Am I doing this right? $2 log_2(x)- log_2(1)-x=3$












0














I know that the logairhtm with base anything of $1$ is $0$, which puts me at $2 log_2(x)-x=3$. Also, I realize $log_2(x)^2=2 log_2(x)$. When it comes to the $x$ standing alone, I also said that $x=log_2(2^x)$. Similarly, I know that $log_2(8)=3$. However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.






algebra-precalculus logarithms







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  • Are you certain you copied the question correctly? Wolfram Alpha (wolframalpha.com/input/?i=solve+for+x+2+log_2(x)+-+x+%3D+3) gives an answer which involves the Lambert $W$ function, which I suppose is material well beyond your level (if my assumption you're in high school is correct). It might also be worth noting if you want to solve for $x$, or are just seeking specific solutions.
    – Eevee Trainer
    Jan 4 at 8:25










  • Minor note. Noticed that Wolfram interpreted my input wrong in my previous comment. Correct input/inverse - wolframalpha.com/input/?i=solve+2+log_2(x)+-+x+%3D+3. Still has the $W$ function though, so my points from my previous comment remain the same despite me being a derp.
    – Eevee Trainer
    Jan 4 at 8:35






  • 4




    Perhaps it should've been $2 log_2(x)- log_2color{blue}{(1-x)}=3$...? That would make more sense if you're supposed to be able to solve it "by hand".
    – StackTD
    Jan 4 at 8:42








  • 1




    The question is probably the one StackTD mentioned. Perhaps you can check the question again and make sure it’s what you’ve mentioned. If so, it may be a typo or something.
    – KM101
    Jan 4 at 8:46






  • 1




    Probably a typo, because $log_2(1) = 0$ anyway.
    – Math_QED
    Jan 4 at 8:58


















0














I know that the logairhtm with base anything of $1$ is $0$, which puts me at $2 log_2(x)-x=3$. Also, I realize $log_2(x)^2=2 log_2(x)$. When it comes to the $x$ standing alone, I also said that $x=log_2(2^x)$. Similarly, I know that $log_2(8)=3$. However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.






algebra-precalculus logarithms







share|cite|improve this question









New contributor




Nick29 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Are you certain you copied the question correctly? Wolfram Alpha (wolframalpha.com/input/?i=solve+for+x+2+log_2(x)+-+x+%3D+3) gives an answer which involves the Lambert $W$ function, which I suppose is material well beyond your level (if my assumption you're in high school is correct). It might also be worth noting if you want to solve for $x$, or are just seeking specific solutions.
    – Eevee Trainer
    Jan 4 at 8:25










  • Minor note. Noticed that Wolfram interpreted my input wrong in my previous comment. Correct input/inverse - wolframalpha.com/input/?i=solve+2+log_2(x)+-+x+%3D+3. Still has the $W$ function though, so my points from my previous comment remain the same despite me being a derp.
    – Eevee Trainer
    Jan 4 at 8:35






  • 4




    Perhaps it should've been $2 log_2(x)- log_2color{blue}{(1-x)}=3$...? That would make more sense if you're supposed to be able to solve it "by hand".
    – StackTD
    Jan 4 at 8:42








  • 1




    The question is probably the one StackTD mentioned. Perhaps you can check the question again and make sure it’s what you’ve mentioned. If so, it may be a typo or something.
    – KM101
    Jan 4 at 8:46






  • 1




    Probably a typo, because $log_2(1) = 0$ anyway.
    – Math_QED
    Jan 4 at 8:58
















0












0








0







I know that the logairhtm with base anything of $1$ is $0$, which puts me at $2 log_2(x)-x=3$. Also, I realize $log_2(x)^2=2 log_2(x)$. When it comes to the $x$ standing alone, I also said that $x=log_2(2^x)$. Similarly, I know that $log_2(8)=3$. However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.






algebra-precalculus logarithms







share|cite|improve this question









New contributor




Nick29 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I know that the logairhtm with base anything of $1$ is $0$, which puts me at $2 log_2(x)-x=3$. Also, I realize $log_2(x)^2=2 log_2(x)$. When it comes to the $x$ standing alone, I also said that $x=log_2(2^x)$. Similarly, I know that $log_2(8)=3$. However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.






algebra-precalculus logarithms




algebra-precalculus logarithms






share|cite|improve this question









New contributor




Nick29 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









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Nick29 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 8:27









Eevee Trainer

5,0211734




5,0211734






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asked Jan 4 at 8:09









Nick29Nick29

81




81




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Nick29 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.












