      3 Answers
      3






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      3 Answers
      3






      active

      oldest

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      active

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      active

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      1












      $begingroup$

      $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



      Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



        Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



          Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.






          share|cite|improve this answer









          $endgroup$



          $$frac{sqrt[4] {125}}{sqrt[4] 5} = frac{sqrt[4] {5^3}}{sqrt[4] 5} = sqrt[4]{5^2} = 5^{frac{2}{4}} = 5^{frac{1}{2}} = sqrt 5$$



          Even quicker, $sqrt[4]{25}$ means $sqrt{sqrt{25}}$, which becomes $sqrt{5}$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 3 at 17:06









          KM101KM101

          6,0601525




          6,0601525























              2












              $begingroup$

              Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$






                  share|cite|improve this answer









                  $endgroup$



                  Write $$sqrt[4]{frac{125}{5}}=sqrt[4]{25}=sqrt{5}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 3 at 16:54









                  Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                  76.8k42866




                  76.8k42866























                      2












                      $begingroup$

                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        $endgroup$
                        – Doug Fir
                        Jan 3 at 17:09
















                      2












                      $begingroup$

                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






                      share|cite|improve this answer









                      $endgroup$













                      • $begingroup$
                        Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        $endgroup$
                        – Doug Fir
                        Jan 3 at 17:09














                      2












                      2








                      2





                      $begingroup$

                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.






                      share|cite|improve this answer









                      $endgroup$



                      It turns out that $sqrt[4]{25}=sqrt{5}$. This is because $25=5^2$, so that $sqrt[4]{25}=sqrt[4]{5^2}=(5^2)^{1/4}=5^{1/2}=sqrt{5}.$ So, you are correct, as is the book.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered Jan 3 at 16:55









                      Antonios-Alexandros RobotisAntonios-Alexandros Robotis

                      10.5k41641




                      10.5k41641












                      • $begingroup$
                        Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        $endgroup$
                        – Doug Fir
                        Jan 3 at 17:09


















                      • $begingroup$
                        Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                        $endgroup$
                        – Doug Fir
                        Jan 3 at 17:09
















                      $begingroup$
                      Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                      $endgroup$
                      – Doug Fir
                      Jan 3 at 17:09




                      $begingroup$
                      Thanks for helping me with your answer. I went with KM101 answer only because, for me, it just read slightly more intuitively between each step. At least for me.
                      $endgroup$
                      – Doug Fir
                      Jan 3 at 17:09


















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