Can i perform this step?











up vote
2
down vote

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a,b $in$ R



$${5^{a-b}}cdot{27^{-(a+b)}}={5^1}cdot {27^1} Rightarrow a^{2}-b^{2}=?$$



My Solution:



$a-b=1$



$a+b=-1$ then $a=0$ ; $b=-1$



Since $a$ and $b$ are not whole numbers, I believe, I cannot perform above solution. Am I right?






algebra-precalculus







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  • 2




    You are right, you cannot do that step since $a$ and $b$ are real numbers.
    – Anurag A
    Nov 22 at 18:41






  • 3




    @idea that’s not true. $5^x=7$ has a solution $x=log_57$.
    – Anurag A
    Nov 22 at 18:43






  • 1




    To debug an incorrect argument replace $27$ by $25 = 5^2$ and notice what goes wrong when you apply the same argument.
    – Bill Dubuque
    Nov 22 at 18:51

















up vote
2
down vote

favorite












a,b $in$ R



$${5^{a-b}}cdot{27^{-(a+b)}}={5^1}cdot {27^1} Rightarrow a^{2}-b^{2}=?$$



My Solution:



$a-b=1$



$a+b=-1$ then $a=0$ ; $b=-1$



Since $a$ and $b$ are not whole numbers, I believe, I cannot perform above solution. Am I right?






algebra-precalculus







share|cite|improve this question


















  • 2




    You are right, you cannot do that step since $a$ and $b$ are real numbers.
    – Anurag A
    Nov 22 at 18:41






  • 3




    @idea that’s not true. $5^x=7$ has a solution $x=log_57$.
    – Anurag A
    Nov 22 at 18:43






  • 1




    To debug an incorrect argument replace $27$ by $25 = 5^2$ and notice what goes wrong when you apply the same argument.
    – Bill Dubuque
    Nov 22 at 18:51















up vote
2
down vote

favorite









up vote
2
down vote

favorite











a,b $in$ R



$${5^{a-b}}cdot{27^{-(a+b)}}={5^1}cdot {27^1} Rightarrow a^{2}-b^{2}=?$$



My Solution:



$a-b=1$



$a+b=-1$ then $a=0$ ; $b=-1$



Since $a$ and $b$ are not whole numbers, I believe, I cannot perform above solution. Am I right?






algebra-precalculus







share|cite|improve this question













a,b $in$ R



$${5^{a-b}}cdot{27^{-(a+b)}}={5^1}cdot {27^1} Rightarrow a^{2}-b^{2}=?$$



My Solution:



$a-b=1$



$a+b=-1$ then $a=0$ ; $b=-1$



Since $a$ and $b$ are not whole numbers, I believe, I cannot perform above solution. Am I right?






algebra-precalculus




algebra-precalculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 18:36









Eldar Rahimli

385




385








  • 2




    You are right, you cannot do that step since $a$ and $b$ are real numbers.
    – Anurag A
    Nov 22 at 18:41






  • 3




    @idea that’s not true. $5^x=7$ has a solution $x=log_57$.
    – Anurag A
    Nov 22 at 18:43






  • 1




    To debug an incorrect argument replace $27$ by $25 = 5^2$ and notice what goes wrong when you apply the same argument.
    – Bill Dubuque
    Nov 22 at 18:51
















  • 2




    You are right, you cannot do that step since $a$ and $b$ are real numbers.
    – Anurag A
    Nov 22 at 18:41






  • 3




    @idea that’s not true. $5^x=7$ has a solution $x=log_57$.
    – Anurag A
    Nov 22 at 18:43






  • 1




    To debug an incorrect argument replace $27$ by $25 = 5^2$ and notice what goes wrong when you apply the same argument.
    – Bill Dubuque
    Nov 22 at 18:51










2




2




You are right, you cannot do that step since $a$ and $b$ are real numbers.
– Anurag A
Nov 22 at 18:41




You are right, you cannot do that step since $a$ and $b$ are real numbers.
– Anurag A
Nov 22 at 18:41




3




3




@idea that’s not true. $5^x=7$ has a solution $x=log_57$.
– Anurag A
Nov 22 at 18:43




