Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$












0












$begingroup$



Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$




I equated their exponents,



That gave me $log_5 x = frac{8}{3}$



But the answer given in my book is $1$.



Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.



When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?



Any help would be appreciated.






functions logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does your calculator show if you plot the range $x=50..100?$
    $endgroup$
    – gammatester
    Dec 6 '18 at 16:36


















0












$begingroup$



Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$




I equated their exponents,



That gave me $log_5 x = frac{8}{3}$



But the answer given in my book is $1$.



Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.



When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?



Any help would be appreciated.






functions logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    What does your calculator show if you plot the range $x=50..100?$
    $endgroup$
    – gammatester
    Dec 6 '18 at 16:36
















0












0








0





$begingroup$



Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$




I equated their exponents,



That gave me $log_5 x = frac{8}{3}$



But the answer given in my book is $1$.



Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.



When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?



Any help would be appreciated.






functions logarithms







share|cite|improve this question











$endgroup$





Find the solution to $x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}$




I equated their exponents,



That gave me $log_5 x = frac{8}{3}$



But the answer given in my book is $1$.



Obviously, 1 satisfies the equation. But my question, how can I get 1 as a solution by actually solving it.



When I tried to graph the function, the graphing calculator showed just 1 as a solution. Why doesn't it show the solution that I have got as well?



Any help would be appreciated.






functions logarithms




functions logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 11:48









N. F. Taussig

43.9k93355




43.9k93355










asked Dec 6 '18 at 16:05









Piano LandPiano Land

379115




379115












  • $begingroup$
    What does your calculator show if you plot the range $x=50..100?$
    $endgroup$
    – gammatester
    Dec 6 '18 at 16:36




















  • $begingroup$
    What does your calculator show if you plot the range $x=50..100?$
    $endgroup$
    – gammatester
    Dec 6 '18 at 16:36


















$begingroup$
What does your calculator show if you plot the range $x=50..100?$
$endgroup$
– gammatester
Dec 6 '18 at 16:36






$begingroup$
What does your calculator show if you plot the range $x=50..100?$
$endgroup$
– gammatester
Dec 6 '18 at 16:36












5 Answers
5






active

oldest

votes


















1












$begingroup$

You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
$$
(log_5(x^2)+log_5x-12)log_5x=-4log_5x
$$
that becomes, setting $y=log_5x$,
$$
(3y-8)y=0
$$
so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.



On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
$$
(y^2+y-8)y=0
$$
with solutions
$$
y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    I suppose you were using
    $$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
    but it is only true when $x>0$ and $xne 1$.



    Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
      $endgroup$
      – Piano Land
      Dec 6 '18 at 16:25










    • $begingroup$
      I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
      $endgroup$
      – Eclipse Sun
      Dec 6 '18 at 16:28










    • $begingroup$
      Perhaps just because $x=5^{8/3}$ is too large.
      $endgroup$
      – Eclipse Sun
      Dec 6 '18 at 16:38



















    1












    $begingroup$

    We need



    $$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$



    that is



    $$x=5^frac83$$



    the other solution $x=1$ is obtained by inspection from the original equation.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @KM101 Opssss...thanks I fix!
      $endgroup$
      – gimusi
      Dec 6 '18 at 16:26










    • $begingroup$
      No problem! :-)
      $endgroup$
      – KM101
      Dec 6 '18 at 16:36





















    1












    $begingroup$

    If the book claims, that $1$ is the only real solution, it is wrong. Your value
    $$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      What you did (probably) was the following:



      $$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$



      By doing so, you removed the possibility of $x = 1$.



      As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.




      Your other solution is valid. Let $x = 5^{frac{8}{3}}$.



      $$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$



      On both sides, you get



      $$big(5^{frac{8}{3}}big)^{-4}$$






      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%2f3028682%2ffind-the-solution-to-x-log-5-x2-log-5-x-12-frac1x4%23new-answer', 'question_page');
        }
        );

        Post as a guest















        Required, but never shown

























        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
        $$
        (log_5(x^2)+log_5x-12)log_5x=-4log_5x
        $$
        that becomes, setting $y=log_5x$,
        $$
        (3y-8)y=0
        $$
        so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.



        On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
        $$
        (y^2+y-8)y=0
        $$
        with solutions
        $$
        y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
        $$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
          $$
          (log_5(x^2)+log_5x-12)log_5x=-4log_5x
          $$
          that becomes, setting $y=log_5x$,
          $$
          (3y-8)y=0
          $$
          so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.



