Why is it that $lnBig(1+frac{x}{y}Big)=ln(1+exp(ln x - ln y))$?












0














Is the relation $lnBig(1+frac{x}{y}Big)=ln(1+exp(ln x - ln y))$ an approximation?



If so, how can I derive this relation?










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    Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
    – Alecos Papadopoulos
    Nov 29 at 19:33


















0














Is the relation $lnBig(1+frac{x}{y}Big)=ln(1+exp(ln x - ln y))$ an approximation?



If so, how can I derive this relation?










share|cite|improve this question




















  • 1




    Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
    – Alecos Papadopoulos
    Nov 29 at 19:33
















0












0








0







Is the relation $lnBig(1+frac{x}{y}Big)=ln(1+exp(ln x - ln y))$ an approximation?



If so, how can I derive this relation?










share|cite|improve this question















Is the relation $lnBig(1+frac{x}{y}Big)=ln(1+exp(ln x - ln y))$ an approximation?



If so, how can I derive this relation?







algebra-precalculus logarithms exponential-function






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edited Nov 29 at 19:29









Foobaz John

20.7k41250




20.7k41250










asked Nov 29 at 19:27









Jeremy Kwak

92




92








  • 1




    Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
    – Alecos Papadopoulos
    Nov 29 at 19:33
















  • 1




    Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
    – Alecos Papadopoulos
    Nov 29 at 19:33










1




1




Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
– Alecos Papadopoulos
Nov 29 at 19:33






Just use a basic property of logarithms, and then the basic relation between the natural logarithm and its base, $exp=e$. The equation holds exactly.
– Alecos Papadopoulos
Nov 29 at 19:33












3 Answers
3






active

oldest

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3














$$ ln(x) - ln(y) = lnleft(frac{x}{y} right) \
expleft(lnleft(frac{x}{y}right)right) = frac{x}{y} \
implies lnleft(1 + frac{x}{y}right)
$$






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  • 3




    Note that you can use exp and ln to format it better.
    – TheSimpliFire
    Nov 29 at 19:37



















0














It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{mathrm{ln} x}=x$$
Replacing $x$ with $frac{x}{y}$, we have $$e^{ln(frac{x}{y})}=frac{x}{y}$$
So, plugging this into the original equation $$lnleft( 1 + frac{x}{y}right)=lnleft(1+e^{lnleft(frac{x}{y}right)}right)$$
We know that a property of all logarithms is that $$log_w(frac{x}{y})=log_w(x)-log_w(y)$$
So, substituting we have $$lnleft(1+e^{lnleft(frac{x}{y}right)}right)= lnleft(1+e^{ln(x)-ln(y)}right)= lnleft(1+exp(ln(x)-ln(y))right)$$






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    0














    Note that



    $$log a-log b = logfrac{a}{b}$$



    Applying that here, you get



    $$ln big(1+e^{ln x-ln y}big) = ln big(1+e^{ln frac{x}{y}}big)$$



    Also, as inverses, $e^{ln a} = a$ (and $ln e^a = a$), meaning the expression simplifies to



    $$ln bigg(1+frac{x}{y}bigg)$$



    Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3














      $$ ln(x) - ln(y) = lnleft(frac{x}{y} right) \
      expleft(lnleft(frac{x}{y}right)right) = frac{x}{y} \
      implies lnleft(1 + frac{x}{y}right)
      $$






      share|cite|improve this answer



















      • 3




        Note that you can use exp and ln to format it better.
        – TheSimpliFire
        Nov 29 at 19:37
















      3














      $$ ln(x) - ln(y) = lnleft(frac{x}{y} right) \
      expleft(lnleft(frac{x}{y}right)right) = frac{x}{y} \
      implies lnleft(1 + frac{x}{y}right)
      $$






      share|cite|improve this answer



















      • 3




        Note that you can use exp and ln to format it better.
        – TheSimpliFire
        Nov 29 at 19:37














      3












      3








      3






      $$ ln(x) - ln(y) = lnleft(frac{x}{y} right) \
      expleft(lnleft(frac{x}{y}right)right) = frac{x}{y} \
      implies lnleft(1 + frac{x}{y}right)
      $$






      share|cite|improve this answer














      $$ ln(x) - ln(y) = lnleft(frac{x}{y} right) \
      expleft(lnleft(frac{x}{y}right)right) = frac{x}{y} \
      implies lnleft(1 + frac{x}{y}right)
      $$







      share|cite|improve this answer














      share|cite|improve this answer



      share|cite|improve this answer








      edited Nov 29 at 20:12









      Eevee Trainer

      3,390225




      3,390225










      answered Nov 29 at 19:34









      Филип Димитровски

      314




      314








      • 3




        Note that you can use exp and ln to format it better.
        – TheSimpliFire
        Nov 29 at 19:37














      • 3




        Note that you can use exp and ln to format it better.
        – TheSimpliFire
        Nov 29 at 19:37








      3




      3




      Note that you can use exp and ln to format it better.
      – TheSimpliFire
      Nov 29 at 19:37




      Note that you can use exp and ln to format it better.
      – TheSimpliFire
      Nov 29 at 19:37











