Why can't I find the value of x using logarithms?

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This is concerning a question in stack exchange : Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^x^2+4x-60=1$. I was actually wondering why the correct result is not obtained when applying log on both sides,like how the second answer does.Why is this so? Is it possible to get this answer using logarithms?






algebra-precalculus proof-writing logarithms exponential-function







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    up vote
    2
    down vote

    favorite












    This is concerning a question in stack exchange : Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^x^2+4x-60=1$. I was actually wondering why the correct result is not obtained when applying log on both sides,like how the second answer does.Why is this so? Is it possible to get this answer using logarithms?






    algebra-precalculus proof-writing logarithms exponential-function







    share|cite|improve this question

























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      This is concerning a question in stack exchange : Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^x^2+4x-60=1$. I was actually wondering why the correct result is not obtained when applying log on both sides,like how the second answer does.Why is this so? Is it possible to get this answer using logarithms?






      algebra-precalculus proof-writing logarithms exponential-function







      share|cite|improve this question















      This is concerning a question in stack exchange : Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^x^2+4x-60=1$. I was actually wondering why the correct result is not obtained when applying log on both sides,like how the second answer does.Why is this so? Is it possible to get this answer using logarithms?






      algebra-precalculus proof-writing logarithms exponential-function




      algebra-precalculus proof-writing logarithms exponential-function






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 14 mins ago









      Michael Rozenberg

      92.9k1585184




      92.9k1585184










      asked 1 hour ago









      Hema

      5611113




      5611113




















          2 Answers
          2






          active

          oldest

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          up vote
          2
          down vote



          accepted










          The solution you have mentioned use indeed logarithm, that is for $A>0$



          $$A^B=1 iff log A^B=log 1 iff Bcdot log A=0 iff B=0 quad lor quad log A=0$$



          Note that usually $A^B$ is (well-)defined only for $A>0$ but we could also extend the solutions to the case



          • $A=-1$


          • $B=2k$ with $kin mathbbZ$

          and include also the case $x=2$ among the solutions since $$(-1)^-52=frac1(-1)^52=1$$



          To summarize in order to address your question:




          • $A^B=1$ can be solved by logarithm only for the case $A>0$


          • $A^B=1$ has also one "special" solution for $Ale 0$ which cannot be find by logarithm but directly by the given conditions





          share|cite|improve this answer





























            up vote
            4
            down vote













            Yes, you are right! We can use logarithm here:
            $$(x^2+4x-60)ln(x^2-5x+5)=0,$$ which gives
            $$x^2-5x+5=1$$ or
            $$x^2+4x-60=0.$$
            By the way, I think the accepted answer on your link is total wrong because if we wrote $$(x^2-5x+5)^x^2+4x-60$$ then $x^2-5x+5>0$ by definition.






            share|cite|improve this answer




















            • Can you please explain '$x^2-5x+5 > 0$ by definition'?
              – Toby Mak
              1 hour ago










            • @Michael Maybe we could also extend the solution to the case A=-1 B=2k.What your thought about that?
              – gimusi
              1 hour ago










            • @Toby Mak If we want to work with expressions like $left(f(x)right)^g(x)$ then we need $f(x)>0.$
              – Michael Rozenberg
              1 hour ago










            • @gimusi Try to calculate $limlimits_xrightarrow2(x^2-5x+5)^x^2+4x-60.$ I think now you'll understand me.
              – Michael Rozenberg
              1 hour ago






            • 1




              @MichaelRozenberg Indeed it is math and $(-1)^-52=1$ wolfy. Just to clarify, I'm not claiming that you are wrong, I'm claiming that other ways cannot be considered absolutely wrong giving the correct context. I think we can agree about that and (+1) for the nice discussion :)
              – gimusi
              24 mins ago











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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            2
            down vote



            accepted










            The solution you have mentioned use indeed logarithm, that is for $A>0$



            $$A^B=1 iff log A^B=log 1 iff Bcdot log A=0 iff B=0 quad lor quad log A=0$$



            Note that usually $A^B$ is (well-)defined only for $A>0$ but we could also extend the solutions to the case



            • $A=-1$


            • $B=2k$ with $kin mathbbZ$

            and include also the case $x=2$ among the solutions since $$(-1)^-52=frac1(-1)^52=1$$



            To summarize in order to address your question:




            • $A^B=1$ can be solved by logarithm only for the case $A>0$


            • $A^B=1$ has also one "special" solution for $Ale 0$ which cannot be find by logarithm but directly by the given conditions





            share|cite|improve this answer


























              up vote
              2
              down vote



              accepted










              The solution you have mentioned use indeed logarithm, that is for $A>0$



              $$A^B=1 iff log A^B=log 1 iff Bcdot log A=0 iff B=0 quad lor quad log A=0$$



              Note that usually $A^B$ is (well-)defined only for $A>0$ but we could also extend the solutions to the case



              • $A=-1$


              • $B=2k$ with $kin mathbbZ$

              and include also the case $x=2$ among the solutions since $$(-1)^-52=frac1(-1)^52=1$$