          On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
          $$
          (y^2+y-8)y=0
          $$
          with solutions
          $$
          y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
          $$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
            $$
            (log_5(x^2)+log_5x-12)log_5x=-4log_5x
            $$
            that becomes, setting $y=log_5x$,
            $$
            (3y-8)y=0
            $$
            so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.



            On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
            $$
            (y^2+y-8)y=0
            $$
            with solutions
            $$
            y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
            $$






            share|cite|improve this answer









            $endgroup$



            You need $x>0$; instead of the logarithm in base $x$, consider the logarithm in base $5$ or any other base:
            $$
            (log_5(x^2)+log_5x-12)log_5x=-4log_5x
            $$
            that becomes, setting $y=log_5x$,
            $$
            (3y-8)y=0
            $$
            so $y=0$ or $3y-8=0$. Thus the solutions are $x=1$ or $x=5^{8/3}$.



            On the other hand, if the first term is $(log_5x)^2$, rather than $log_5(x^2)$, the equation would become
            $$
            (y^2+y-8)y=0
            $$
            with solutions
            $$
            y=0,quad y=frac{-1+sqrt{33}}{2},quad y=frac{-1-sqrt{33}}{2}
            $$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 6 '18 at 16:49









            egregegreg

            180k1485202




            180k1485202























                1












                $begingroup$

                I suppose you were using
                $$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
                but it is only true when $x>0$ and $xne 1$.



                Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                  $endgroup$
                  – Piano Land
                  Dec 6 '18 at 16:25










                • $begingroup$
                  I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:28










                • $begingroup$
                  Perhaps just because $x=5^{8/3}$ is too large.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:38
















                1












                $begingroup$

                I suppose you were using
                $$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
                but it is only true when $x>0$ and $xne 1$.



                Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.






                share|cite|improve this answer









                $endgroup$













                • $begingroup$
                  But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                  $endgroup$
                  – Piano Land
                  Dec 6 '18 at 16:25










                • $begingroup$
                  I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:28










                • $begingroup$
                  Perhaps just because $x=5^{8/3}$ is too large.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:38














                1












                1








                1





                $begingroup$

                I suppose you were using
                $$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
                but it is only true when $x>0$ and $xne 1$.



                Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.






                share|cite|improve this answer









                $endgroup$



                I suppose you were using
                $$x^{(log_5 x^2 + log _5 x-12)}=frac{1}{x^4}implies log_5 x^2 + log _5 x-12 =-4,$$
                but it is only true when $x>0$ and $xne 1$.



                Of course $x>0$ holds because it is the input of a logarithmic function. But you don't have $xne 1$. That's why you missed a solution.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 6 '18 at 16:11









                Eclipse SunEclipse Sun

                6,9841437




                6,9841437












                • $begingroup$
                  But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                  $endgroup$
                  – Piano Land
                  Dec 6 '18 at 16:25










                • $begingroup$
                  I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:28










                • $begingroup$
                  Perhaps just because $x=5^{8/3}$ is too large.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:38


















                • $begingroup$
                  But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                  $endgroup$
                  – Piano Land
                  Dec 6 '18 at 16:25










                • $begingroup$
                  I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:28










                • $begingroup$
                  Perhaps just because $x=5^{8/3}$ is too large.
                  $endgroup$
                  – Eclipse Sun
                  Dec 6 '18 at 16:38
















                $begingroup$
                But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                $endgroup$
                – Piano Land
                Dec 6 '18 at 16:25




                $begingroup$
                But then why wasn't the other solution also given by the graphing calculator? When I plugged the equation in it, it just showed the graph of x=1.
                $endgroup$
                – Piano Land
                Dec 6 '18 at 16:25












                $begingroup$
                I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                $endgroup$
                – Eclipse Sun
                Dec 6 '18 at 16:28




                $begingroup$
                I think $x=5^{8/3}$ is also a solution. I'm not sure why the calculator didn't give you that one.
                $endgroup$
                – Eclipse Sun
                Dec 6 '18 at 16:28












                $begingroup$
                Perhaps just because $x=5^{8/3}$ is too large.
                $endgroup$
                – Eclipse Sun
                Dec 6 '18 at 16:38




                $begingroup$
                Perhaps just because $x=5^{8/3}$ is too large.
                $endgroup$
                – Eclipse Sun
                Dec 6 '18 at 16:38











                1












                $begingroup$

                We need



                $$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$



                that is



                $$x=5^frac83$$



                the other solution $x=1$ is obtained by inspection from the original equation.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  @KM101 Opssss...thanks I fix!
                  $endgroup$
                  – gimusi
                  Dec 6 '18 at 16:26










