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







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$endgroup$

















    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
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          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









          $endgroup$













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            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






























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