Proof of algebraic equation

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I have been trying to prove that this expression is true, but I don't think I have an adequate grasp of the rules of logarithmic expressions. Here is the expression:



$$a^log_b c = c^log_b a$$



I understand that $a^log_a b = b$ (and vice versa), but I must be missing something.






algebra-precalculus logarithms







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  • While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
    – Brevan Ellefsen
    5 hours ago










  • I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
    – Lubin
    5 hours ago










  • HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
    – Arturo Magidin
    5 hours ago






  • 1




    @NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
    – Brevan Ellefsen
    4 hours ago







  • 1




    I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
    – Nat Porter
    4 hours ago














up vote
4
down vote

favorite












I have been trying to prove that this expression is true, but I don't think I have an adequate grasp of the rules of logarithmic expressions. Here is the expression:



$$a^log_b c = c^log_b a$$



I understand that $a^log_a b = b$ (and vice versa), but I must be missing something.






algebra-precalculus logarithms







share|cite|improve this question























  • While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
    – Brevan Ellefsen
    5 hours ago










  • I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
    – Lubin
    5 hours ago










  • HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
    – Arturo Magidin
    5 hours ago






  • 1




    @NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
    – Brevan Ellefsen
    4 hours ago







  • 1




    I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
    – Nat Porter
    4 hours ago












up vote
4
down vote

favorite









up vote
4
down vote

favorite











I have been trying to prove that this expression is true, but I don't think I have an adequate grasp of the rules of logarithmic expressions. Here is the expression:



$$a^log_b c = c^log_b a$$



I understand that $a^log_a b = b$ (and vice versa), but I must be missing something.






algebra-precalculus logarithms







share|cite|improve this question















I have been trying to prove that this expression is true, but I don't think I have an adequate grasp of the rules of logarithmic expressions. Here is the expression:



$$a^log_b c = c^log_b a$$



I understand that $a^log_a b = b$ (and vice versa), but I must be missing something.






algebra-precalculus logarithms




algebra-precalculus logarithms






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 5 hours ago









José Carlos Santos

127k17102189




127k17102189










asked 5 hours ago









Nat Porter

271




271











  • While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
    – Brevan Ellefsen
    5 hours ago










  • I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
    – Lubin
    5 hours ago










  • HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
    – Arturo Magidin
    5 hours ago






  • 1




    @NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
    – Brevan Ellefsen
    4 hours ago







  • 1




    I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
    – Nat Porter
    4 hours ago
















  • While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
    – Brevan Ellefsen
    5 hours ago










  • I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
    – Lubin
    5 hours ago










  • HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
    – Arturo Magidin
    5 hours ago






  • 1




    @NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
    – Brevan Ellefsen
    4 hours ago







  • 1




    I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
    – Nat Porter
    4 hours ago















While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
– Brevan Ellefsen
5 hours ago




While perhaps counterintuitive, I would take a logarithm on both sides to get rid of all exponents and get a product of logs which you can simplify using logarithm rules and basic algebra.
– Brevan Ellefsen
5 hours ago












I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
– Lubin
5 hours ago




I think you want to write $a=b^log_ba$ and $c=b^log_bc$.
– Lubin
5 hours ago












HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
– Arturo Magidin
5 hours ago




HINT: $r^s = b^log_b(r^s) = b^slog_b(r)$.
– Arturo Magidin
5 hours ago




1




1




@NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
– Brevan Ellefsen
4 hours ago





@NatPorter that's the right idea, but not quite correct the way you wrote it. It's more like $a^log_b c = b^log_b(c) log_b(a) $. Do that to the other side ad well and use $log_b(c) log_b(a) = log_b(a) log_b(c)$
– Brevan Ellefsen
4 hours ago





1




1




I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
– Nat Porter
4 hours ago




I'm sorry, I don't quite understand the algebra there... do you mind if I shoot you a private message?
– Nat Porter
4 hours ago










3 Answers
3






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up vote
2
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Two positive numbers are equal if and only if their logarithms (in the same base) are equal.



Compute the logarithm in base $b$ of both:



  1. $log_b(a^log_bc)=log_b clog_b a$

  2. $log_b(c^log_ba)=dotsb$

Done.






share|cite|improve this answer



























    up vote
    1
    down vote













    First Prove and convince yourself that for $x> 0; y > 0; n > 0; n ne 1$ then $x = y iff log_n x = log_n y$.



    Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $log_n x = log_n y$



    And if $log_n x = log_n y = k$ then $n^k = x$... and $n^k = y$. So $x = y$.



    (For this to be acceptable it is ESSENTIAL that you accept for all $n > 0$ and $n ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$. That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)



    .....



    Okays so $a^log_b c = c^log_b a iff log_b (a^log_b c) = log_b (c^log_b a)$.



    And $log_b (a^log_b c) = log_b clog_b a$ (because $log_n a^m = mlog_n a$).



    And $log_b (c^log_b a)=log_b alog_b c$



    And that's that.



    ....



    Another way of thinking about it is:



    $a^log_b c = (b^log_b a)^log_b c = b^log_b acdot log_b c=(b^log_b c)^log_b a = c^log_b a$.






    share|cite|improve this answer





























      up vote
      0
      down vote













      $$a^log_bc=a^fraclog_aclog_ab\ = c^frac1log_ab\=c^frac1left(fraclog blog aright)\=c^fraclog alog b\=c^log_ba$$



      The key to the answer is based on the rule that $$log_xy = fraclog_kylog_kx$$
      This is the Change-of-Base Formula.






      share|cite|improve this answer






















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        3 Answers
        3






        active

        oldest

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        3 Answers
        3






        active

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        active

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        active

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        up vote
        2
        down vote













        Two positive numbers are equal if and only if their logarithms (in the same base) are equal.



        Compute the logarithm in base $b$ of both:



        1. $log_b(a^log_bc)=log_b clog_b a$

        2. $log_b(c^log_ba)=dotsb$

        Done.






        share|cite|improve this answer
























          up vote
          2
          down vote













          Two positive numbers are equal if and only if their logarithms (in the same base) are equal.



          Compute the logarithm in base $b$ of both:



          1. $log_b(a^log_bc)=log_b clog_b a$

          2. $log_b(c^log_ba)=dotsb$

          Done.






          share|cite|improve this answer






















            up vote
            2
            down vote










            up vote
            2
            down vote









            Two positive numbers are equal if and only if their logarithms (in the same base) are equal.



            Compute the logarithm in base $b$ of both:



            1. $log_b(a^log_bc)=log_b clog_b a$

            2. $log_b(c^log_ba)=dotsb$

            Done.






            share|cite|improve this answer












            Two positive numbers are equal if and only if their logarithms (in the same base) are equal.



            Compute the logarithm in base $b$ of both:



            1. $log_b(a^log_bc)=log_b clog_b a$

            2. $log_b(c^log_ba)=dotsb$

            Done.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 4 hours ago









            egreg

            169k1283191




            169k1283191




















                up vote
                1
                down vote













                First Prove and convince yourself that for $x> 0; y > 0; n > 0; n ne 1$ then $x = y iff log_n x = log_n y$.



                Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $log_n x = log_n y$



                And if $log_n x = log_n y = k$ then $n^k = x$... and $n^k = y$. So $x = y$.



                (For this to be acceptable it is ESSENTIAL that you accept for all $n > 0$ and $n ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$. That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)



                .....



                Okays so $a^log_b c = c^log_b a iff log_b (a^log_b c) = log_b (c^log_b a)$.



                And $log_b (a^log_b c) = log_b clog_b a$ (because $log_n a^m = mlog_n a$).



                And $log_b (c^log_b a)=log_b alog_b c$



                And that's that.



                ....



                Another way of thinking about it is:



                $a^log_b c = (b^log_b a)^log_b c = b^log_b acdot log_b c=(b^log_b c)^log_b a = c^log_b a$.






                share|cite|improve this answer


























                  up vote
                  1
                  down vote













                  First Prove and convince yourself that for $x> 0; y > 0; n > 0; n ne 1$ then $x = y iff log_n x = log_n y$.



                  Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $log_n x = log_n y$



                  And if $log_n x = log_n y = k$ then $n^k = x$... and $n^k = y$. So $x = y$.



                  (For this to be acceptable it is ESSENTIAL that you accept for all $n > 0$ and $n ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$. That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)



                  .....



                  Okays so $a^log_b c = c^log_b a iff log_b (a^log_b c) = log_b (c^log_b a)$.



                  And $log_b (a^log_b c) = log_b clog_b a$ (because $log_n a^m = mlog_n a$).



