Solving logarithmic equation $2log _{2}(x-6)-log _{2}(x)=3$












0












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This is the question: $$2log_{2} (x-6)-log_{2} (x)=3$$



I think I would combine the two on the left to make $2log_{2}big({x-6over x}big) = 3$ but I'm stuck at what to do with the $2$ in front of the log. Would I divide it out to get $log_{2}big({x-6over x}big) = tfrac{3}{2}$ or change to equation to exponential form?



Any help would be greatly appreciated as I've been stuck on this question for a while.






logarithms







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  • 1




    $begingroup$
    This is a bit hard to read. Do you mean $2log_2(x-6)-log_2(x)=3$?
    $endgroup$
    – lulu
    Jan 22 at 17:52










  • $begingroup$
    @lulu Yes, I mean that. Sorry, I'm not sure how to format it correctly
    $endgroup$
    – Grimestock
    Jan 22 at 17:53






  • 1




    $begingroup$
    Then note that $nlog a= log a^n$.
    $endgroup$
    – lulu
    Jan 22 at 17:55
















0












$begingroup$


This is the question: $$2log_{2} (x-6)-log_{2} (x)=3$$



I think I would combine the two on the left to make $2log_{2}big({x-6over x}big) = 3$ but I'm stuck at what to do with the $2$ in front of the log. Would I divide it out to get $log_{2}big({x-6over x}big) = tfrac{3}{2}$ or change to equation to exponential form?



Any help would be greatly appreciated as I've been stuck on this question for a while.






logarithms







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    This is a bit hard to read. Do you mean $2log_2(x-6)-log_2(x)=3$?
    $endgroup$
    – lulu
    Jan 22 at 17:52










  • $begingroup$
    @lulu Yes, I mean that. Sorry, I'm not sure how to format it correctly
    $endgroup$
    – Grimestock
    Jan 22 at 17:53






  • 1




    $begingroup$
    Then note that $nlog a= log a^n$.
    $endgroup$
    – lulu
    Jan 22 at 17:55














0












0








0





$begingroup$


This is the question: $$2log_{2} (x-6)-log_{2} (x)=3$$



I think I would combine the two on the left to make $2log_{2}big({x-6over x}big) = 3$ but I'm stuck at what to do with the $2$ in front of the log. Would I divide it out to get $log_{2}big({x-6over x}big) = tfrac{3}{2}$ or change to equation to exponential form?



Any help would be greatly appreciated as I've been stuck on this question for a while.






logarithms







share|cite|improve this question











$endgroup$




This is the question: $$2log_{2} (x-6)-log_{2} (x)=3$$



I think I would combine the two on the left to make $2log_{2}big({x-6over x}big) = 3$ but I'm stuck at what to do with the $2$ in front of the log. Would I divide it out to get $log_{2}big({x-6over x}big) = tfrac{3}{2}$ or change to equation to exponential form?



Any help would be greatly appreciated as I've been stuck on this question for a while.






logarithms




logarithms






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share|cite|improve this question













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edited Jan 22 at 19:12









saz

81.6k861127




81.6k861127










asked Jan 22 at 17:46









GrimestockGrimestock

869




869








  • 1




    $begingroup$
    This is a bit hard to read. Do you mean $2log_2(x-6)-log_2(x)=3$?
    $endgroup$
    – lulu
    Jan 22 at 17:52










  • $begingroup$
    @lulu Yes, I mean that. Sorry, I'm not sure how to format it correctly
    $endgroup$
    – Grimestock
    Jan 22 at 17:53






  • 1




    $begingroup$
    Then note that $nlog a= log a^n$.
    $endgroup$
    – lulu
    Jan 22 at 17:55














  • 1




    $begingroup$
    This is a bit hard to read. Do you mean $2log_2(x-6)-log_2(x)=3$?
    $endgroup$
    – lulu
    Jan 22 at 17:52










