Which of the following has the greatest value












2












$begingroup$


Which of the following has the greatest value?



a) $2^{64}$ b) $4^{63}$ c) $8^{34}$ d) $16^{17}$



I tried finding a pattern among exponents and their is none. but there is a pattern in base, but I'm unable to find the common power through which I'll compare the base and figure the answer. What is the best possible option to solve this question within 1.5 minutes?






algebra-precalculus arithmetic exponentiation







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$endgroup$








  • 5




    $begingroup$
    Express all of them as exponents of two
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Nov 30 '18 at 10:46






  • 4




    $begingroup$
    There's a very clear pattern in the bases: $4 = 2^2$, for a start.
    $endgroup$
    – user3482749
    Nov 30 '18 at 10:46










  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Nov 30 '18 at 12:41
















2












$begingroup$


Which of the following has the greatest value?



a) $2^{64}$ b) $4^{63}$ c) $8^{34}$ d) $16^{17}$



I tried finding a pattern among exponents and their is none. but there is a pattern in base, but I'm unable to find the common power through which I'll compare the base and figure the answer. What is the best possible option to solve this question within 1.5 minutes?






algebra-precalculus arithmetic exponentiation







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    Express all of them as exponents of two
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Nov 30 '18 at 10:46






  • 4




    $begingroup$
    There's a very clear pattern in the bases: $4 = 2^2$, for a start.
    $endgroup$
    – user3482749
    Nov 30 '18 at 10:46










  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Nov 30 '18 at 12:41














2












2








2





$begingroup$


Which of the following has the greatest value?



a) $2^{64}$ b) $4^{63}$ c) $8^{34}$ d) $16^{17}$



I tried finding a pattern among exponents and their is none. but there is a pattern in base, but I'm unable to find the common power through which I'll compare the base and figure the answer. What is the best possible option to solve this question within 1.5 minutes?






algebra-precalculus arithmetic exponentiation







share|cite|improve this question











$endgroup$




Which of the following has the greatest value?



a) $2^{64}$ b) $4^{63}$ c) $8^{34}$ d) $16^{17}$



I tried finding a pattern among exponents and their is none. but there is a pattern in base, but I'm unable to find the common power through which I'll compare the base and figure the answer. What is the best possible option to solve this question within 1.5 minutes?






algebra-precalculus arithmetic exponentiation




algebra-precalculus arithmetic exponentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 12:41









N. F. Taussig

43.7k93355




43.7k93355










asked Nov 30 '18 at 10:44









shawn kshawn k

396




396








  • 5




    $begingroup$
    Express all of them as exponents of two
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Nov 30 '18 at 10:46






  • 4




    $begingroup$
    There's a very clear pattern in the bases: $4 = 2^2$, for a start.
    $endgroup$
    – user3482749
    Nov 30 '18 at 10:46










  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Nov 30 '18 at 12:41














  • 5




    $begingroup$
    Express all of them as exponents of two
    $endgroup$
    – GNUSupporter 8964民主女神 地下教會
    Nov 30 '18 at 10:46






  • 4




    $begingroup$
    There's a very clear pattern in the bases: $4 = 2^2$, for a start.
    $endgroup$
    – user3482749
    Nov 30 '18 at 10:46










  • $begingroup$
    Please read this tutorial on how to typeset mathematics on this site.
    $endgroup$
    – N. F. Taussig
    Nov 30 '18 at 12:41








5




5




$begingroup$
Express all of them as exponents of two
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:46




$begingroup$
Express all of them as exponents of two
$endgroup$
– GNUSupporter 8964民主女神 地下教會
Nov 30 '18 at 10:46




4




4




$begingroup$
There's a very clear pattern in the bases: $4 = 2^2$, for a start.
$endgroup$
– user3482749
Nov 30 '18 at 10:46




$begingroup$
There's a very clear pattern in the bases: $4 = 2^2$, for a start.
$endgroup$
– user3482749
Nov 30 '18 at 10:46












$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Nov 30 '18 at 12:41




$begingroup$
Please read this tutorial on how to typeset mathematics on this site.
$endgroup$
– N. F. Taussig
Nov 30 '18 at 12:41










2 Answers
2






active

oldest

votes


















4












$begingroup$

Hint:




  • $16^{17}=left(2^4right)^{17}=2^{4times17}=2^{68}$


and similarly for the other powers of $2$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Ohh I got it, It was so right in the front, I was being stupid. Thank you so much. Correct answer is B right?
    $endgroup$
    – shawn k
    Nov 30 '18 at 10:52










  • $begingroup$
    Can you please help me with this question as well? I'm stuck and not getting any replies math.stackexchange.com/questions/3019103/…
    $endgroup$
    – shawn k
    Nov 30 '18 at 10:56






  • 1




    $begingroup$
    @shawn k I’ve answered your question.
    $endgroup$
    – KM101
    Nov 30 '18 at 11:37










  • $begingroup$
    @KM101 Thanks a-lot!!! Really appreciated.
    $endgroup$
    – shawn k
    Nov 30 '18 at 11:44





















2












$begingroup$

We know that $$16^{17}=4^{34}<4^{63}\2^{64}=4^{32}<4^{63}\8^{34}=4^{34times {3over 2}}=4^{51}<4^{63}$$ therefore $4^{63}$ has the greatest value among all.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Hint:




    • $16^{17}=left(2^4right)^{17}=2^{4times17}=2^{68}$


    and similarly for the other powers of $2$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ohh I got it, It was so right in the front, I was being stupid. Thank you so much. Correct answer is B right?
      $endgroup$
      – shawn k
      Nov 30 '18 at 10:52










    • $begingroup$
      Can you please help me with this question as well? I'm stuck and not getting any replies math.stackexchange.com/questions/3019103/…
      $endgroup$
      – shawn k
      Nov 30 '18 at 10:56






