Can i perform this step?











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












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

435




435








  • 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

votes

















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





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009488%2fcan-i-perform-this-step%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    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













      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


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3009488%2fcan-i-perform-this-step%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Quarter-circle Tiles

            Image
            up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2
            Read more

            build a pushdown automaton that recognizes the reverse language of a given pushdown automaton?

            Image
            0 I think it's similiar to NFAs. I replace the finite states of the given automaton for start-states for my new automaton. I do it with $epsilon$ -transition from the start state to the actual finite states of the given automaton. The start state of the given automaton will be my accepting(finite) state of my new one. All transitions will be reversed of course to recognize the reversed word. But how do I have to change the stack of the pushdown automaton? That's tricky. formal-languages automata context-free-grammar share | cite | improve this question edited Nov 29 '18 at 14:20 joee
            Read more

            Mont Emei

            Image
            Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog
            Read more