  • $begingroup$
    @lulu Yes, I mean that. Sorry, I'm not sure how to format it correctly
    $endgroup$
    – Grimestock
    Jan 22 at 17:53






  • 1




    $begingroup$
    Then note that $nlog a= log a^n$.
    $endgroup$
    – lulu
    Jan 22 at 17:55








1




1




$begingroup$
This is a bit hard to read. Do you mean $2log_2(x-6)-log_2(x)=3$?
$endgroup$
– lulu
Jan 22 at 17:52




$begingroup$
This is a bit hard to read. Do you mean $2log_2(x-6)-log_2(x)=3$?
$endgroup$
– lulu
Jan 22 at 17:52












$begingroup$
@lulu Yes, I mean that. Sorry, I'm not sure how to format it correctly
$endgroup$
– Grimestock
Jan 22 at 17:53




$begingroup$
@lulu Yes, I mean that. Sorry, I'm not sure how to format it correctly
$endgroup$
– Grimestock
Jan 22 at 17:53




1




1




$begingroup$
Then note that $nlog a= log a^n$.
$endgroup$
– lulu
Jan 22 at 17:55




$begingroup$
Then note that $nlog a= log a^n$.
$endgroup$
– lulu
Jan 22 at 17:55










4 Answers
4






active

oldest

votes


















1












$begingroup$

You can't quite use your first step. First you should convert $2log_{2}(x-6)$ to $log_{2}(x-6)^2$, and then you can apply the subtraction of logs property to get $log_{2}frac{(x-6)^2}{x} = 3$. Then exponentiate to get rid of the logs and you should soon find yourself with a quadratic equation that you should be able to solve.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    With this would I get x^(3 - 6^6)/x^3?
    $endgroup$
    – Grimestock
    Jan 22 at 18:05








  • 1




    $begingroup$
    Raise $2$ to the power of each side. You should get $frac{(x-6)^2}{x} = 8$.
    $endgroup$
    – kcborys
    Jan 22 at 18:06






  • 1




    $begingroup$
    $(x-6)^2 = (x-6)(x-6) not= x^2 - 36$
    $endgroup$
    – kcborys
    Jan 22 at 18:10






  • 1




    $begingroup$
    Correct. Then you will have $x^2 - 12x + 36 = 8x$.
    $endgroup$
    – kcborys
    Jan 22 at 18:16






  • 1




    $begingroup$
    Correct, because if you plug $x=2$ back in, you have a term of $log_{2}(-4)$ which is not possible
    $endgroup$
    – kcborys
    Jan 22 at 18:18



















1












$begingroup$

Hint: You can write $$log_{2}{(x-6)^2}-log_{2}{6}=3$$ and by the hint above
$$log_{2}frac{(x-6)^2}{x}=3$$






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Use the rules $logbig(frac{a}{b}big) = log(a) - log(b)$ and $log(a^n) = nlog(a)$ to write everything as one logarithm. Then exponentiate.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      With the basics rules:
      $$ log_a(b)=x iff a^x=b label{1}tag{1}$$
      $$ log_a(b^n)=nlog_a(b) label{2}tag{2}$$
      $$ log_a(b) + log_a(c)=log_a(bc) label{3}tag{3}$$
      You can solve this equation:
      $$ 2log_{2} (x-6) - log_2(x)=3$$
      from eqref{2}:
      $$ log_{2} big[(x-6)^2big] - log_2(x)=3$$
      from eqref{2}:
      $$ log_{2} big[(x-6)^2big] + log_2big(x^{-1}big)=3$$
      $$ log_{2} big[(x-6)^2big] + log_2Big(frac{1}{x}Big)=3$$
      from eqref{3}:
      $$ log_{2} Bigg[frac{(x-6)^2}{x}Bigg]=3$$
      from eqref{1}:
      $$ 2^3 = frac{(x-6)^2}{x} $$
      $$ 8 = frac{(x-6)^2}{x} $$
      $$ 8x = (x-6)^2 $$
      $$ 8x = x^2-12x+36 $$
      $$ 0 = x^2-20x+36 $$
      Here, $-20 = -18-2$ and $36=(-18)cdot(-2)$, then:
      $$ 0 = (x-18)(x-2) $$
      Thus $x=18$ or $x=2$