@idea that’s not true. $5^x=7$ has a solution $x=log_57$.
– Anurag A
Nov 22 at 18:43




1




1




To debug an incorrect argument replace $27$ by $25 = 5^2$ and notice what goes wrong when you apply the same argument.
– Bill Dubuque
Nov 22 at 18:51






To debug an incorrect argument replace $27$ by $25 = 5^2$ and notice what goes wrong when you apply the same argument.
– Bill Dubuque
Nov 22 at 18:51












2 Answers
2






active

oldest

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up vote
0
down vote













Note that for $(a,b)=(0,-1)$ we have
$$5^{a-b}cdot27^{-(a+b)}=5^1cdot27^1,$$
and $a^2-b^2=-1$, and that for $(a,b)=left(frac{log(5^1cdot27^1)}{log(5/27)},0right)$ we have
$$5^{a-b}cdot27^{-(a+b)}=5^acdot27^{-a}=left(frac{5}{27}right)^{frac{log(5^1cdot27^1)}{log(5/27)}}=5^1cdot27^1,$$
and $a^2-b^2=left(frac{log(5^1cdot27^1)}{log(5/27)}right)^2geq0$. So the value of $a^2-b^2$ is not uniquely determined by the given equation. In particular, this shows that the step you took is not valid.






share|cite|improve this answer




























    up vote
    -2
    down vote













    In the above equation a and b are not whole numbers but real numbers
    $a-b=1$



    $a+b=-1$



    We know that
    $a^2-b^2=(a+b)(a-b)$



    $a^2-b^2=(-1)(1)$
    $a^2-b^2=-1$






    share|cite|improve this answer

















    • 4




      This is incorrect, see my comment.
      – Anurag A
      Nov 22 at 18:43










    • This is correct answer you r right about powers but still
      – Noone Noone
      Nov 22 at 18:46






    • 1




      This answer is incorrect - see the comments on the question.
      – Bill Dubuque
      Nov 22 at 18:54











    Your Answer





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    2 Answers
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    2 Answers
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    up vote
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    down vote













    Note that for $(a,b)=(0,-1)$ we have
    $$5^{a-b}cdot27^{-(a+b)}=5^1cdot27^1,$$
    and $a^2-b^2=-1$, and that for $(a,b)=left(frac{log(5^1cdot27^1)}{log(5/27)},0right)$ we have
    $$5^{a-b}cdot27^{-(a+b)}=5^acdot27^{-a}=left(frac{5}{27}right)^{frac{log(5^1cdot27^1)}{log(5/27)}}=5^1cdot27^1,$$
    and $a^2-b^2=left(frac{log(5^1cdot27^1)}{log(5/27)}right)^2geq0$. So the value of $a^2-b^2$ is not uniquely determined by the given equation. In particular, this shows that the step you took is not valid.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Note that for $(a,b)=(0,-1)$ we have
      $$5^{a-b}cdot27^{-(a+b)}=5^1cdot27^1,$$
      and $a^2-b^2=-1$, and that for $(a,b)=left(frac{log(5^1cdot27^1)}{log(5/27)},0right)$ we have
      $$5^{a-b}cdot27^{-(a+b)}=5^acdot27^{-a}=left(frac{5}{27}right)^{frac{log(5^1cdot27^1)}{log(5/27)}}=5^1cdot27^1,$$
      and $a^2-b^2=left(frac{log(5^1cdot27^1)}{log(5/27)}right)^2geq0$. So the value of $a^2-b^2$ is not uniquely determined by the given equation. In particular, this shows that the step you took is not valid.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Note that for $(a,b)=(0,-1)$ we have
        $$5^{a-b}cdot27^{-(a+b)}=5^1cdot27^1,$$
        and $a^2-b^2=-1$, and that for $(a,b)=left(frac{log(5^1cdot27^1)}{log(5/27)},0right)$ we have
        $$5^{a-b}cdot27^{-(a+b)}=5^acdot27^{-a}=left(frac{5}{27}right)^{frac{log(5^1cdot27^1)}{log(5/27)}}=5^1cdot27^1,$$
        and $a^2-b^2=left(frac{log(5^1cdot27^1)}{log(5/27)}right)^2geq0$. So the value of $a^2-b^2$ is not uniquely determined by the given equation. In particular, this shows that the step you took is not valid.






