Can i perform this step?












2












$begingroup$



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 real numbers, I believe, I cannot perform above solution. Am I right?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




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






  • 3




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






  • 1




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


















2












$begingroup$



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 real numbers, I believe, I cannot perform above solution. Am I right?






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 2




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






  • 3




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






  • 1




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
















2












2








2





$begingroup$



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 real numbers, I believe, I cannot perform above solution. Am I right?






algebra-precalculus







share|cite|improve this question











$endgroup$





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 real 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








edited Jan 21 at 14:53







Eldar Rahimli

















asked Nov 22 '18 at 18:36









Eldar RahimliEldar Rahimli

1679




1679








  • 2




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






  • 3




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






  • 1




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
















  • 2




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






  • 3




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






  • 1




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










2




2




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




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




3




3




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




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




1




1




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






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












2 Answers
2






active

oldest

votes


















1












$begingroup$

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









$endgroup$





















    -2












    $begingroup$

    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









    $endgroup$









    • 4




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










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






    • 1




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











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    2 Answers
    2






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    2 Answers
    2






    active

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    active

    oldest

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    active

    oldest

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    1












    $begingroup$

    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









    $endgroup$


















      1












      $begingroup$

      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









      $endgroup$
















        1












        1








        1





        $begingroup$

        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









        $endgroup$



        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 '18 at 20:33









        ServaesServaes

        24k33893




        24k33893























            -2












            $begingroup$

            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









            $endgroup$









            • 4




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










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






            • 1




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
















            -2












            $begingroup$

            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









            $endgroup$









            • 4




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










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






            • 1




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














            -2












            -2








            -2





            $begingroup$

            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









            $endgroup$



            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 '18 at 18:43









            Noone NooneNoone Noone

            11




            11








            • 4




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










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






            • 1




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














            • 4




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










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






            • 1




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








            4




            4




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




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












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




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




            1




            1




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




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


















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