      share|cite|improve this answer











      $endgroup$













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        4 Answers
        4






        active

        oldest

        votes








        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        1












        $begingroup$

        You can't quite use your first step. First you should convert $2log_{2}(x-6)$ to $log_{2}(x-6)^2$, and then you can apply the subtraction of logs property to get $log_{2}frac{(x-6)^2}{x} = 3$. Then exponentiate to get rid of the logs and you should soon find yourself with a quadratic equation that you should be able to solve.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          With this would I get x^(3 - 6^6)/x^3?
          $endgroup$
          – Grimestock
          Jan 22 at 18:05








        • 1




          $begingroup$
          Raise $2$ to the power of each side. You should get $frac{(x-6)^2}{x} = 8$.
          $endgroup$
          – kcborys
          Jan 22 at 18:06






        • 1




          $begingroup$
          $(x-6)^2 = (x-6)(x-6) not= x^2 - 36$
          $endgroup$
          – kcborys
          Jan 22 at 18:10






        • 1




          $begingroup$
          Correct. Then you will have $x^2 - 12x + 36 = 8x$.
          $endgroup$
          – kcborys
          Jan 22 at 18:16






        • 1




          $begingroup$
          Correct, because if you plug $x=2$ back in, you have a term of $log_{2}(-4)$ which is not possible
          $endgroup$
          – kcborys
          Jan 22 at 18:18
















        1












        $begingroup$

        You can't quite use your first step. First you should convert $2log_{2}(x-6)$ to $log_{2}(x-6)^2$, and then you can apply the subtraction of logs property to get $log_{2}frac{(x-6)^2}{x} = 3$. Then exponentiate to get rid of the logs and you should soon find yourself with a quadratic equation that you should be able to solve.






        share|cite|improve this answer









        $endgroup$













        • $begingroup$
          With this would I get x^(3 - 6^6)/x^3?
          $endgroup$
          – Grimestock
          Jan 22 at 18:05








        • 1




          $begingroup$
          Raise $2$ to the power of each side. You should get $frac{(x-6)^2}{x} = 8$.
          $endgroup$
          – kcborys
          Jan 22 at 18:06






        • 1




          $begingroup$
          $(x-6)^2 = (x-6)(x-6) not= x^2 - 36$
          $endgroup$
          – kcborys
          Jan 22 at 18:10






        • 1




          $begingroup$
          Correct. Then you will have $x^2 - 12x + 36 = 8x$.
          $endgroup$
          – kcborys
          Jan 22 at 18:16






        • 1




          $begingroup$
          Correct, because if you plug $x=2$ back in, you have a term of $log_{2}(-4)$ which is not possible
          $endgroup$
          – kcborys
          Jan 22 at 18:18














        1












        1








        1





        $begingroup$

        You can't quite use your first step. First you should convert $2log_{2}(x-6)$ to $log_{2}(x-6)^2$, and then you can apply the subtraction of logs property to get $log_{2}frac{(x-6)^2}{x} = 3$. Then exponentiate to get rid of the logs and you should soon find yourself with a quadratic equation that you should be able to solve.






        share|cite|improve this answer









        $endgroup$



        You can't quite use your first step. First you should convert $2log_{2}(x-6)$ to $log_{2}(x-6)^2$, and then you can apply the subtraction of logs property to get $log_{2}frac{(x-6)^2}{x} = 3$. Then exponentiate to get rid of the logs and you should soon find yourself with a quadratic equation that you should be able to solve.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 22 at 17:57









        kcboryskcborys

        49429




        49429












        • $begingroup$
          With this would I get x^(3 - 6^6)/x^3?
          $endgroup$
          – Grimestock
          Jan 22 at 18:05