  • Are you certain you copied the question correctly? Wolfram Alpha (wolframalpha.com/input/?i=solve+for+x+2+log_2(x)+-+x+%3D+3) gives an answer which involves the Lambert $W$ function, which I suppose is material well beyond your level (if my assumption you're in high school is correct). It might also be worth noting if you want to solve for $x$, or are just seeking specific solutions.
    – Eevee Trainer
    Jan 4 at 8:25










  • Minor note. Noticed that Wolfram interpreted my input wrong in my previous comment. Correct input/inverse - wolframalpha.com/input/?i=solve+2+log_2(x)+-+x+%3D+3. Still has the $W$ function though, so my points from my previous comment remain the same despite me being a derp.
    – Eevee Trainer
    Jan 4 at 8:35






  • 4




    Perhaps it should've been $2 log_2(x)- log_2color{blue}{(1-x)}=3$...? That would make more sense if you're supposed to be able to solve it "by hand".
    – StackTD
    Jan 4 at 8:42








  • 1




    The question is probably the one StackTD mentioned. Perhaps you can check the question again and make sure it’s what you’ve mentioned. If so, it may be a typo or something.
    – KM101
    Jan 4 at 8:46






  • 1




    Probably a typo, because $log_2(1) = 0$ anyway.
    – Math_QED
    Jan 4 at 8:58




















  • Are you certain you copied the question correctly? Wolfram Alpha (wolframalpha.com/input/?i=solve+for+x+2+log_2(x)+-+x+%3D+3) gives an answer which involves the Lambert $W$ function, which I suppose is material well beyond your level (if my assumption you're in high school is correct). It might also be worth noting if you want to solve for $x$, or are just seeking specific solutions.
    – Eevee Trainer
    Jan 4 at 8:25










  • Minor note. Noticed that Wolfram interpreted my input wrong in my previous comment. Correct input/inverse - wolframalpha.com/input/?i=solve+2+log_2(x)+-+x+%3D+3. Still has the $W$ function though, so my points from my previous comment remain the same despite me being a derp.
    – Eevee Trainer
    Jan 4 at 8:35






  • 4




    Perhaps it should've been $2 log_2(x)- log_2color{blue}{(1-x)}=3$...? That would make more sense if you're supposed to be able to solve it "by hand".
    – StackTD
    Jan 4 at 8:42








  • 1




    The question is probably the one StackTD mentioned. Perhaps you can check the question again and make sure it’s what you’ve mentioned. If so, it may be a typo or something.
    – KM101
    Jan 4 at 8:46






  • 1




    Probably a typo, because $log_2(1) = 0$ anyway.
    – Math_QED
    Jan 4 at 8:58


















Are you certain you copied the question correctly? Wolfram Alpha (wolframalpha.com/input/?i=solve+for+x+2+log_2(x)+-+x+%3D+3) gives an answer which involves the Lambert $W$ function, which I suppose is material well beyond your level (if my assumption you're in high school is correct). It might also be worth noting if you want to solve for $x$, or are just seeking specific solutions.
– Eevee Trainer
Jan 4 at 8:25




Are you certain you copied the question correctly? Wolfram Alpha (wolframalpha.com/input/?i=solve+for+x+2+log_2(x)+-+x+%3D+3) gives an answer which involves the Lambert $W$ function, which I suppose is material well beyond your level (if my assumption you're in high school is correct). It might also be worth noting if you want to solve for $x$, or are just seeking specific solutions.
– Eevee Trainer
Jan 4 at 8:25












Minor note. Noticed that Wolfram interpreted my input wrong in my previous comment. Correct input/inverse - wolframalpha.com/input/?i=solve+2+log_2(x)+-+x+%3D+3. Still has the $W$ function though, so my points from my previous comment remain the same despite me being a derp.
– Eevee Trainer
Jan 4 at 8:35




