Solving this logarithmic problemStuck on an 'advanced logarithm problem': $2 log_2 x - log_2 (x - tfrac1 2) =...

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Solving this logarithmic problem

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Solving this logarithmic problem


Stuck on an 'advanced logarithm problem': $2 log_2 x - log_2 (x - tfrac1 2) = log_3 3$Solve logarithmic simultaneous equations.Logarithmic EquationFind the roots of the quadratics functionLinear Equation Problem answerSimple Logarithmic question.Solving exponential equation: $-6+3e^x=8$Laws of logarithms: why isn't $frac{1}{4}log_2(8x - 56)^{16 }- 12 = log_2((8x-56)^{16})^{frac 1 4} - 12$?solving this linear equation by substitution?Factorization of combinatorics equation













5












$begingroup$


Given this equation: $$ln(x^2-1)-3=ln(x+1)$$



Evaluate x.



Applying the natural logarithmic rule



$$ln(x+1)=3$$



$$x+1=e^3$$



$$x=e^3-1$$



The answer was different from the book. Where did I when wrong?






algebra-precalculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
    $endgroup$
    – fleablood
    1 hour ago
















5












$begingroup$


Given this equation: $$ln(x^2-1)-3=ln(x+1)$$



Evaluate x.



Applying the natural logarithmic rule



$$ln(x+1)=3$$



$$x+1=e^3$$



$$x=e^3-1$$



The answer was different from the book. Where did I when wrong?






algebra-precalculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
    $endgroup$
    – fleablood
    1 hour ago














5












5








5





$begingroup$


Given this equation: $$ln(x^2-1)-3=ln(x+1)$$



Evaluate x.



Applying the natural logarithmic rule



$$ln(x+1)=3$$



$$x+1=e^3$$



$$x=e^3-1$$



The answer was different from the book. Where did I when wrong?






algebra-precalculus







share|cite|improve this question









$endgroup$




Given this equation: $$ln(x^2-1)-3=ln(x+1)$$



Evaluate x.



Applying the natural logarithmic rule



$$ln(x+1)=3$$



$$x+1=e^3$$



$$x=e^3-1$$



The answer was different from the book. Where did I when wrong?






algebra-precalculus




algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 8 hours ago









TriangleTriangle

1174




1174












  • $begingroup$
    How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
    $endgroup$
    – fleablood
    1 hour ago


















  • $begingroup$
    How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
    $endgroup$
    – fleablood
    1 hour ago
















$begingroup$
How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
$endgroup$
– fleablood
1 hour ago




$begingroup$
How did you get $ln(x+1) = 3$??? There's no reason to think $ln(x^2 -1) -3 = 3$ so why does $ln(x+1)=3$.
$endgroup$
– fleablood
1 hour ago










2 Answers
2






active

oldest

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10












$begingroup$

You wrote $ln(x+1)=3$ instead of $ln(x-1)=3$.






share|cite|improve this answer









$endgroup$





















    5












    $begingroup$

    $ln(x^2-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)=ln(x+1)+3$



    implies $ln(x-1)=3$, $x=e^3+1$






    share|cite|improve this answer











    $endgroup$














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      2 Answers
      2






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      2 Answers
      2






      active

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      active

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      active

      oldest

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      10












      $begingroup$

      You wrote $ln(x+1)=3$ instead of $ln(x-1)=3$.






      share|cite|improve this answer









      $endgroup$


















        10












        $begingroup$

        You wrote $ln(x+1)=3$ instead of $ln(x-1)=3$.






        share|cite|improve this answer









        $endgroup$
















          10












          10








          10





          $begingroup$

          You wrote $ln(x+1)=3$ instead of $ln(x-1)=3$.






          share|cite|improve this answer









          $endgroup$



          You wrote $ln(x+1)=3$ instead of $ln(x-1)=3$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 8 hours ago









          Spencer KraislerSpencer Kraisler

          666312




          666312























              5












              $begingroup$

              $ln(x^2-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)=ln(x+1)+3$



              implies $ln(x-1)=3$, $x=e^3+1$






              share|cite|improve this answer











              $endgroup$


















                5












                $begingroup$

                $ln(x^2-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)=ln(x+1)+3$



                implies $ln(x-1)=3$, $x=e^3+1$






                share|cite|improve this answer











                $endgroup$
















                  5












                  5








                  5





                  $begingroup$

                  $ln(x^2-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)=ln(x+1)+3$



                  implies $ln(x-1)=3$, $x=e^3+1$






                  share|cite|improve this answer











                  $endgroup$



                  $ln(x^2-1)=ln((x+1)(x-1))=ln(x+1)+ln(x-1)=ln(x+1)+3$



                  implies $ln(x-1)=3$, $x=e^3+1$







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited 5 hours ago









                  Andrew Li

                  4,19621128




                  4,19621128










                  answered 8 hours ago









                  Tsemo AristideTsemo Aristide

                  63.1k11547




                  63.1k11547






























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