                • $begingroup$
                  No problem! :-)
                  $endgroup$
                  – KM101
                  Dec 6 '18 at 16:36


















                1












                $begingroup$

                We need



                $$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$



                that is



                $$x=5^frac83$$



                the other solution $x=1$ is obtained by inspection from the original equation.






                share|cite|improve this answer











                $endgroup$













                • $begingroup$
                  @KM101 Opssss...thanks I fix!
                  $endgroup$
                  – gimusi
                  Dec 6 '18 at 16:26










                • $begingroup$
                  No problem! :-)
                  $endgroup$
                  – KM101
                  Dec 6 '18 at 16:36
















                1












                1








                1





                $begingroup$

                We need



                $$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$



                that is



                $$x=5^frac83$$



                the other solution $x=1$ is obtained by inspection from the original equation.






                share|cite|improve this answer











                $endgroup$



                We need



                $$log_5 x^2 + log _5 x-12=-4 iff log_5 (x^3)=8$$



                that is



                $$x=5^frac83$$



                the other solution $x=1$ is obtained by inspection from the original equation.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 6 '18 at 16:27

























                answered Dec 6 '18 at 16:24









                gimusigimusi

                92.9k94494




                92.9k94494












                • $begingroup$
                  @KM101 Opssss...thanks I fix!
                  $endgroup$
                  – gimusi
                  Dec 6 '18 at 16:26










                • $begingroup$
                  No problem! :-)
                  $endgroup$
                  – KM101
                  Dec 6 '18 at 16:36




















                • $begingroup$
                  @KM101 Opssss...thanks I fix!
                  $endgroup$
                  – gimusi
                  Dec 6 '18 at 16:26










                • $begingroup$
                  No problem! :-)
                  $endgroup$
                  – KM101
                  Dec 6 '18 at 16:36


















                $begingroup$
                @KM101 Opssss...thanks I fix!
                $endgroup$
                – gimusi
                Dec 6 '18 at 16:26




                $begingroup$
                @KM101 Opssss...thanks I fix!
                $endgroup$
                – gimusi
                Dec 6 '18 at 16:26












                $begingroup$
                No problem! :-)
                $endgroup$
                – KM101
                Dec 6 '18 at 16:36






                $begingroup$
                No problem! :-)
                $endgroup$
                – KM101
                Dec 6 '18 at 16:36













                1












                $begingroup$

                If the book claims, that $1$ is the only real solution, it is wrong. Your value
                $$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  If the book claims, that $1$ is the only real solution, it is wrong. Your value
                  $$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    If the book claims, that $1$ is the only real solution, it is wrong. Your value
                    $$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.






                    share|cite|improve this answer











                    $endgroup$



                    If the book claims, that $1$ is the only real solution, it is wrong. Your value
                    $$x = 5^{8/3} = e^{frac{8}{3} ln 5} approx 73.1$$ is indeed a solution of the equation.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 6 '18 at 16:28

























                    answered Dec 6 '18 at 16:22









                    gammatestergammatester

                    16.7k21632




                    16.7k21632























                        1












                        $begingroup$

                        What you did (probably) was the following:



                        $$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$



                        By doing so, you removed the possibility of $x = 1$.



                        As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.




                        Your other solution is valid. Let $x = 5^{frac{8}{3}}$.



                        $$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$



                        On both sides, you get



                        $$big(5^{frac{8}{3}}big)^{-4}$$






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          What you did (probably) was the following:



                          $$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$



                          By doing so, you removed the possibility of $x = 1$.



                          As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.




                          Your other solution is valid. Let $x = 5^{frac{8}{3}}$.



                          $$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$



                          On both sides, you get



                          $$big(5^{frac{8}{3}}big)^{-4}$$






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            What you did (probably) was the following:



                            $$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$



                            By doing so, you removed the possibility of $x = 1$.



                            As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.




                            Your other solution is valid. Let $x = 5^{frac{8}{3}}$.



                            $$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$



                            On both sides, you get



                            $$big(5^{frac{8}{3}}big)^{-4}$$






                            share|cite|improve this answer











                            $endgroup$



                            What you did (probably) was the following:



                            $$x^{log_5 x^2+log_5 x-12} = frac{1}{x^4} = x^{-4} implies log_5 x^2+log_5 x-12 = -4$$



                            By doing so, you removed the possibility of $x = 1$.



                            As you know, $1$ raised to any power is simply one, so $x = 1$ is a trivial solution and doesn’t really require solving. Just note that $x$ is valid for the domain of $log_5 x^2$ and $log_5 x$.