Minor note. Noticed that Wolfram interpreted my input wrong in my previous comment. Correct input/inverse - wolframalpha.com/input/?i=solve+2+log_2(x)+-+x+%3D+3. Still has the $W$ function though, so my points from my previous comment remain the same despite me being a derp.
– Eevee Trainer
Jan 4 at 8:35




4




4




Perhaps it should've been $2 log_2(x)- log_2color{blue}{(1-x)}=3$...? That would make more sense if you're supposed to be able to solve it "by hand".
– StackTD
Jan 4 at 8:42






Perhaps it should've been $2 log_2(x)- log_2color{blue}{(1-x)}=3$...? That would make more sense if you're supposed to be able to solve it "by hand".
– StackTD
Jan 4 at 8:42






1




1




The question is probably the one StackTD mentioned. Perhaps you can check the question again and make sure it’s what you’ve mentioned. If so, it may be a typo or something.
– KM101
Jan 4 at 8:46




The question is probably the one StackTD mentioned. Perhaps you can check the question again and make sure it’s what you’ve mentioned. If so, it may be a typo or something.
– KM101
Jan 4 at 8:46




1




1




Probably a typo, because $log_2(1) = 0$ anyway.
– Math_QED
Jan 4 at 8:58






Probably a typo, because $log_2(1) = 0$ anyway.
– Math_QED
Jan 4 at 8:58












2 Answers
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3















However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.




As pointed out in the comments, it's no surprise you get stuck because the equation
$$2 log_2(x)- log_2(1)-x=3$$
doesn't have a "nice" solution which you can easily find manually. If you just began learning about logarithms in an algebra-precalculus context, it's very unlikely that the equation was supposed to be like this.



So there is probably an error in the equation, or you miscopied or misinterpreted a part. My best guess is you (or the question setter) missed brackets and that it was supposed to be:
$$2 log_2(x)- log_2color{blue}{(1-x)}=3$$



Perhaps you can take it from there already and if not, here's a start:




$$log_2(x^2)- log_2color{blue}{(1-x)}=3ifflog_2left(frac{x^2}{1-x}right)=3iff ldots$$







share|cite|improve this answer





















  • Thank you so much. The source I was looking at for this problem did not even have parenthesis around any of the arguments.
    – Nick29
    Jan 5 at 4:59



















0














Well, solving a more general problem:



$$text{n}cdotlog_text{n}left(xright)-log_text{n}left(text{n}-1right)-x=1+text{n}tag1$$



Now, we know that:




  • $$log_alphaleft(betaright)=frac{lnleft(betaright)}{lnleft(alpharight)}tag2$$

  • $$alphacdotlnleft(betaright)=lnleft(beta^alpharight)tag3$$

  • $$gamma=log_alphaleft(alpha^gammaright)tag4$$

  • $$lnleft(alpharight)+lnleft(betaright)=lnleft(alphacdotbetaright)tag5$$

  • $$lnleft(alpharight)-lnleft(betaright)=lnleft(frac{alpha}{beta}right)tag6$$


So, we can write:



$$frac{lnleft(x^text{n}right)}{lnleft(text{n}right)}-frac{lnleft(text{n}-1right)}{lnleft(text{n}right)}-frac{lnleft(text{n}^xright)}{lnleft(text{n}right)}=frac{lnleft(text{n}^{1+text{n}}right)}{lnleft(text{n}right)}tag7$$



Now, formula $(7)$ gives (when we assume $lnleft(text{n}right)ne0$):



$$lnleft(x^text{n}right)-lnleft(text{n}-1right)-lnleft(text{n}^xright)=lnleft(text{n}^{1+text{n}}right)tag8$$



Now, using rule $(5)$ and $(6)$ we can write:



$$lnleft(frac{x^text{n}}{text{n}^xcdotleft(text{n}-1right)}right)=lnleft(text{n}^{1+text{n}}right)spaceLongleftrightarrowspacefrac{x^text{n}}{text{n}^xcdotleft(text{n}-1right)}=text{n}^{1+text{n}}spaceLongleftrightarrowspace$$
$$frac{x^text{n}}{text{n}^x}=text{n}^{1+text{n}}cdotleft(text{n}-1right)tag9$$






share|cite|improve this answer























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    2 Answers
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    2 Answers
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    3















    However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.