                  And $log_b (c^log_b a)=log_b alog_b c$



                  And that's that.



                  ....



                  Another way of thinking about it is:



                  $a^log_b c = (b^log_b a)^log_b c = b^log_b acdot log_b c=(b^log_b c)^log_b a = c^log_b a$.






                  share|cite|improve this answer
























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    First Prove and convince yourself that for $x> 0; y > 0; n > 0; n ne 1$ then $x = y iff log_n x = log_n y$.



                    Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $log_n x = log_n y$



                    And if $log_n x = log_n y = k$ then $n^k = x$... and $n^k = y$. So $x = y$.



                    (For this to be acceptable it is ESSENTIAL that you accept for all $n > 0$ and $n ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$. That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)



                    .....



                    Okays so $a^log_b c = c^log_b a iff log_b (a^log_b c) = log_b (c^log_b a)$.



                    And $log_b (a^log_b c) = log_b clog_b a$ (because $log_n a^m = mlog_n a$).



                    And $log_b (c^log_b a)=log_b alog_b c$



                    And that's that.



                    ....



                    Another way of thinking about it is:



                    $a^log_b c = (b^log_b a)^log_b c = b^log_b acdot log_b c=(b^log_b c)^log_b a = c^log_b a$.






                    share|cite|improve this answer














                    First Prove and convince yourself that for $x> 0; y > 0; n > 0; n ne 1$ then $x = y iff log_n x = log_n y$.



                    Pf: If $x = y$ then $f(x) = f(y)$ for all functions $f$ so $log_n x = log_n y$



                    And if $log_n x = log_n y = k$ then $n^k = x$... and $n^k = y$. So $x = y$.



                    (For this to be acceptable it is ESSENTIAL that you accept for all $n > 0$ and $n ne 1$ and $x$ then there does exist a unique $k$ so that $n^k = x$. That's actually not trivial and should not be taken as obvious but it is essential that this be shown and verified for the mere definition of logarithms to even make sense.)



                    .....



                    Okays so $a^log_b c = c^log_b a iff log_b (a^log_b c) = log_b (c^log_b a)$.



                    And $log_b (a^log_b c) = log_b clog_b a$ (because $log_n a^m = mlog_n a$).



                    And $log_b (c^log_b a)=log_b alog_b c$



                    And that's that.



                    ....



                    Another way of thinking about it is:



                    $a^log_b c = (b^log_b a)^log_b c = b^log_b acdot log_b c=(b^log_b c)^log_b a = c^log_b a$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 4 hours ago

























                    answered 4 hours ago









                    fleablood

                    63k22679




                    63k22679




















                        up vote
                        0
                        down vote













                        $$a^log_bc=a^fraclog_aclog_ab\ = c^frac1log_ab\=c^frac1left(fraclog blog aright)\=c^fraclog alog b\=c^log_ba$$



                        The key to the answer is based on the rule that $$log_xy = fraclog_kylog_kx$$
                        This is the Change-of-Base Formula.






                        share|cite|improve this answer


























                          up vote
                          0
                          down vote













                          $$a^log_bc=a^fraclog_aclog_ab\ = c^frac1log_ab\=c^frac1left(fraclog blog aright)\=c^fraclog alog b\=c^log_ba$$



                          The key to the answer is based on the rule that $$log_xy = fraclog_kylog_kx$$
                          This is the Change-of-Base Formula.






                          share|cite|improve this answer
























                            up vote
                            0
                            down vote










                            up vote
                            0
                            down vote









                            $$a^log_bc=a^fraclog_aclog_ab\ = c^frac1log_ab\=c^frac1left(fraclog blog aright)\=c^fraclog alog b\=c^log_ba$$



                            The key to the answer is based on the rule that $$log_xy = fraclog_kylog_kx$$
                            This is the Change-of-Base Formula.






                            share|cite|improve this answer














                            $$a^log_bc=a^fraclog_aclog_ab\ = c^frac1log_ab\=c^frac1left(fraclog blog aright)\=c^fraclog alog b\=c^log_ba$$



                            The key to the answer is based on the rule that $$log_xy = fraclog_kylog_kx$$
                            This is the Change-of-Base Formula.







                            share|cite|improve this answer














                            share|cite|improve this answer



                            share|cite|improve this answer








                            edited 4 hours ago

























                            answered 4 hours ago









                            Larry

                            622318




                            622318



























                                 

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