      0














      It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{mathrm{ln} x}=x$$
      Replacing $x$ with $frac{x}{y}$, we have $$e^{ln(frac{x}{y})}=frac{x}{y}$$
      So, plugging this into the original equation $$lnleft( 1 + frac{x}{y}right)=lnleft(1+e^{lnleft(frac{x}{y}right)}right)$$
      We know that a property of all logarithms is that $$log_w(frac{x}{y})=log_w(x)-log_w(y)$$
      So, substituting we have $$lnleft(1+e^{lnleft(frac{x}{y}right)}right)= lnleft(1+e^{ln(x)-ln(y)}right)= lnleft(1+exp(ln(x)-ln(y))right)$$






      share|cite|improve this answer




























        0














        It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{mathrm{ln} x}=x$$
        Replacing $x$ with $frac{x}{y}$, we have $$e^{ln(frac{x}{y})}=frac{x}{y}$$
        So, plugging this into the original equation $$lnleft( 1 + frac{x}{y}right)=lnleft(1+e^{lnleft(frac{x}{y}right)}right)$$
        We know that a property of all logarithms is that $$log_w(frac{x}{y})=log_w(x)-log_w(y)$$
        So, substituting we have $$lnleft(1+e^{lnleft(frac{x}{y}right)}right)= lnleft(1+e^{ln(x)-ln(y)}right)= lnleft(1+exp(ln(x)-ln(y))right)$$






        share|cite|improve this answer


























          0












          0








          0






          It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{mathrm{ln} x}=x$$
          Replacing $x$ with $frac{x}{y}$, we have $$e^{ln(frac{x}{y})}=frac{x}{y}$$
          So, plugging this into the original equation $$lnleft( 1 + frac{x}{y}right)=lnleft(1+e^{lnleft(frac{x}{y}right)}right)$$
          We know that a property of all logarithms is that $$log_w(frac{x}{y})=log_w(x)-log_w(y)$$
          So, substituting we have $$lnleft(1+e^{lnleft(frac{x}{y}right)}right)= lnleft(1+e^{ln(x)-ln(y)}right)= lnleft(1+exp(ln(x)-ln(y))right)$$






          share|cite|improve this answer














          It’s not an approximation; it’s exact. Since the natural logarithm and $e^x$ are inverse functions, we know that, since $f^{-1}(f(x)=x$, $$e^{mathrm{ln} x}=x$$
          Replacing $x$ with $frac{x}{y}$, we have $$e^{ln(frac{x}{y})}=frac{x}{y}$$
          So, plugging this into the original equation $$lnleft( 1 + frac{x}{y}right)=lnleft(1+e^{lnleft(frac{x}{y}right)}right)$$
          We know that a property of all logarithms is that $$log_w(frac{x}{y})=log_w(x)-log_w(y)$$
          So, substituting we have $$lnleft(1+e^{lnleft(frac{x}{y}right)}right)= lnleft(1+e^{ln(x)-ln(y)}right)= lnleft(1+exp(ln(x)-ln(y))right)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 29 at 19:46

























          answered Nov 29 at 19:43









          Tesseract

          21615




          21615























              0














              Note that



              $$log a-log b = logfrac{a}{b}$$



              Applying that here, you get



              $$ln big(1+e^{ln x-ln y}big) = ln big(1+e^{ln frac{x}{y}}big)$$



              Also, as inverses, $e^{ln a} = a$ (and $ln e^a = a$), meaning the expression simplifies to



              $$ln bigg(1+frac{x}{y}bigg)$$



              Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.






              share|cite|improve this answer


























                0














                Note that



                $$log a-log b = logfrac{a}{b}$$



                Applying that here, you get



                $$ln big(1+e^{ln x-ln y}big) = ln big(1+e^{ln frac{x}{y}}big)$$



                Also, as inverses, $e^{ln a} = a$ (and $ln e^a = a$), meaning the expression simplifies to



                $$ln bigg(1+frac{x}{y}bigg)$$



                Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.






                share|cite|improve this answer
























                  0












                  0








                  0






                  Note that



                  $$log a-log b = logfrac{a}{b}$$



                  Applying that here, you get



                  $$ln big(1+e^{ln x-ln y}big) = ln big(1+e^{ln frac{x}{y}}big)$$



                  Also, as inverses, $e^{ln a} = a$ (and $ln e^a = a$), meaning the expression simplifies to



                  $$ln bigg(1+frac{x}{y}bigg)$$



                  Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.






                  share|cite|improve this answer












                  Note that



                  $$log a-log b = logfrac{a}{b}$$



                  Applying that here, you get



                  $$ln big(1+e^{ln x-ln y}big) = ln big(1+e^{ln frac{x}{y}}big)$$



                  Also, as inverses, $e^{ln a} = a$ (and $ln e^a = a$), meaning the expression simplifies to



                  $$ln bigg(1+frac{x}{y}bigg)$$



                  Also, to point out, this isn’t an approximation. The two expressions are exactly equivalent.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 at 20:07









                  KM101

                  3,958417




                  3,958417






























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