                            Your other solution is valid. Let $x = 5^{frac{8}{3}}$.



                            $$log_5 big(5^{frac{8}{3}}big)^2+log_5 5^{frac{8}{3}}-12 = log_5 big(5^{frac{8}{3}}big)^3-12 = log_5 5^8-12 = 8-12 = -4$$



                            On both sides, you get



                            $$big(5^{frac{8}{3}}big)^{-4}$$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Dec 6 '18 at 16:36

























                            answered Dec 6 '18 at 16:22









                            KM101KM101

                            5,9431523




                            5,9431523






























                                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%2f3028682%2ffind-the-solution-to-x-log-5-x2-log-5-x-12-frac1x4%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

                                Wiesbaden

                                Image
                                Wappen Deutschlandkarte 50.0825 8.24 117 Koordinaten: 50° 5′  N , 8° 14′  O Basisdaten Bundesland: Hessen Regierungsbezirk: Darmstadt Höhe: 117 m ü. NHN Fläche: 203,93 km 2 Einwohner: 278.654 (31. Dez. 2017) [1] Bevölkerungsdichte: 1366 Einwohner je km 2 Postleitzahlen: 65183–65207, 55246, 55252 Vorlage:Infobox Gemeinde in Deutschland/Wartung/PLZ enthält Text Vorwahlen: 0611, 06122, 06127, 06134 Kfz-Kennzeichen: WI Gemeindeschlüssel: 06 4 14 000 LOCODE: DE WIB NUTS: DE714 Stadtgliederung: 26 Ortsbezirke Adresse der Stadtverwaltung: Schlossplatz 6 65183 Wiesbaden Webpräsenz: www.wiesbaden.de Oberbürgermeister: Sven Gerich (SPD) Lage der Landeshauptstadt Wiesbaden in Hessen und im Regierungsbezirk Darmstadt Offizielles Logo der Landeshauptstadt Wiesbaden Flagge der Landeshauptstadt Wiesbaden Wiesbaden is
                                Read more

                                Marschland

                                Image
                                Kilometerweite Marsch in den Vier- und Marschlanden am Elbdeich Die Wedeler Marschlandschaft in Schleswig-Holstein Der Hadelner Kanal zur Entwässerung der Marsch Die tief gelegene Marsch und ein Entwässerungskanal Das Schöpfwerk Otterndorf im Land Hadeln zur Entwässerung der Medem verfügte damals bereits über die größte Kreiselpumpe Europas. Das Schöpfwerk in Neuhaus für die Aue und des Neuhaus-Bülkauer Kanals Als Marsch(land) (v. niederdt., altsächs. mersc ) – auch Masch , Mersch oder Schwemmland genannt – bezeichnet man eine nacheiszeitlich entstandene geomorphologische Landform im Gebiet der nordwestdeutschen Küsten und Flüsse sowie vergleichbare Landformen weltweit. Inhaltsverzeichnis 1 Landschaft 2 Entstehung 3 Marschböden 4 Nutzung 5 Marschgebiete in Deutschland 6 Siehe auch 7 Literatur 8 Weblinks 9 Einzelnachweise Landschaft | Marschen sind generell flache Landstriche ohne natürliche Erhebungen. Si
                                Read more

                                Dieringhausen

                                Image
                                Dieringhausen Stadt Gummersbach 50.982777777778 7.5316666666667 165 Koordinaten: 50° 58′ 58″  N , 7° 31′ 54″  O Höhe: 165  (160–245)  m Einwohner: 5055  (30. Jun. 2016) Postleitzahl: 51645 Vorwahl: 02261 Lage von Dieringhausen in Gummersbach Im Hohl ca. 1927–1930 Ansicht von Südwesten bei Bünghausen Dieringhausen (im örtlichen Dialekt Dierkesen ) ist ein Ortsteil von Gummersbach im Oberbergischen Kreis, Regierungsbezirk Köln, Nordrhein-Westfalen (Deutschland). Inhaltsverzeichnis 1 Geographie 2 Geschichte 3 Kultur 3.1 Sehenswürdigkeiten 3.2 Regelmäßige Veranstaltungen 4 Infrastruktur und Wirtschaft 4.1 Verkehr 4.1.1 Schienen- und Busverkehr 4.1.2 Straßen 4.2 Öffentliche Einrichtungen 4.2.1 Schulen und Bildungseinrichtungen 4.2.2 Kirchliche Einrichtungen 5 Persönlichkeiten 5.1 In Dieringhausen geboren 5.2 In Dieringhausen gelebt 6 Einzelnachweise 7
                                Read more