        • 1




          $begingroup$
          Raise $2$ to the power of each side. You should get $frac{(x-6)^2}{x} = 8$.
          $endgroup$
          – kcborys
          Jan 22 at 18:06






        • 1




          $begingroup$
          $(x-6)^2 = (x-6)(x-6) not= x^2 - 36$
          $endgroup$
          – kcborys
          Jan 22 at 18:10






        • 1




          $begingroup$
          Correct. Then you will have $x^2 - 12x + 36 = 8x$.
          $endgroup$
          – kcborys
          Jan 22 at 18:16






        • 1




          $begingroup$
          Correct, because if you plug $x=2$ back in, you have a term of $log_{2}(-4)$ which is not possible
          $endgroup$
          – kcborys
          Jan 22 at 18:18


















        • $begingroup$
          With this would I get x^(3 - 6^6)/x^3?
          $endgroup$
          – Grimestock
          Jan 22 at 18:05








        • 1




          $begingroup$
          Raise $2$ to the power of each side. You should get $frac{(x-6)^2}{x} = 8$.
          $endgroup$
          – kcborys
          Jan 22 at 18:06






        • 1




          $begingroup$
          $(x-6)^2 = (x-6)(x-6) not= x^2 - 36$
          $endgroup$
          – kcborys
          Jan 22 at 18:10






        • 1




          $begingroup$
          Correct. Then you will have $x^2 - 12x + 36 = 8x$.
          $endgroup$
          – kcborys
          Jan 22 at 18:16






        • 1




          $begingroup$
          Correct, because if you plug $x=2$ back in, you have a term of $log_{2}(-4)$ which is not possible
          $endgroup$
          – kcborys
          Jan 22 at 18:18
















        $begingroup$
        With this would I get x^(3 - 6^6)/x^3?
        $endgroup$
        – Grimestock
        Jan 22 at 18:05






        $begingroup$
        With this would I get x^(3 - 6^6)/x^3?
        $endgroup$
        – Grimestock
        Jan 22 at 18:05






        1




        1




        $begingroup$
        Raise $2$ to the power of each side. You should get $frac{(x-6)^2}{x} = 8$.
        $endgroup$
        – kcborys
        Jan 22 at 18:06




        $begingroup$
        Raise $2$ to the power of each side. You should get $frac{(x-6)^2}{x} = 8$.
        $endgroup$
        – kcborys
        Jan 22 at 18:06




        1




        1




        $begingroup$
        $(x-6)^2 = (x-6)(x-6) not= x^2 - 36$
        $endgroup$
        – kcborys
        Jan 22 at 18:10




        $begingroup$
        $(x-6)^2 = (x-6)(x-6) not= x^2 - 36$
        $endgroup$
        – kcborys
        Jan 22 at 18:10




        1




        1




        $begingroup$
        Correct. Then you will have $x^2 - 12x + 36 = 8x$.
        $endgroup$
        – kcborys
        Jan 22 at 18:16




        $begingroup$
        Correct. Then you will have $x^2 - 12x + 36 = 8x$.
        $endgroup$
        – kcborys
        Jan 22 at 18:16




        1




        1




        $begingroup$
        Correct, because if you plug $x=2$ back in, you have a term of $log_{2}(-4)$ which is not possible
        $endgroup$
        – kcborys
        Jan 22 at 18:18




        $begingroup$
        Correct, because if you plug $x=2$ back in, you have a term of $log_{2}(-4)$ which is not possible
        $endgroup$
        – kcborys
        Jan 22 at 18:18











