Expressing logarithmic equations without logs












3












$begingroup$


I've been self-studying from the amazing "Engineering Mathematics" by Stroud and Booth, and am currently learning about algebra, particularly logarithms.



There is a question which I don't understand who they've solved. Namely, I'm supposed to express the following equations without logs:



$$ln A = ln P + rn$$



The solution they provide is:



$$A = Pe^{rn}$$



But I absolutely have no idea how they got to these solutions. (I managed to "decipher" some of the similar ones piece by piece by studying the rules of logarithms).






algebra-precalculus logarithms







share|cite|improve this question











$endgroup$

















    3












    $begingroup$


    I've been self-studying from the amazing "Engineering Mathematics" by Stroud and Booth, and am currently learning about algebra, particularly logarithms.



    There is a question which I don't understand who they've solved. Namely, I'm supposed to express the following equations without logs:



    $$ln A = ln P + rn$$



    The solution they provide is:



    $$A = Pe^{rn}$$



    But I absolutely have no idea how they got to these solutions. (I managed to "decipher" some of the similar ones piece by piece by studying the rules of logarithms).






    algebra-precalculus logarithms







    share|cite|improve this question











    $endgroup$















      3












      3








      3





      $begingroup$


      I've been self-studying from the amazing "Engineering Mathematics" by Stroud and Booth, and am currently learning about algebra, particularly logarithms.



      There is a question which I don't understand who they've solved. Namely, I'm supposed to express the following equations without logs:



      $$ln A = ln P + rn$$



      The solution they provide is:



      $$A = Pe^{rn}$$



      But I absolutely have no idea how they got to these solutions. (I managed to "decipher" some of the similar ones piece by piece by studying the rules of logarithms).






      algebra-precalculus logarithms







      share|cite|improve this question











      $endgroup$




      I've been self-studying from the amazing "Engineering Mathematics" by Stroud and Booth, and am currently learning about algebra, particularly logarithms.



      There is a question which I don't understand who they've solved. Namely, I'm supposed to express the following equations without logs:



      $$ln A = ln P + rn$$



      The solution they provide is:



      $$A = Pe^{rn}$$



      But I absolutely have no idea how they got to these solutions. (I managed to "decipher" some of the similar ones piece by piece by studying the rules of logarithms).






      algebra-precalculus logarithms




      algebra-precalculus logarithms






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited 3 hours ago









      Mutantoe

      619513




      619513










      asked 5 hours ago









      neuronneuron

      1402




      1402






















          3 Answers
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          $begingroup$

          The basic idea behind all basic algebraic manipulations is that you are trying to isolate some variable or expression from the rest of the equation (that is, you are trying to "solve" for $A$ in this equation by putting it on one side of the equality by itself).



          For this particular example (and indeed, most questions involving logarithms), you will have to know that the logarithm is "invertible"; just like multiplying and dividing by the same non-zero number changes nothing, taking a logarithm and then an exponential of a positive number changes nothing.



          So, when we see $ln(A)=ln(P)+rn$, we can "undo" the logarithm by taking an exponential. However, what we do to one side must also be done to the other, so we are left with the following after recalling our basic rules of exponentiation:
          $$
          A=e^{ln(A)}=e^{ln(P)+rn}=e^{ln(P)}cdot e^{rn}=Pe^{rn}
          $$






          share|cite|improve this answer










          New contributor




          ItsJustVennDiagramsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$





















            4












            $begingroup$

            Hint: Write $$e^{ln(A)}=e^{ln(P)+rn}$$ and use that $$e^{ln(x)}=x$$






            share|cite|improve this answer









            $endgroup$





















              0












              $begingroup$

              There are a couple of things you must know about logarithms to understand that piece of mathematics. First, you need to know that $1=ln{e}$ and secondly you need to know the fact that you can slide whatever you've got in front of the logarithm up and make it an exponent on the thing that's inside the logarithm: $xln{y}=ln{y^x}$. And the last fact you're going to need is the fact that $ln{x}=ln{y}implies x=y$.



              $$
              ln{A}=ln{P}+rncdot 1\
              ln{A}=ln{P}+rncdot ln{e}\
              ln{A}=ln{P}+ln{e^{rn}}\
              ln{A}=ln{(Pe^{rn})}\
              A=Pe^{rn}
              $$






              share|cite|improve this answer









              $endgroup$













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                3 Answers
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                3 Answers
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                active

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                8












                $begingroup$

                The basic idea behind all basic algebraic manipulations is that you are trying to isolate some variable or expression from the rest of the equation (that is, you are trying to "solve" for $A$ in this equation by putting it on one side of the equality by itself).



                For this particular example (and indeed, most questions involving logarithms), you will have to know that the logarithm is "invertible"; just like multiplying and dividing by the same non-zero number changes nothing, taking a logarithm and then an exponential of a positive number changes nothing.



