Smallest integer power for an inequality to hold












1












$begingroup$


so I have this inequality:
Given integers $m, kgeq1$.



$$2^{m/k} > frac{3}{2}$$



I'm interested in finding the smallest integer power $m$, as a function of $k$, that will make this inequality hold true.



I've made a few calculations and came up with $m$ as:
$$m(k) = [ 6k/10 + 0.5 ] - [k/70]$$



I'm almost certain that this $m$, for any given $k$, will satisfy the inequality. I believe it is the smallest integer (I don't know how to prove it, a proof will be appreciated!), but I wonder if there's a way to find another m that's a lot simpler... or maybe a way to simplify the m I found so I can use it in power additions and other things.



Or in general, is there a way/technique for such problems on how to find the smallest integer power?



Thank you for your time!






algebra-precalculus inequality exponentiation integers







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    so I have this inequality:
    Given integers $m, kgeq1$.



    $$2^{m/k} > frac{3}{2}$$



    I'm interested in finding the smallest integer power $m$, as a function of $k$, that will make this inequality hold true.



    I've made a few calculations and came up with $m$ as:
    $$m(k) = [ 6k/10 + 0.5 ] - [k/70]$$



    I'm almost certain that this $m$, for any given $k$, will satisfy the inequality. I believe it is the smallest integer (I don't know how to prove it, a proof will be appreciated!), but I wonder if there's a way to find another m that's a lot simpler... or maybe a way to simplify the m I found so I can use it in power additions and other things.



    Or in general, is there a way/technique for such problems on how to find the smallest integer power?



    Thank you for your time!






    algebra-precalculus inequality exponentiation integers







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      so I have this inequality:
      Given integers $m, kgeq1$.



      $$2^{m/k} > frac{3}{2}$$



      I'm interested in finding the smallest integer power $m$, as a function of $k$, that will make this inequality hold true.



      I've made a few calculations and came up with $m$ as:
      $$m(k) = [ 6k/10 + 0.5 ] - [k/70]$$



      I'm almost certain that this $m$, for any given $k$, will satisfy the inequality. I believe it is the smallest integer (I don't know how to prove it, a proof will be appreciated!), but I wonder if there's a way to find another m that's a lot simpler... or maybe a way to simplify the m I found so I can use it in power additions and other things.



      Or in general, is there a way/technique for such problems on how to find the smallest integer power?



      Thank you for your time!






      algebra-precalculus inequality exponentiation integers







      share|cite|improve this question











      $endgroup$




      so I have this inequality:
      Given integers $m, kgeq1$.



      $$2^{m/k} > frac{3}{2}$$



      I'm interested in finding the smallest integer power $m$, as a function of $k$, that will make this inequality hold true.



      I've made a few calculations and came up with $m$ as:
      $$m(k) = [ 6k/10 + 0.5 ] - [k/70]$$



      I'm almost certain that this $m$, for any given $k$, will satisfy the inequality. I believe it is the smallest integer (I don't know how to prove it, a proof will be appreciated!), but I wonder if there's a way to find another m that's a lot simpler... or maybe a way to simplify the m I found so I can use it in power additions and other things.



      Or in general, is there a way/technique for such problems on how to find the smallest integer power?



      Thank you for your time!






      algebra-precalculus inequality exponentiation integers




      algebra-precalculus inequality exponentiation integers






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 2 '18 at 22:21









      Servaes

      23.3k33893




      23.3k33893










      asked Dec 2 '18 at 22:06









      Buk LauBuk Lau

      1878




      1878






















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          Given a positive integer $k$ you want to find the least positive integer $m$ such that $2^{m/k}>frac{3}{2}$, or equivalently
          $$frac{m}{k}>frac{log(frac{3}{2})}{log2}=frac{log3-log2}{log2}=frac{log3}{log2}-1,$$
          or equivalently $m>kcdotleft(frac{log3}{log2}-1right)$. The latter is a constant; an online calculator tells me that
          $$frac{log3}{log2}-1approx0.5849625,$$
          so $m=lceil0.5849625ldotstimes krceil$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Thank you for your quick answer! So m should be [k*0.5849625 + 0.5] correct?
            $endgroup$
            – Buk Lau
            Dec 2 '18 at 22:12





















          1












          $begingroup$

          HINT



          We have that since $log$ function is strictly increasing



          $$2^{m/k} > 3/2 iff log (2^{m/k}) > log (3/2)$$






          share|cite|improve this answer









          $endgroup$













            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',
            autoActivateHeartbeat: false,
            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%2f3023280%2fsmallest-integer-power-for-an-inequality-to-hold%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