        1












        $begingroup$

        Hint: You can write $$log_{2}{(x-6)^2}-log_{2}{6}=3$$ and by the hint above
        $$log_{2}frac{(x-6)^2}{x}=3$$






        share|cite|improve this answer









        $endgroup$


















          1












          $begingroup$

          Hint: You can write $$log_{2}{(x-6)^2}-log_{2}{6}=3$$ and by the hint above
          $$log_{2}frac{(x-6)^2}{x}=3$$






          share|cite|improve this answer









          $endgroup$
















            1












            1








            1





            $begingroup$

            Hint: You can write $$log_{2}{(x-6)^2}-log_{2}{6}=3$$ and by the hint above
            $$log_{2}frac{(x-6)^2}{x}=3$$






            share|cite|improve this answer









            $endgroup$



            Hint: You can write $$log_{2}{(x-6)^2}-log_{2}{6}=3$$ and by the hint above
            $$log_{2}frac{(x-6)^2}{x}=3$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 22 at 17:56









            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

            77.4k42866




            77.4k42866























                1












                $begingroup$

                Use the rules $logbig(frac{a}{b}big) = log(a) - log(b)$ and $log(a^n) = nlog(a)$ to write everything as one logarithm. Then exponentiate.






                share|cite|improve this answer











                $endgroup$


















                  1












                  $begingroup$

                  Use the rules $logbig(frac{a}{b}big) = log(a) - log(b)$ and $log(a^n) = nlog(a)$ to write everything as one logarithm. Then exponentiate.






                  share|cite|improve this answer











                  $endgroup$
















                    1












                    1








                    1





                    $begingroup$

                    Use the rules $logbig(frac{a}{b}big) = log(a) - log(b)$ and $log(a^n) = nlog(a)$ to write everything as one logarithm. Then exponentiate.






                    share|cite|improve this answer











                    $endgroup$



                    Use the rules $logbig(frac{a}{b}big) = log(a) - log(b)$ and $log(a^n) = nlog(a)$ to write everything as one logarithm. Then exponentiate.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Jan 22 at 17:58









                    El borito

                    666216




                    666216










                    answered Jan 22 at 17:52









                    KlausKlaus

                    2,12711




                    2,12711























                        1












                        $begingroup$

                        With the basics rules:
                        $$ log_a(b)=x iff a^x=b label{1}tag{1}$$
                        $$ log_a(b^n)=nlog_a(b) label{2}tag{2}$$
                        $$ log_a(b) + log_a(c)=log_a(bc) label{3}tag{3}$$
                        You can solve this equation:
                        $$ 2log_{2} (x-6) - log_2(x)=3$$
                        from eqref{2}:
                        $$ log_{2} big[(x-6)^2big] - log_2(x)=3$$
                        from eqref{2}:
                        $$ log_{2} big[(x-6)^2big] + log_2big(x^{-1}big)=3$$
                        $$ log_{2} big[(x-6)^2big] + log_2Big(frac{1}{x}Big)=3$$
                        from eqref{3}:
                        $$ log_{2} Bigg[frac{(x-6)^2}{x}Bigg]=3$$
                        from eqref{1}:
                        $$ 2^3 = frac{(x-6)^2}{x} $$
                        $$ 8 = frac{(x-6)^2}{x} $$
                        $$ 8x = (x-6)^2 $$
                        $$ 8x = x^2-12x+36 $$
                        $$ 0 = x^2-20x+36 $$
                        Here, $-20 = -18-2$ and $36=(-18)cdot(-2)$, then:
                        $$ 0 = (x-18)(x-2) $$
                        Thus $x=18$ or $x=2$