        share|cite|improve this answer












        Note that for $(a,b)=(0,-1)$ we have
        $$5^{a-b}cdot27^{-(a+b)}=5^1cdot27^1,$$
        and $a^2-b^2=-1$, and that for $(a,b)=left(frac{log(5^1cdot27^1)}{log(5/27)},0right)$ we have
        $$5^{a-b}cdot27^{-(a+b)}=5^acdot27^{-a}=left(frac{5}{27}right)^{frac{log(5^1cdot27^1)}{log(5/27)}}=5^1cdot27^1,$$
        and $a^2-b^2=left(frac{log(5^1cdot27^1)}{log(5/27)}right)^2geq0$. So the value of $a^2-b^2$ is not uniquely determined by the given equation. In particular, this shows that the step you took is not valid.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 22 at 20:33









        Servaes

        22.2k33793




        22.2k33793






















            up vote
            -2
            down vote













            In the above equation a and b are not whole numbers but real numbers
            $a-b=1$



            $a+b=-1$



            We know that
            $a^2-b^2=(a+b)(a-b)$



            $a^2-b^2=(-1)(1)$
            $a^2-b^2=-1$






            share|cite|improve this answer

















            • 4




              This is incorrect, see my comment.
              – Anurag A
              Nov 22 at 18:43










            • This is correct answer you r right about powers but still
              – Noone Noone
              Nov 22 at 18:46






            • 1




              This answer is incorrect - see the comments on the question.
              – Bill Dubuque
              Nov 22 at 18:54















            up vote
            -2
            down vote













            In the above equation a and b are not whole numbers but real numbers
            $a-b=1$



            $a+b=-1$



            We know that
            $a^2-b^2=(a+b)(a-b)$



            $a^2-b^2=(-1)(1)$
            $a^2-b^2=-1$






            share|cite|improve this answer

















            • 4




              This is incorrect, see my comment.
              – Anurag A
              Nov 22 at 18:43










            • This is correct answer you r right about powers but still
              – Noone Noone
              Nov 22 at 18:46






            • 1




              This answer is incorrect - see the comments on the question.
              – Bill Dubuque
              Nov 22 at 18:54













            up vote
            -2
            down vote










            up vote
            -2
            down vote









            In the above equation a and b are not whole numbers but real numbers
            $a-b=1$



            $a+b=-1$



            We know that
            $a^2-b^2=(a+b)(a-b)$



            $a^2-b^2=(-1)(1)$
            $a^2-b^2=-1$






            share|cite|improve this answer












            In the above equation a and b are not whole numbers but real numbers
            $a-b=1$



            $a+b=-1$



            We know that
            $a^2-b^2=(a+b)(a-b)$



            $a^2-b^2=(-1)(1)$
            $a^2-b^2=-1$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 22 at 18:43









            Noone Noone

            11




            11








            • 4




              This is incorrect, see my comment.
              – Anurag A
              Nov 22 at 18:43










            • This is correct answer you r right about powers but still
              – Noone Noone
              Nov 22 at 18:46






            • 1




              This answer is incorrect - see the comments on the question.
              – Bill Dubuque
              Nov 22 at 18:54














            • 4




              This is incorrect, see my comment.
              – Anurag A
              Nov 22 at 18:43










            • This is correct answer you r right about powers but still
              – Noone Noone
              Nov 22 at 18:46






            • 1




              This answer is incorrect - see the comments on the question.
              – Bill Dubuque
              Nov 22 at 18:54








            4




            4




            This is incorrect, see my comment.
            – Anurag A
            Nov 22 at 18:43




            This is incorrect, see my comment.
            – Anurag A
            Nov 22 at 18:43












            This is correct answer you r right about powers but still
            – Noone Noone
            Nov 22 at 18:46




            This is correct answer you r right about powers but still
            – Noone Noone
            Nov 22 at 18:46




            1




            1




            This answer is incorrect - see the comments on the question.
            – Bill Dubuque
            Nov 22 at 18:54




            This answer is incorrect - see the comments on the question.
            – Bill Dubuque
            Nov 22 at 18:54


















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