    • 1




      $begingroup$
      @shawn k I’ve answered your question.
      $endgroup$
      – KM101
      Nov 30 '18 at 11:37










    • $begingroup$
      @KM101 Thanks a-lot!!! Really appreciated.
      $endgroup$
      – shawn k
      Nov 30 '18 at 11:44


















    4












    $begingroup$

    Hint:




    • $16^{17}=left(2^4right)^{17}=2^{4times17}=2^{68}$


    and similarly for the other powers of $2$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Ohh I got it, It was so right in the front, I was being stupid. Thank you so much. Correct answer is B right?
      $endgroup$
      – shawn k
      Nov 30 '18 at 10:52










    • $begingroup$
      Can you please help me with this question as well? I'm stuck and not getting any replies math.stackexchange.com/questions/3019103/…
      $endgroup$
      – shawn k
      Nov 30 '18 at 10:56






    • 1




      $begingroup$
      @shawn k I’ve answered your question.
      $endgroup$
      – KM101
      Nov 30 '18 at 11:37










    • $begingroup$
      @KM101 Thanks a-lot!!! Really appreciated.
      $endgroup$
      – shawn k
      Nov 30 '18 at 11:44
















    4












    4








    4





    $begingroup$

    Hint:




    • $16^{17}=left(2^4right)^{17}=2^{4times17}=2^{68}$


    and similarly for the other powers of $2$






    share|cite|improve this answer









    $endgroup$



    Hint:




    • $16^{17}=left(2^4right)^{17}=2^{4times17}=2^{68}$


    and similarly for the other powers of $2$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 30 '18 at 10:47









    HenryHenry

    98.5k476163




    98.5k476163












    • $begingroup$
      Ohh I got it, It was so right in the front, I was being stupid. Thank you so much. Correct answer is B right?
      $endgroup$
      – shawn k
      Nov 30 '18 at 10:52










    • $begingroup$
      Can you please help me with this question as well? I'm stuck and not getting any replies math.stackexchange.com/questions/3019103/…
      $endgroup$
      – shawn k
      Nov 30 '18 at 10:56






    • 1




      $begingroup$
      @shawn k I’ve answered your question.
      $endgroup$
      – KM101
      Nov 30 '18 at 11:37










    • $begingroup$
      @KM101 Thanks a-lot!!! Really appreciated.
      $endgroup$
      – shawn k
      Nov 30 '18 at 11:44




















    • $begingroup$
      Ohh I got it, It was so right in the front, I was being stupid. Thank you so much. Correct answer is B right?
      $endgroup$
      – shawn k
      Nov 30 '18 at 10:52










    • $begingroup$
      Can you please help me with this question as well? I'm stuck and not getting any replies math.stackexchange.com/questions/3019103/…
      $endgroup$
      – shawn k
      Nov 30 '18 at 10:56






    • 1




      $begingroup$
      @shawn k I’ve answered your question.
      $endgroup$
      – KM101
      Nov 30 '18 at 11:37










    • $begingroup$
      @KM101 Thanks a-lot!!! Really appreciated.
      $endgroup$
      – shawn k
      Nov 30 '18 at 11:44


















    $begingroup$
    Ohh I got it, It was so right in the front, I was being stupid. Thank you so much. Correct answer is B right?
    $endgroup$
    – shawn k
    Nov 30 '18 at 10:52




    $begingroup$
    Ohh I got it, It was so right in the front, I was being stupid. Thank you so much. Correct answer is B right?
    $endgroup$
    – shawn k
    Nov 30 '18 at 10:52












    $begingroup$
    Can you please help me with this question as well? I'm stuck and not getting any replies math.stackexchange.com/questions/3019103/…
    $endgroup$
    – shawn k
    Nov 30 '18 at 10:56




    $begingroup$
    Can you please help me with this question as well? I'm stuck and not getting any replies math.stackexchange.com/questions/3019103/…
    $endgroup$
    – shawn k
    Nov 30 '18 at 10:56




    1




    1




    $begingroup$
    @shawn k I’ve answered your question.
    $endgroup$
    – KM101
    Nov 30 '18 at 11:37




    $begingroup$
    @shawn k I’ve answered your question.
    $endgroup$
    – KM101
    Nov 30 '18 at 11:37












    $begingroup$
    @KM101 Thanks a-lot!!! Really appreciated.
    $endgroup$
    – shawn k
    Nov 30 '18 at 11:44






    $begingroup$
    @KM101 Thanks a-lot!!! Really appreciated.
    $endgroup$
    – shawn k
    Nov 30 '18 at 11:44













    2












    $begingroup$

    We know that $$16^{17}=4^{34}<4^{63}\2^{64}=4^{32}<4^{63}\8^{34}=4^{34times {3over 2}}=4^{51}<4^{63}$$ therefore $4^{63}$ has the greatest value among all.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      We know that $$16^{17}=4^{34}<4^{63}\2^{64}=4^{32}<4^{63}\8^{34}=4^{34times {3over 2}}=4^{51}<4^{63}$$ therefore $4^{63}$ has the greatest value among all.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        We know that $$16^{17}=4^{34}<4^{63}\2^{64}=4^{32}<4^{63}\8^{34}=4^{34times {3over 2}}=4^{51}<4^{63}$$ therefore $4^{63}$ has the greatest value among all.






        share|cite|improve this answer









        $endgroup$



        We know that $$16^{17}=4^{34}<4^{63}\2^{64}=4^{32}<4^{63}\8^{34}=4^{34times {3over 2}}=4^{51}<4^{63}$$ therefore $4^{63}$ has the greatest value among all.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 10:59









        Mostafa AyazMostafa Ayaz

        14.7k3938




        14.7k3938






























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