                        share|cite|improve this answer











                        $endgroup$


















                          1












                          $begingroup$

                          With the basics rules:
                          $$ log_a(b)=x iff a^x=b label{1}tag{1}$$
                          $$ log_a(b^n)=nlog_a(b) label{2}tag{2}$$
                          $$ log_a(b) + log_a(c)=log_a(bc) label{3}tag{3}$$
                          You can solve this equation:
                          $$ 2log_{2} (x-6) - log_2(x)=3$$
                          from eqref{2}:
                          $$ log_{2} big[(x-6)^2big] - log_2(x)=3$$
                          from eqref{2}:
                          $$ log_{2} big[(x-6)^2big] + log_2big(x^{-1}big)=3$$
                          $$ log_{2} big[(x-6)^2big] + log_2Big(frac{1}{x}Big)=3$$
                          from eqref{3}:
                          $$ log_{2} Bigg[frac{(x-6)^2}{x}Bigg]=3$$
                          from eqref{1}:
                          $$ 2^3 = frac{(x-6)^2}{x} $$
                          $$ 8 = frac{(x-6)^2}{x} $$
                          $$ 8x = (x-6)^2 $$
                          $$ 8x = x^2-12x+36 $$
                          $$ 0 = x^2-20x+36 $$
                          Here, $-20 = -18-2$ and $36=(-18)cdot(-2)$, then:
                          $$ 0 = (x-18)(x-2) $$
                          Thus $x=18$ or $x=2$






                          share|cite|improve this answer











                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            With the basics rules:
                            $$ log_a(b)=x iff a^x=b label{1}tag{1}$$
                            $$ log_a(b^n)=nlog_a(b) label{2}tag{2}$$
                            $$ log_a(b) + log_a(c)=log_a(bc) label{3}tag{3}$$
                            You can solve this equation:
                            $$ 2log_{2} (x-6) - log_2(x)=3$$
                            from eqref{2}:
                            $$ log_{2} big[(x-6)^2big] - log_2(x)=3$$
                            from eqref{2}:
                            $$ log_{2} big[(x-6)^2big] + log_2big(x^{-1}big)=3$$
                            $$ log_{2} big[(x-6)^2big] + log_2Big(frac{1}{x}Big)=3$$
                            from eqref{3}:
                            $$ log_{2} Bigg[frac{(x-6)^2}{x}Bigg]=3$$
                            from eqref{1}:
                            $$ 2^3 = frac{(x-6)^2}{x} $$
                            $$ 8 = frac{(x-6)^2}{x} $$
                            $$ 8x = (x-6)^2 $$
                            $$ 8x = x^2-12x+36 $$
                            $$ 0 = x^2-20x+36 $$
                            Here, $-20 = -18-2$ and $36=(-18)cdot(-2)$, then:
                            $$ 0 = (x-18)(x-2) $$
                            Thus $x=18$ or $x=2$






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                            $endgroup$



                            With the basics rules:
                            $$ log_a(b)=x iff a^x=b label{1}tag{1}$$
                            $$ log_a(b^n)=nlog_a(b) label{2}tag{2}$$
                            $$ log_a(b) + log_a(c)=log_a(bc) label{3}tag{3}$$
                            You can solve this equation:
                            $$ 2log_{2} (x-6) - log_2(x)=3$$
                            from eqref{2}:
                            $$ log_{2} big[(x-6)^2big] - log_2(x)=3$$
                            from eqref{2}:
                            $$ log_{2} big[(x-6)^2big] + log_2big(x^{-1}big)=3$$
                            $$ log_{2} big[(x-6)^2big] + log_2Big(frac{1}{x}Big)=3$$
                            from eqref{3}:
                            $$ log_{2} Bigg[frac{(x-6)^2}{x}Bigg]=3$$
                            from eqref{1}:
                            $$ 2^3 = frac{(x-6)^2}{x} $$
                            $$ 8 = frac{(x-6)^2}{x} $$
                            $$ 8x = (x-6)^2 $$
                            $$ 8x = x^2-12x+36 $$
                            $$ 0 = x^2-20x+36 $$
                            Here, $-20 = -18-2$ and $36=(-18)cdot(-2)$, then:
                            $$ 0 = (x-18)(x-2) $$
                            Thus $x=18$ or $x=2$







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited Feb 3 at 5:20

























                            answered Jan 22 at 18:11









                            El boritoEl borito

                            666216




                            666216






























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