Expressing logarithmic equations without logs












4












$begingroup$


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











$endgroup$

















    4












    $begingroup$


    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











    $endgroup$















      4












      4








      4





      $begingroup$


      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













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      edited 4 hours ago









      Mutantoe

      619513




      619513










      asked 6 hours ago









      neuronneuron

      1452




      1452






















          4 Answers
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          11












          $begingroup$

          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









              $endgroup$





















                0












                $begingroup$

                A key intuition behind logarithms is that multiplication translates to addition, i.e.



                $ln(A*B)=ln(A)+ln(P)qquad$ and
                $qquadln(A/B)=ln(A)-ln(P)$



                We can use this to solve your equation



                $begin{align}
                ln(A)&=ln(P)+rnnewline
                ln(A)-ln(P)&=rnnewline
                ln(A/P)&=rnnewline
                A/P&=e^{rn}newline
                A&=Pe^{rn}
                end{align}$






                share|cite|improve this answer








                New contributor




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






                $endgroup$













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                  4 Answers
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                  11












                  $begingroup$

                  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.






                  $endgroup$


















                    11












                    $begingroup$

                    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.






                    $endgroup$
















                      11












                      11








                      11





                      $begingroup$

                      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




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                      answered 5 hours ago









                      ItsJustVennDiagramsBroItsJustVennDiagramsBro

                      1514




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                          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

                                      3,710420




                                      3,710420























                                          0












                                          $begingroup$

                                          A key intuition behind logarithms is that multiplication translates to addition, i.e.



                                          $ln(A*B)=ln(A)+ln(P)qquad$ and
                                          $qquadln(A/B)=ln(A)-ln(P)$



                                          We can use this to solve your equation



                                          $begin{align}
                                          ln(A)&=ln(P)+rnnewline
                                          ln(A)-ln(P)&=rnnewline
                                          ln(A/P)&=rnnewline
                                          A/P&=e^{rn}newline
                                          A&=Pe^{rn}
                                          end{align}$






                                          share|cite|improve this answer








                                          New contributor




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






                                          $endgroup$


















                                            0












                                            $begingroup$

                                            A key intuition behind logarithms is that multiplication translates to addition, i.e.



                                            $ln(A*B)=ln(A)+ln(P)qquad$ and
                                            $qquadln(A/B)=ln(A)-ln(P)$



                                            We can use this to solve your equation



                                            $begin{align}
                                            ln(A)&=ln(P)+rnnewline
                                            ln(A)-ln(P)&=rnnewline
                                            ln(A/P)&=rnnewline
                                            A/P&=e^{rn}newline
                                            A&=Pe^{rn}
                                            end{align}$






                                            share|cite|improve this answer








                                            New contributor




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






                                            $endgroup$
















                                              0












                                              0








                                              0





                                              $begingroup$

                                              A key intuition behind logarithms is that multiplication translates to addition, i.e.



                                              $ln(A*B)=ln(A)+ln(P)qquad$ and
                                              $qquadln(A/B)=ln(A)-ln(P)$



                                              We can use this to solve your equation



                                              $begin{align}
                                              ln(A)&=ln(P)+rnnewline
                                              ln(A)-ln(P)&=rnnewline
                                              ln(A/P)&=rnnewline
                                              A/P&=e^{rn}newline
                                              A&=Pe^{rn}
                                              end{align}$






                                              share|cite|improve this answer








                                              New contributor




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






                                              $endgroup$



                                              A key intuition behind logarithms is that multiplication translates to addition, i.e.



                                              $ln(A*B)=ln(A)+ln(P)qquad$ and
                                              $qquadln(A/B)=ln(A)-ln(P)$



                                              We can use this to solve your equation



                                              $begin{align}
                                              ln(A)&=ln(P)+rnnewline
                                              ln(A)-ln(P)&=rnnewline
                                              ln(A/P)&=rnnewline
                                              A/P&=e^{rn}newline
                                              A&=Pe^{rn}
                                              end{align}$







                                              share|cite|improve this answer








                                              New contributor




                                              t_d_milan 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






                                              New contributor




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









                                              answered 34 mins ago









                                              t_d_milant_d_milan

                                              1




                                              1




                                              New contributor




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                                              New contributor





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






                                              t_d_milan is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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