                So, when we see $ln(A)=ln(P)+rn$, we can "undo" the logarithm by taking an exponential. However, what we do to one side must also be done to the other, so we are left with the following after recalling our basic rules of exponentiation:
                $$
                A=e^{ln(A)}=e^{ln(P)+rn}=e^{ln(P)}cdot e^{rn}=Pe^{rn}
                $$






                share|cite|improve this answer










                New contributor




                ItsJustVennDiagramsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$


















                  8












                  $begingroup$

                  The basic idea behind all basic algebraic manipulations is that you are trying to isolate some variable or expression from the rest of the equation (that is, you are trying to "solve" for $A$ in this equation by putting it on one side of the equality by itself).



                  For this particular example (and indeed, most questions involving logarithms), you will have to know that the logarithm is "invertible"; just like multiplying and dividing by the same non-zero number changes nothing, taking a logarithm and then an exponential of a positive number changes nothing.



                  So, when we see $ln(A)=ln(P)+rn$, we can "undo" the logarithm by taking an exponential. However, what we do to one side must also be done to the other, so we are left with the following after recalling our basic rules of exponentiation:
                  $$
                  A=e^{ln(A)}=e^{ln(P)+rn}=e^{ln(P)}cdot e^{rn}=Pe^{rn}
                  $$






                  share|cite|improve this answer










                  New contributor




                  ItsJustVennDiagramsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  $endgroup$
















                    8












                    8








                    8





                    $begingroup$

                    The basic idea behind all basic algebraic manipulations is that you are trying to isolate some variable or expression from the rest of the equation (that is, you are trying to "solve" for $A$ in this equation by putting it on one side of the equality by itself).



                    For this particular example (and indeed, most questions involving logarithms), you will have to know that the logarithm is "invertible"; just like multiplying and dividing by the same non-zero number changes nothing, taking a logarithm and then an exponential of a positive number changes nothing.



                    So, when we see $ln(A)=ln(P)+rn$, we can "undo" the logarithm by taking an exponential. However, what we do to one side must also be done to the other, so we are left with the following after recalling our basic rules of exponentiation:
                    $$
                    A=e^{ln(A)}=e^{ln(P)+rn}=e^{ln(P)}cdot e^{rn}=Pe^{rn}
                    $$






                    share|cite|improve this answer










                    New contributor




                    ItsJustVennDiagramsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.






                    $endgroup$



                    The basic idea behind all basic algebraic manipulations is that you are trying to isolate some variable or expression from the rest of the equation (that is, you are trying to "solve" for $A$ in this equation by putting it on one side of the equality by itself).



                    For this particular example (and indeed, most questions involving logarithms), you will have to know that the logarithm is "invertible"; just like multiplying and dividing by the same non-zero number changes nothing, taking a logarithm and then an exponential of a positive number changes nothing.



                    So, when we see $ln(A)=ln(P)+rn$, we can "undo" the logarithm by taking an exponential. However, what we do to one side must also be done to the other, so we are left with the following after recalling our basic rules of exponentiation:
                    $$
                    A=e^{ln(A)}=e^{ln(P)+rn}=e^{ln(P)}cdot e^{rn}=Pe^{rn}
                    $$







                    share|cite|improve this answer










                    New contributor




                    ItsJustVennDiagramsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.









                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 5 hours ago





















                    New contributor




                    ItsJustVennDiagramsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    answered 5 hours ago









                    ItsJustVennDiagramsBroItsJustVennDiagramsBro

                    1213




                    1213




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                    New contributor





                    ItsJustVennDiagramsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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                    ItsJustVennDiagramsBro is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                    Check out our Code of Conduct.























                        4












                        $begingroup$

                        Hint: Write $$e^{ln(A)}=e^{ln(P)+rn}$$ and use that $$e^{ln(x)}=x$$






                        share|cite|improve this answer









                        $endgroup$


















                          4












                          $begingroup$

                          Hint: Write $$e^{ln(A)}=e^{ln(P)+rn}$$ and use that $$e^{ln(x)}=x$$






                          share|cite|improve this answer









                          $endgroup$
















                            4












                            4








                            4





                            $begingroup$

                            Hint: Write $$e^{ln(A)}=e^{ln(P)+rn}$$ and use that $$e^{ln(x)}=x$$






                            share|cite|improve this answer









                            $endgroup$



                            Hint: Write $$e^{ln(A)}=e^{ln(P)+rn}$$ and use that $$e^{ln(x)}=x$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 5 hours ago









                            Dr. Sonnhard GraubnerDr. Sonnhard Graubner

                            77.4k42866




                            77.4k42866























                                0












                                $begingroup$

                                There are a couple of things you must know about logarithms to understand that piece of mathematics. First, you need to know that $1=ln{e}$ and secondly you need to know the fact that you can slide whatever you've got in front of the logarithm up and make it an exponent on the thing that's inside the logarithm: $xln{y}=ln{y^x}$. And the last fact you're going to need is the fact that $ln{x}=ln{y}implies x=y$.



