Simplify $log(α+ b(x- t)+ k)$












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$begingroup$


I am trying to simplify this natural log expression the best that I can, but I am unsure what to do in order to separate $b(x-t)$. Would it be $log(b) + log (x/t)$? so would the whole thing be:



$$ log (alpha + b) + log (x/t) + log k quad?$$



Thank you!






logarithms







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  • $begingroup$
    Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
    $endgroup$
    – KM101
    Dec 8 '18 at 15:52


















-1












$begingroup$


I am trying to simplify this natural log expression the best that I can, but I am unsure what to do in order to separate $b(x-t)$. Would it be $log(b) + log (x/t)$? so would the whole thing be:



$$ log (alpha + b) + log (x/t) + log k quad?$$



Thank you!






logarithms







share|cite|improve this question











$endgroup$












  • $begingroup$
    Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
    $endgroup$
    – KM101
    Dec 8 '18 at 15:52
















-1












-1








-1





$begingroup$


I am trying to simplify this natural log expression the best that I can, but I am unsure what to do in order to separate $b(x-t)$. Would it be $log(b) + log (x/t)$? so would the whole thing be:



$$ log (alpha + b) + log (x/t) + log k quad?$$



Thank you!






logarithms







share|cite|improve this question











$endgroup$




I am trying to simplify this natural log expression the best that I can, but I am unsure what to do in order to separate $b(x-t)$. Would it be $log(b) + log (x/t)$? so would the whole thing be:



$$ log (alpha + b) + log (x/t) + log k quad?$$



Thank you!






logarithms




logarithms






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edited Dec 8 '18 at 16:01









Dando18

4,67741235




4,67741235










asked Dec 8 '18 at 15:43









Beary1213Beary1213

1




1












  • $begingroup$
    Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
    $endgroup$
    – KM101
    Dec 8 '18 at 15:52




















  • $begingroup$
    Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
    $endgroup$
    – KM101
    Dec 8 '18 at 15:52


















$begingroup$
Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
$endgroup$
– KM101
Dec 8 '18 at 15:52






$begingroup$
Logs don’t work well with sums. $$log(ab) = log a+log b neq log(a+b)$$ $$log frac{a}{b} = log a-log b neq log(a-b)$$
$endgroup$
– KM101
Dec 8 '18 at 15:52












2 Answers
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No. You can't separate $log(A+B)$ into two logarithms, no matter what $A$ and $B$ are. The only things you can separate are $log(AB)=log(A)+log(B)$, $log_B(A)=frac{log(A)}{log(B)}$ and $log(A^B)=Blog(A)$, and their inverse operations






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    You can only write $$log(alpha+bx-bt+k)$$ it holds only
    $$log(ab)=log(a)+log(b)$$ and $$logfrac{a}{b}=log(a)-log(b)$$ for positive variables $a,b$






    share|cite|improve this answer









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      2 Answers
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      2 Answers
      2






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      active

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      0












      $begingroup$

      No. You can't separate $log(A+B)$ into two logarithms, no matter what $A$ and $B$ are. The only things you can separate are $log(AB)=log(A)+log(B)$, $log_B(A)=frac{log(A)}{log(B)}$ and $log(A^B)=Blog(A)$, and their inverse operations






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        No. You can't separate $log(A+B)$ into two logarithms, no matter what $A$ and $B$ are. The only things you can separate are $log(AB)=log(A)+log(B)$, $log_B(A)=frac{log(A)}{log(B)}$ and $log(A^B)=Blog(A)$, and their inverse operations






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          No. You can't separate $log(A+B)$ into two logarithms, no matter what $A$ and $B$ are. The only things you can separate are $log(AB)=log(A)+log(B)$, $log_B(A)=frac{log(A)}{log(B)}$ and $log(A^B)=Blog(A)$, and their inverse operations






          share|cite|improve this answer









          $endgroup$



          No. You can't separate $log(A+B)$ into two logarithms, no matter what $A$ and $B$ are. The only things you can separate are $log(AB)=log(A)+log(B)$, $log_B(A)=frac{log(A)}{log(B)}$ and $log(A^B)=Blog(A)$, and their inverse operations







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 8 '18 at 15:50









          MoKo19MoKo19

          1914




          1914























              0












              $begingroup$

              You can only write $$log(alpha+bx-bt+k)$$ it holds only
              $$log(ab)=log(a)+log(b)$$ and $$logfrac{a}{b}=log(a)-log(b)$$ for positive variables $a,b$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                You can only write $$log(alpha+bx-bt+k)$$ it holds only
                $$log(ab)=log(a)+log(b)$$ and $$logfrac{a}{b}=log(a)-log(b)$$ for positive variables $a,b$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  You can only write $$log(alpha+bx-bt+k)$$ it holds only
                  $$log(ab)=log(a)+log(b)$$ and $$logfrac{a}{b}=log(a)-log(b)$$ for positive variables $a,b$






                  share|cite|improve this answer









                  $endgroup$



                  You can only write $$log(alpha+bx-bt+k)$$ it holds only
                  $$log(ab)=log(a)+log(b)$$ and $$logfrac{a}{b}=log(a)-log(b)$$ for positive variables $a,b$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 8 '18 at 15:51









                  Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                  74.6k42865




                  74.6k42865






























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