              To summarize in order to address your question:




              • $A^B=1$ can be solved by logarithm only for the case $A>0$


              • $A^B=1$ has also one "special" solution for $Ale 0$ which cannot be find by logarithm but directly by the given conditions





              share|cite|improve this answer
























                up vote
                2
                down vote



                accepted







                up vote
                2
                down vote



                accepted






                The solution you have mentioned use indeed logarithm, that is for $A>0$



                $$A^B=1 iff log A^B=log 1 iff Bcdot log A=0 iff B=0 quad lor quad log A=0$$



                Note that usually $A^B$ is (well-)defined only for $A>0$ but we could also extend the solutions to the case



                • $A=-1$


                • $B=2k$ with $kin mathbbZ$

                and include also the case $x=2$ among the solutions since $$(-1)^-52=frac1(-1)^52=1$$



                To summarize in order to address your question:




                • $A^B=1$ can be solved by logarithm only for the case $A>0$


                • $A^B=1$ has also one "special" solution for $Ale 0$ which cannot be find by logarithm but directly by the given conditions





                share|cite|improve this answer














                The solution you have mentioned use indeed logarithm, that is for $A>0$



                $$A^B=1 iff log A^B=log 1 iff Bcdot log A=0 iff B=0 quad lor quad log A=0$$



                Note that usually $A^B$ is (well-)defined only for $A>0$ but we could also extend the solutions to the case



                • $A=-1$


                • $B=2k$ with $kin mathbbZ$

                and include also the case $x=2$ among the solutions since $$(-1)^-52=frac1(-1)^52=1$$



                To summarize in order to address your question:




                • $A^B=1$ can be solved by logarithm only for the case $A>0$


                • $A^B=1$ has also one "special" solution for $Ale 0$ which cannot be find by logarithm but directly by the given conditions






                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 29 mins ago

























                answered 1 hour ago









                gimusi

                79.5k73990




                79.5k73990




















                    up vote
                    4
                    down vote













                    Yes, you are right! We can use logarithm here:
                    $$(x^2+4x-60)ln(x^2-5x+5)=0,$$ which gives
                    $$x^2-5x+5=1$$ or
                    $$x^2+4x-60=0.$$
                    By the way, I think the accepted answer on your link is total wrong because if we wrote $$(x^2-5x+5)^x^2+4x-60$$ then $x^2-5x+5>0$ by definition.






                    share|cite|improve this answer




















                    • Can you please explain '$x^2-5x+5 > 0$ by definition'?
                      – Toby Mak
                      1 hour ago










                    • @Michael Maybe we could also extend the solution to the case A=-1 B=2k.What your thought about that?
                      – gimusi
                      1 hour ago










                    • @Toby Mak If we want to work with expressions like $left(f(x)right)^g(x)$ then we need $f(x)>0.$
                      – Michael Rozenberg
                      1 hour ago










                    • @gimusi Try to calculate $limlimits_xrightarrow2(x^2-5x+5)^x^2+4x-60.$ I think now you'll understand me.
                      – Michael Rozenberg
                      1 hour ago






                    • 1




                      @MichaelRozenberg Indeed it is math and $(-1)^-52=1$ wolfy. Just to clarify, I'm not claiming that you are wrong, I'm claiming that other ways cannot be considered absolutely wrong giving the correct context. I think we can agree about that and (+1) for the nice discussion :)
                      – gimusi
                      24 mins ago















                    up vote
                    4
                    down vote













                    Yes, you are right! We can use logarithm here:
                    $$(x^2+4x-60)ln(x^2-5x+5)=0,$$ which gives
                    $$x^2-5x+5=1$$ or
                    $$x^2+4x-60=0.$$
                    By the way, I think the accepted answer on your link is total wrong because if we wrote $$(x^2-5x+5)^x^2+4x-60$$ then $x^2-5x+5>0$ by definition.






                    share|cite|improve this answer




















                    • Can you please explain '$x^2-5x+5 > 0$ by definition'?
                      – Toby Mak
                      1 hour ago










                    • @Michael Maybe we could also extend the solution to the case A=-1 B=2k.What your thought about that?
                      – gimusi
                      1 hour ago










                    • @Toby Mak If we want to work with expressions like $left(f(x)right)^g(x)$ then we need $f(x)>0.$
                      – Michael Rozenberg
                      1 hour ago










                    • @gimusi Try to calculate $limlimits_xrightarrow2(x^2-5x+5)^x^2+4x-60.$ I think now you'll understand me.
                      – Michael Rozenberg
                      1 hour ago