                                $$
                                ln{A}=ln{P}+rncdot 1\
                                ln{A}=ln{P}+rncdot ln{e}\
                                ln{A}=ln{P}+ln{e^{rn}}\
                                ln{A}=ln{(Pe^{rn})}\
                                A=Pe^{rn}
                                $$






                                share|cite|improve this answer









                                $endgroup$


















                                  0












                                  $begingroup$

                                  There are a couple of things you must know about logarithms to understand that piece of mathematics. First, you need to know that $1=ln{e}$ and secondly you need to know the fact that you can slide whatever you've got in front of the logarithm up and make it an exponent on the thing that's inside the logarithm: $xln{y}=ln{y^x}$. And the last fact you're going to need is the fact that $ln{x}=ln{y}implies x=y$.



                                  $$
                                  ln{A}=ln{P}+rncdot 1\
                                  ln{A}=ln{P}+rncdot ln{e}\
                                  ln{A}=ln{P}+ln{e^{rn}}\
                                  ln{A}=ln{(Pe^{rn})}\
                                  A=Pe^{rn}
                                  $$






                                  share|cite|improve this answer









                                  $endgroup$
















                                    0












                                    0








                                    0





                                    $begingroup$

                                    There are a couple of things you must know about logarithms to understand that piece of mathematics. First, you need to know that $1=ln{e}$ and secondly you need to know the fact that you can slide whatever you've got in front of the logarithm up and make it an exponent on the thing that's inside the logarithm: $xln{y}=ln{y^x}$. And the last fact you're going to need is the fact that $ln{x}=ln{y}implies x=y$.



                                    $$
                                    ln{A}=ln{P}+rncdot 1\
                                    ln{A}=ln{P}+rncdot ln{e}\
                                    ln{A}=ln{P}+ln{e^{rn}}\
                                    ln{A}=ln{(Pe^{rn})}\
                                    A=Pe^{rn}
                                    $$






                                    share|cite|improve this answer









                                    $endgroup$



                                    There are a couple of things you must know about logarithms to understand that piece of mathematics. First, you need to know that $1=ln{e}$ and secondly you need to know the fact that you can slide whatever you've got in front of the logarithm up and make it an exponent on the thing that's inside the logarithm: $xln{y}=ln{y^x}$. And the last fact you're going to need is the fact that $ln{x}=ln{y}implies x=y$.



                                    $$
                                    ln{A}=ln{P}+rncdot 1\
                                    ln{A}=ln{P}+rncdot ln{e}\
                                    ln{A}=ln{P}+ln{e^{rn}}\
                                    ln{A}=ln{(Pe^{rn})}\
                                    A=Pe^{rn}
                                    $$







                                    share|cite|improve this answer












                                    share|cite|improve this answer



                                    share|cite|improve this answer










                                    answered 5 hours ago









                                    Michael RybkinMichael Rybkin

                                    3,695420




                                    3,695420






























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                                        シテ島

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                                        シテ島 Île de la Cité シテ島（トゥルネル橋からの眺望） 所在地 フランス･パリ 座標 北緯48度51分17秒 東経2度20分45秒 座標: 北緯48度51分17秒 東経2度20分45秒 面積 0.225 km² 最高標高 48.854722 m  プロジェクト 地形 テンプレートを表示 シテ島 （シテとう、仏 : Île de la Cité）は、サン＝ルイ島と並ぶ、パリの中心部を流れるセーヌ川の中州である。パリ市内で最も古い歴史をもつパリ1区及び4区に属し、“パリ発祥の地”とも称される。ユリウス・カエサルの『ガリア戦記』では、すでに紀元前1世紀にパリシイ族 (Parisii) が住んでいたことが述べられている。 目次 1 概要 2 歴史ギャラリー 3 名所 4 シテ島の交通 4.1 橋 4.2 鉄道 5 ギャラリー 6 関連項目 7 脚注 8 外部リンク 概要 東半分が4区に、西半分は1区に属する。 ノートルダム大聖堂とパリ警視庁とパリ市立病院（ オテル・デュー・ド・パリ （フランス語版） 、直訳では「パリの神の館」）がある。病院の名前は、旅の病人を宿泊させ、病になれば看病した教会、巡礼教会からの発展の名残である。近代外科医学の開拓者で、床屋医者の出身でもある外科医アンブロワーズ・パレが勤務していた病院としてもその名が知られている。位置関係は大聖堂が警視庁庁舎と向き合い、警視庁の隣に市立病院。 また、この島はとても狭いので、ここの住民たちは夜も一定の時間を過ぎるとトイレの水は流さない等、互いに気配りしつつ一緒に住んでいくことの不便さを味わってきた。ここから「シテ島に住む者」を表す古仏語：citeain （現代仏語：シトワイヤン citoyen 、現代英語：シティズン citizen）の意味が転じ、「 市民 」という言葉が成立したという（仏語：cité は英語：city に相当する）。 中世以来からの狭く入り組んだ路地は、19世紀に入りセーヌ川左岸と共に貧民窟化していた。そうした狭く不衛生で入り組んだ空間は、2月革命に見るようなバリケード戦といった民衆騒...
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