            1












            $begingroup$

            Given a positive integer $k$ you want to find the least positive integer $m$ such that $2^{m/k}>frac{3}{2}$, or equivalently
            $$frac{m}{k}>frac{log(frac{3}{2})}{log2}=frac{log3-log2}{log2}=frac{log3}{log2}-1,$$
            or equivalently $m>kcdotleft(frac{log3}{log2}-1right)$. The latter is a constant; an online calculator tells me that
            $$frac{log3}{log2}-1approx0.5849625,$$
            so $m=lceil0.5849625ldotstimes krceil$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for your quick answer! So m should be [k*0.5849625 + 0.5] correct?
              $endgroup$
              – Buk Lau
              Dec 2 '18 at 22:12


















            1












            $begingroup$

            Given a positive integer $k$ you want to find the least positive integer $m$ such that $2^{m/k}>frac{3}{2}$, or equivalently
            $$frac{m}{k}>frac{log(frac{3}{2})}{log2}=frac{log3-log2}{log2}=frac{log3}{log2}-1,$$
            or equivalently $m>kcdotleft(frac{log3}{log2}-1right)$. The latter is a constant; an online calculator tells me that
            $$frac{log3}{log2}-1approx0.5849625,$$
            so $m=lceil0.5849625ldotstimes krceil$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you for your quick answer! So m should be [k*0.5849625 + 0.5] correct?
              $endgroup$
              – Buk Lau
              Dec 2 '18 at 22:12
















            1












            1








            1





            $begingroup$

            Given a positive integer $k$ you want to find the least positive integer $m$ such that $2^{m/k}>frac{3}{2}$, or equivalently
            $$frac{m}{k}>frac{log(frac{3}{2})}{log2}=frac{log3-log2}{log2}=frac{log3}{log2}-1,$$
            or equivalently $m>kcdotleft(frac{log3}{log2}-1right)$. The latter is a constant; an online calculator tells me that
            $$frac{log3}{log2}-1approx0.5849625,$$
            so $m=lceil0.5849625ldotstimes krceil$.






            share|cite|improve this answer











            $endgroup$



            Given a positive integer $k$ you want to find the least positive integer $m$ such that $2^{m/k}>frac{3}{2}$, or equivalently
            $$frac{m}{k}>frac{log(frac{3}{2})}{log2}=frac{log3-log2}{log2}=frac{log3}{log2}-1,$$
            or equivalently $m>kcdotleft(frac{log3}{log2}-1right)$. The latter is a constant; an online calculator tells me that
            $$frac{log3}{log2}-1approx0.5849625,$$
            so $m=lceil0.5849625ldotstimes krceil$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 2 '18 at 22:13

























            answered Dec 2 '18 at 22:09









            ServaesServaes

            23.3k33893




            23.3k33893












            • $begingroup$
              Thank you for your quick answer! So m should be [k*0.5849625 + 0.5] correct?
              $endgroup$
              – Buk Lau
              Dec 2 '18 at 22:12




















            • $begingroup$
              Thank you for your quick answer! So m should be [k*0.5849625 + 0.5] correct?
              $endgroup$
              – Buk Lau
              Dec 2 '18 at 22:12


















            $begingroup$
            Thank you for your quick answer! So m should be [k*0.5849625 + 0.5] correct?
            $endgroup$
            – Buk Lau
            Dec 2 '18 at 22:12






            $begingroup$
            Thank you for your quick answer! So m should be [k*0.5849625 + 0.5] correct?
            $endgroup$
            – Buk Lau
            Dec 2 '18 at 22:12













            1












            $begingroup$

            HINT



            We have that since $log$ function is strictly increasing



            $$2^{m/k} > 3/2 iff log (2^{m/k}) > log (3/2)$$






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              HINT



              We have that since $log$ function is strictly increasing



              $$2^{m/k} > 3/2 iff log (2^{m/k}) > log (3/2)$$






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                HINT



                We have that since $log$ function is strictly increasing



                $$2^{m/k} > 3/2 iff log (2^{m/k}) > log (3/2)$$






                share|cite|improve this answer









                $endgroup$



                HINT



                We have that since $log$ function is strictly increasing



                $$2^{m/k} > 3/2 iff log (2^{m/k}) > log (3/2)$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 '18 at 22:08









                gimusigimusi

                92.8k84494




                92.8k84494






























                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3023280%2fsmallest-integer-power-for-an-inequality-to-hold%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

                    Plaza Victoria

                    Image
                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                    Read more

                    Puebla de Zaragoza

                    Image
                    Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                    Read more

                    Musa

                    Image
                    Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                    Read more