    As pointed out in the comments, it's no surprise you get stuck because the equation
    $$2 log_2(x)- log_2(1)-x=3$$
    doesn't have a "nice" solution which you can easily find manually. If you just began learning about logarithms in an algebra-precalculus context, it's very unlikely that the equation was supposed to be like this.



    So there is probably an error in the equation, or you miscopied or misinterpreted a part. My best guess is you (or the question setter) missed brackets and that it was supposed to be:
    $$2 log_2(x)- log_2color{blue}{(1-x)}=3$$



    Perhaps you can take it from there already and if not, here's a start:




    $$log_2(x^2)- log_2color{blue}{(1-x)}=3ifflog_2left(frac{x^2}{1-x}right)=3iff ldots$$







    share|cite|improve this answer





















    • Thank you so much. The source I was looking at for this problem did not even have parenthesis around any of the arguments.
      – Nick29
      Jan 5 at 4:59
















    3















    However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.




    As pointed out in the comments, it's no surprise you get stuck because the equation
    $$2 log_2(x)- log_2(1)-x=3$$
    doesn't have a "nice" solution which you can easily find manually. If you just began learning about logarithms in an algebra-precalculus context, it's very unlikely that the equation was supposed to be like this.



    So there is probably an error in the equation, or you miscopied or misinterpreted a part. My best guess is you (or the question setter) missed brackets and that it was supposed to be:
    $$2 log_2(x)- log_2color{blue}{(1-x)}=3$$



    Perhaps you can take it from there already and if not, here's a start:




    $$log_2(x^2)- log_2color{blue}{(1-x)}=3ifflog_2left(frac{x^2}{1-x}right)=3iff ldots$$







    share|cite|improve this answer





















    • Thank you so much. The source I was looking at for this problem did not even have parenthesis around any of the arguments.
      – Nick29
      Jan 5 at 4:59














    3












    3








    3







    However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.




    As pointed out in the comments, it's no surprise you get stuck because the equation
    $$2 log_2(x)- log_2(1)-x=3$$
    doesn't have a "nice" solution which you can easily find manually. If you just began learning about logarithms in an algebra-precalculus context, it's very unlikely that the equation was supposed to be like this.



    So there is probably an error in the equation, or you miscopied or misinterpreted a part. My best guess is you (or the question setter) missed brackets and that it was supposed to be:
    $$2 log_2(x)- log_2color{blue}{(1-x)}=3$$



    Perhaps you can take it from there already and if not, here's a start:




    $$log_2(x^2)- log_2color{blue}{(1-x)}=3ifflog_2left(frac{x^2}{1-x}right)=3iff ldots$$







    share|cite|improve this answer













    However, I end up with $x^2/2^x=8$. Since the bases aren't the same, I can't move forward.




    As pointed out in the comments, it's no surprise you get stuck because the equation
    $$2 log_2(x)- log_2(1)-x=3$$
    doesn't have a "nice" solution which you can easily find manually. If you just began learning about logarithms in an algebra-precalculus context, it's very unlikely that the equation was supposed to be like this.



    So there is probably an error in the equation, or you miscopied or misinterpreted a part. My best guess is you (or the question setter) missed brackets and that it was supposed to be:
    $$2 log_2(x)- log_2color{blue}{(1-x)}=3$$



    Perhaps you can take it from there already and if not, here's a start:




    $$log_2(x^2)- log_2color{blue}{(1-x)}=3ifflog_2left(frac{x^2}{1-x}right)=3iff ldots$$








    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 4 at 8:50









    StackTDStackTD

    22.3k2049




    22.3k2049












    • Thank you so much. The source I was looking at for this problem did not even have parenthesis around any of the arguments.
      – Nick29
      Jan 5 at 4:59


















    • Thank you so much. The source I was looking at for this problem did not even have parenthesis around any of the arguments.
      – Nick29
      Jan 5 at 4:59
