                    • 1




                      @MichaelRozenberg Indeed it is math and $(-1)^-52=1$ wolfy. Just to clarify, I'm not claiming that you are wrong, I'm claiming that other ways cannot be considered absolutely wrong giving the correct context. I think we can agree about that and (+1) for the nice discussion :)
                      – gimusi
                      24 mins ago













                    up vote
                    4
                    down vote










                    up vote
                    4
                    down vote









                    Yes, you are right! We can use logarithm here:
                    $$(x^2+4x-60)ln(x^2-5x+5)=0,$$ which gives
                    $$x^2-5x+5=1$$ or
                    $$x^2+4x-60=0.$$
                    By the way, I think the accepted answer on your link is total wrong because if we wrote $$(x^2-5x+5)^x^2+4x-60$$ then $x^2-5x+5>0$ by definition.






                    share|cite|improve this answer












                    Yes, you are right! We can use logarithm here:
                    $$(x^2+4x-60)ln(x^2-5x+5)=0,$$ which gives
                    $$x^2-5x+5=1$$ or
                    $$x^2+4x-60=0.$$
                    By the way, I think the accepted answer on your link is total wrong because if we wrote $$(x^2-5x+5)^x^2+4x-60$$ then $x^2-5x+5>0$ by definition.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered 1 hour ago









                    Michael Rozenberg

                    92.9k1585184




                    92.9k1585184











                    • Can you please explain '$x^2-5x+5 > 0$ by definition'?
                      – Toby Mak
                      1 hour ago










                    • @Michael Maybe we could also extend the solution to the case A=-1 B=2k.What your thought about that?
                      – gimusi
                      1 hour ago










                    • @Toby Mak If we want to work with expressions like $left(f(x)right)^g(x)$ then we need $f(x)>0.$
                      – Michael Rozenberg
                      1 hour ago










                    • @gimusi Try to calculate $limlimits_xrightarrow2(x^2-5x+5)^x^2+4x-60.$ I think now you'll understand me.
                      – Michael Rozenberg
                      1 hour ago






                    • 1




                      @MichaelRozenberg Indeed it is math and $(-1)^-52=1$ wolfy. Just to clarify, I'm not claiming that you are wrong, I'm claiming that other ways cannot be considered absolutely wrong giving the correct context. I think we can agree about that and (+1) for the nice discussion :)
                      – gimusi
                      24 mins ago

















                    • Can you please explain '$x^2-5x+5 > 0$ by definition'?
                      – Toby Mak
                      1 hour ago










                    • @Michael Maybe we could also extend the solution to the case A=-1 B=2k.What your thought about that?
                      – gimusi
                      1 hour ago










                    • @Toby Mak If we want to work with expressions like $left(f(x)right)^g(x)$ then we need $f(x)>0.$
                      – Michael Rozenberg
                      1 hour ago










                    • @gimusi Try to calculate $limlimits_xrightarrow2(x^2-5x+5)^x^2+4x-60.$ I think now you'll understand me.
                      – Michael Rozenberg
                      1 hour ago






                    • 1




                      @MichaelRozenberg Indeed it is math and $(-1)^-52=1$ wolfy. Just to clarify, I'm not claiming that you are wrong, I'm claiming that other ways cannot be considered absolutely wrong giving the correct context. I think we can agree about that and (+1) for the nice discussion :)
                      – gimusi
                      24 mins ago
















                    Can you please explain '$x^2-5x+5 > 0$ by definition'?
                    – Toby Mak
                    1 hour ago




                    Can you please explain '$x^2-5x+5 > 0$ by definition'?
                    – Toby Mak
                    1 hour ago












                    @Michael Maybe we could also extend the solution to the case A=-1 B=2k.What your thought about that?
                    – gimusi
                    1 hour ago




                    @Michael Maybe we could also extend the solution to the case A=-1 B=2k.What your thought about that?
                    – gimusi
                    1 hour ago












                    @Toby Mak If we want to work with expressions like $left(f(x)right)^g(x)$ then we need $f(x)>0.$
                    – Michael Rozenberg
                    1 hour ago




                    @Toby Mak If we want to work with expressions like $left(f(x)right)^g(x)$ then we need $f(x)>0.$
                    – Michael Rozenberg
                    1 hour ago












                    @gimusi Try to calculate $limlimits_xrightarrow2(x^2-5x+5)^x^2+4x-60.$ I think now you'll understand me.
                    – Michael Rozenberg
                    1 hour ago




                    @gimusi Try to calculate $limlimits_xrightarrow2(x^2-5x+5)^x^2+4x-60.$ I think now you'll understand me.
                    – Michael Rozenberg
                    1 hour ago




                    1




                    1




                    @MichaelRozenberg Indeed it is math and $(-1)^-52=1$ wolfy. Just to clarify, I'm not claiming that you are wrong, I'm claiming that other ways cannot be considered absolutely wrong giving the correct context. I think we can agree about that and (+1) for the nice discussion :)
                    – gimusi
                    24 mins ago





                    @MichaelRozenberg Indeed it is math and $(-1)^-52=1$ wolfy. Just to clarify, I'm not claiming that you are wrong, I'm claiming that other ways cannot be considered absolutely wrong giving the correct context. I think we can agree about that and (+1) for the nice discussion :)
                    – gimusi
                    24 mins ago


















                     

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