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












0












$begingroup$


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











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 8:25










  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 8:35






  • 4




    $begingroup$
    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".
    $endgroup$
    – StackTD
    Jan 4 at 8:42








  • 1




    $begingroup$
    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.
    $endgroup$
    – KM101
    Jan 4 at 8:46






  • 1




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


















0












$begingroup$


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











$endgroup$












  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 8:25










  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 8:35






  • 4




    $begingroup$
    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".
    $endgroup$
    – StackTD
    Jan 4 at 8:42








  • 1




    $begingroup$
    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.
    $endgroup$
    – KM101
    Jan 4 at 8:46






  • 1




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
















0












0








0





$begingroup$


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











$endgroup$




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















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 8:27









Eevee Trainer

7,23821337




7,23821337










asked Jan 4 at 8:09









Nick29Nick29

31




31












  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 8:25










  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 8:35






  • 4




    $begingroup$
    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".
    $endgroup$
    – StackTD
    Jan 4 at 8:42








  • 1




    $begingroup$
    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.
    $endgroup$
    – KM101
    Jan 4 at 8:46






  • 1




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




















  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 8:25










  • $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    Jan 4 at 8:35






  • 4




    $begingroup$
    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".
    $endgroup$
    – StackTD
    Jan 4 at 8:42








  • 1




    $begingroup$
    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.
    $endgroup$
    – KM101
    Jan 4 at 8:46






  • 1




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


















$begingroup$
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.
$endgroup$
– Eevee Trainer
Jan 4 at 8:25




$begingroup$
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.
$endgroup$
– Eevee Trainer
Jan 4 at 8:25












$begingroup$
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.
$endgroup$
– Eevee Trainer
Jan 4 at 8:35




$begingroup$
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.
$endgroup$
– Eevee Trainer
Jan 4 at 8:35




4




4




$begingroup$
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".
$endgroup$
– StackTD
Jan 4 at 8:42






$begingroup$
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".
$endgroup$
– StackTD
Jan 4 at 8:42






1




1




$begingroup$
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.
$endgroup$
– KM101
Jan 4 at 8:46




$begingroup$
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.
$endgroup$
– KM101
Jan 4 at 8:46




1




1




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






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












2 Answers
2






active

oldest

votes


















3












$begingroup$


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









$endgroup$













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



















0












$begingroup$

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











$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061413%2fam-i-doing-this-right-2-log-2x-log-21-x-3%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$


    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









    $endgroup$













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
















    3












    $begingroup$


    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









    $endgroup$













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














    3












    3








    3





    $begingroup$


    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









    $endgroup$




    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

    22.9k2152




    22.9k2152












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


















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
















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




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











    0












    $begingroup$

    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











    $endgroup$


















      0












      $begingroup$

      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











      $endgroup$
















        0












        0








        0





        $begingroup$

        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











        $endgroup$



        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

        22k31340




        22k31340






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061413%2fam-i-doing-this-right-2-log-2x-log-21-x-3%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more