    Thank you so much. The source I was looking at for this problem did not even have parenthesis around any of the arguments.
    – Nick29
    Jan 5 at 4:59




    Thank you so much. The source I was looking at for this problem did not even have parenthesis around any of the arguments.
    – Nick29
    Jan 5 at 4:59











    0














    Well, solving a more general problem:



    $$text{n}cdotlog_text{n}left(xright)-log_text{n}left(text{n}-1right)-x=1+text{n}tag1$$



    Now, we know that:




    • $$log_alphaleft(betaright)=frac{lnleft(betaright)}{lnleft(alpharight)}tag2$$

    • $$alphacdotlnleft(betaright)=lnleft(beta^alpharight)tag3$$

    • $$gamma=log_alphaleft(alpha^gammaright)tag4$$

    • $$lnleft(alpharight)+lnleft(betaright)=lnleft(alphacdotbetaright)tag5$$

    • $$lnleft(alpharight)-lnleft(betaright)=lnleft(frac{alpha}{beta}right)tag6$$


    So, we can write:



    $$frac{lnleft(x^text{n}right)}{lnleft(text{n}right)}-frac{lnleft(text{n}-1right)}{lnleft(text{n}right)}-frac{lnleft(text{n}^xright)}{lnleft(text{n}right)}=frac{lnleft(text{n}^{1+text{n}}right)}{lnleft(text{n}right)}tag7$$



    Now, formula $(7)$ gives (when we assume $lnleft(text{n}right)ne0$):



    $$lnleft(x^text{n}right)-lnleft(text{n}-1right)-lnleft(text{n}^xright)=lnleft(text{n}^{1+text{n}}right)tag8$$



    Now, using rule $(5)$ and $(6)$ we can write:



    $$lnleft(frac{x^text{n}}{text{n}^xcdotleft(text{n}-1right)}right)=lnleft(text{n}^{1+text{n}}right)spaceLongleftrightarrowspacefrac{x^text{n}}{text{n}^xcdotleft(text{n}-1right)}=text{n}^{1+text{n}}spaceLongleftrightarrowspace$$
    $$frac{x^text{n}}{text{n}^x}=text{n}^{1+text{n}}cdotleft(text{n}-1right)tag9$$






    share|cite|improve this answer




























      0














      Well, solving a more general problem:



      $$text{n}cdotlog_text{n}left(xright)-log_text{n}left(text{n}-1right)-x=1+text{n}tag1$$



      Now, we know that:




      • $$log_alphaleft(betaright)=frac{lnleft(betaright)}{lnleft(alpharight)}tag2$$

      • $$alphacdotlnleft(betaright)=lnleft(beta^alpharight)tag3$$

      • $$gamma=log_alphaleft(alpha^gammaright)tag4$$

      • $$lnleft(alpharight)+lnleft(betaright)=lnleft(alphacdotbetaright)tag5$$

      • $$lnleft(alpharight)-lnleft(betaright)=lnleft(frac{alpha}{beta}right)tag6$$


      So, we can write:



      $$frac{lnleft(x^text{n}right)}{lnleft(text{n}right)}-frac{lnleft(text{n}-1right)}{lnleft(text{n}right)}-frac{lnleft(text{n}^xright)}{lnleft(text{n}right)}=frac{lnleft(text{n}^{1+text{n}}right)}{lnleft(text{n}right)}tag7$$



      Now, formula $(7)$ gives (when we assume $lnleft(text{n}right)ne0$):



      $$lnleft(x^text{n}right)-lnleft(text{n}-1right)-lnleft(text{n}^xright)=lnleft(text{n}^{1+text{n}}right)tag8$$



      Now, using rule $(5)$ and $(6)$ we can write:



      $$lnleft(frac{x^text{n}}{text{n}^xcdotleft(text{n}-1right)}right)=lnleft(text{n}^{1+text{n}}right)spaceLongleftrightarrowspacefrac{x^text{n}}{text{n}^xcdotleft(text{n}-1right)}=text{n}^{1+text{n}}spaceLongleftrightarrowspace$$
      $$frac{x^text{n}}{text{n}^x}=text{n}^{1+text{n}}cdotleft(text{n}-1right)tag9$$






      share|cite|improve this answer


























        0












        0








        0






        Well, solving a more general problem:



        $$text{n}cdotlog_text{n}left(xright)-log_text{n}left(text{n}-1right)-x=1+text{n}tag1$$



        Now, we know that:




        • $$log_alphaleft(betaright)=frac{lnleft(betaright)}{lnleft(alpharight)}tag2$$

        • $$alphacdotlnleft(betaright)=lnleft(beta^alpharight)tag3$$

        • $$gamma=log_alphaleft(alpha^gammaright)tag4$$

        • $$lnleft(alpharight)+lnleft(betaright)=lnleft(alphacdotbetaright)tag5$$

        • $$lnleft(alpharight)-lnleft(betaright)=lnleft(frac{alpha}{beta}right)tag6$$


        So, we can write:



        $$frac{lnleft(x^text{n}right)}{lnleft(text{n}right)}-frac{lnleft(text{n}-1right)}{lnleft(text{n}right)}-frac{lnleft(text{n}^xright)}{lnleft(text{n}right)}=frac{lnleft(text{n}^{1+text{n}}right)}{lnleft(text{n}right)}tag7$$



        Now, formula $(7)$ gives (when we assume $lnleft(text{n}right)ne0$):



        $$lnleft(x^text{n}right)-lnleft(text{n}-1right)-lnleft(text{n}^xright)=lnleft(text{n}^{1+text{n}}right)tag8$$



        Now, using rule $(5)$ and $(6)$ we can write:



        $$lnleft(frac{x^text{n}}{text{n}^xcdotleft(text{n}-1right)}right)=lnleft(text{n}^{1+text{n}}right)spaceLongleftrightarrowspacefrac{x^text{n}}{text{n}^xcdotleft(text{n}-1right)}=text{n}^{1+text{n}}spaceLongleftrightarrowspace$$
        $$frac{x^text{n}}{text{n}^x}=text{n}^{1+text{n}}cdotleft(text{n}-1right)tag9$$






        share|cite|improve this answer














        Well, solving a more general problem:



        $$text{n}cdotlog_text{n}left(xright)-log_text{n}left(text{n}-1right)-x=1+text{n}tag1$$



        Now, we know that:




        • $$log_alphaleft(betaright)=frac{lnleft(betaright)}{lnleft(alpharight)}tag2$$

        • $$alphacdotlnleft(betaright)=lnleft(beta^alpharight)tag3$$

        • $$gamma=log_alphaleft(alpha^gammaright)tag4$$

        • $$lnleft(alpharight)+lnleft(betaright)=lnleft(alphacdotbetaright)tag5$$

        • $$lnleft(alpharight)-lnleft(betaright)=lnleft(frac{alpha}{beta}right)tag6$$


        So, we can write:



        $$frac{lnleft(x^text{n}right)}{lnleft(text{n}right)}-frac{lnleft(text{n}-1right)}{lnleft(text{n}right)}-frac{lnleft(text{n}^xright)}{lnleft(text{n}right)}=frac{lnleft(text{n}^{1+text{n}}right)}{lnleft(text{n}right)}tag7$$



        Now, formula $(7)$ gives (when we assume $lnleft(text{n}right)ne0$):



        $$lnleft(x^text{n}right)-lnleft(text{n}-1right)-lnleft(text{n}^xright)=lnleft(text{n}^{1+text{n}}right)tag8$$



        Now, using rule $(5)$ and $(6)$ we can write:



        $$lnleft(frac{x^text{n}}{text{n}^xcdotleft(text{n}-1right)}right)=lnleft(text{n}^{1+text{n}}right)spaceLongleftrightarrowspacefrac{x^text{n}}{text{n}^xcdotleft(text{n}-1right)}=text{n}^{1+text{n}}spaceLongleftrightarrowspace$$
        $$frac{x^text{n}}{text{n}^x}=text{n}^{1+text{n}}cdotleft(text{n}-1right)tag9$$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 4 at 10:47

























        answered Jan 4 at 10:42









        JanJan

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