How to simplify $2^{(3+(4+(5+(6+(7+(8))))))}$? [closed]












0












$begingroup$


Thanks to the rules of parenthesis in exponents, $2^{(3+(4+(5+(6+(7+(8))))))}$ has a huge number of parenthesis.



How do you simplify this so you wouldn't have to write $))))))$ at the end of a math problem?






algebra-precalculus exponentiation







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$endgroup$



closed as unclear what you're asking by Did, Don Thousand, José Carlos Santos, metamorphy, Adrian Keister Jan 26 at 19:22


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


















  • $begingroup$
    If you are working in a normal context, You don't need any of the parentheses thanks to commutativity. If you are working in a weird group, that's another story...
    $endgroup$
    – Don Thousand
    Jan 25 at 23:43










  • $begingroup$
    What is this "rule of parentheses in exponents"?
    $endgroup$
    – kccu
    Jan 25 at 23:44
















0












$begingroup$


Thanks to the rules of parenthesis in exponents, $2^{(3+(4+(5+(6+(7+(8))))))}$ has a huge number of parenthesis.



How do you simplify this so you wouldn't have to write $))))))$ at the end of a math problem?






algebra-precalculus exponentiation







share|cite|improve this question











$endgroup$



closed as unclear what you're asking by Did, Don Thousand, José Carlos Santos, metamorphy, Adrian Keister Jan 26 at 19:22


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.


















  • $begingroup$
    If you are working in a normal context, You don't need any of the parentheses thanks to commutativity. If you are working in a weird group, that's another story...
    $endgroup$
    – Don Thousand
    Jan 25 at 23:43










  • $begingroup$
    What is this "rule of parentheses in exponents"?
    $endgroup$
    – kccu
    Jan 25 at 23:44














0












0








0





$begingroup$


Thanks to the rules of parenthesis in exponents, $2^{(3+(4+(5+(6+(7+(8))))))}$ has a huge number of parenthesis.



How do you simplify this so you wouldn't have to write $))))))$ at the end of a math problem?






algebra-precalculus exponentiation







share|cite|improve this question











$endgroup$




Thanks to the rules of parenthesis in exponents, $2^{(3+(4+(5+(6+(7+(8))))))}$ has a huge number of parenthesis.



How do you simplify this so you wouldn't have to write $))))))$ at the end of a math problem?






algebra-precalculus exponentiation




algebra-precalculus exponentiation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 26 at 0:10







LukeyBear

















asked Jan 25 at 23:38









LukeyBearLukeyBear

84




84




closed as unclear what you're asking by Did, Don Thousand, José Carlos Santos, metamorphy, Adrian Keister Jan 26 at 19:22


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.









closed as unclear what you're asking by Did, Don Thousand, José Carlos Santos, metamorphy, Adrian Keister Jan 26 at 19:22


Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.














  • $begingroup$
    If you are working in a normal context, You don't need any of the parentheses thanks to commutativity. If you are working in a weird group, that's another story...
    $endgroup$
    – Don Thousand
    Jan 25 at 23:43










  • $begingroup$
    What is this "rule of parentheses in exponents"?
    $endgroup$
    – kccu
    Jan 25 at 23:44


















  • $begingroup$
    If you are working in a normal context, You don't need any of the parentheses thanks to commutativity. If you are working in a weird group, that's another story...
    $endgroup$
    – Don Thousand
    Jan 25 at 23:43










  • $begingroup$
    What is this "rule of parentheses in exponents"?
    $endgroup$
    – kccu
    Jan 25 at 23:44
















$begingroup$
If you are working in a normal context, You don't need any of the parentheses thanks to commutativity. If you are working in a weird group, that's another story...
$endgroup$
– Don Thousand
Jan 25 at 23:43




$begingroup$
If you are working in a normal context, You don't need any of the parentheses thanks to commutativity. If you are working in a weird group, that's another story...
$endgroup$
– Don Thousand
Jan 25 at 23:43












$begingroup$
What is this "rule of parentheses in exponents"?
$endgroup$
– kccu
Jan 25 at 23:44




$begingroup$
What is this "rule of parentheses in exponents"?
$endgroup$
– kccu
Jan 25 at 23:44










5 Answers
5






active

oldest

votes


















3












$begingroup$

$2^{(3+(4+(5+(6+(7+(8))))))} = 2^{(3+4+5+6+7+8)} = 2^{33}$. The only set of parantheses needed is the outermost one, to imply that you do the addition before you do the exponentiation.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    And even the outer ones aren't really needed since they're implied by the fact the whole exponent is one superscript!
    $endgroup$
    – timtfj
    Jan 26 at 1:07



















2












$begingroup$

$3+(4+(5+(6+(7+(8))))) = 3+4+5+6+7+8 = 33$



Therefore, $2^{(3+(4+(5+(6+(7+(8))))))} = 2^{33}$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    Addition is associative so in your case, we have:



    $$2^{(3+(4+(5+(6))))}=2^{(3+4+5+6)}$$



    No need for an inordinate number of parentheses here.






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      This would be identical to simply simplifying the exponent.



      So what would an equivalent expression be for the exponent, i.e. what does



      $$(3+(4+(5+(6+(7+(8)))))))$$



      equal? (Or the sum with the numbers up to $6$, you have two different expressions in the question.) Obviously it would just be the sum of these numbers, right? Thus, you can just substitute the sum into the exponent. (This applies to either one.)






      share|cite|improve this answer









      $endgroup$





















        1












        $begingroup$

        To combat this kind of problem when typesetting documents, we often either just evaluate the exponent (like Mike suggested), or use dots, like $2(3+(4+(5+cdots+20(21))))$. Alternatively, I've seen people use $)cdots)$, but that's exceedingly rare. If working in a context when you have commutativity and associativity though, the first option is probably the way to go.






        share|cite|improve this answer









        $endgroup$




















          5 Answers
          5






          active

          oldest

          votes








          5 Answers
          5






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          3












          $begingroup$

          $2^{(3+(4+(5+(6+(7+(8))))))} = 2^{(3+4+5+6+7+8)} = 2^{33}$. The only set of parantheses needed is the outermost one, to imply that you do the addition before you do the exponentiation.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            And even the outer ones aren't really needed since they're implied by the fact the whole exponent is one superscript!
            $endgroup$
            – timtfj
            Jan 26 at 1:07
















          3












          $begingroup$

          $2^{(3+(4+(5+(6+(7+(8))))))} = 2^{(3+4+5+6+7+8)} = 2^{33}$. The only set of parantheses needed is the outermost one, to imply that you do the addition before you do the exponentiation.






          share|cite|improve this answer











          $endgroup$









          • 1




            $begingroup$
            And even the outer ones aren't really needed since they're implied by the fact the whole exponent is one superscript!
            $endgroup$
            – timtfj
            Jan 26 at 1:07














          3












          3








          3





          $begingroup$

          $2^{(3+(4+(5+(6+(7+(8))))))} = 2^{(3+4+5+6+7+8)} = 2^{33}$. The only set of parantheses needed is the outermost one, to imply that you do the addition before you do the exponentiation.






          share|cite|improve this answer











          $endgroup$



          $2^{(3+(4+(5+(6+(7+(8))))))} = 2^{(3+4+5+6+7+8)} = 2^{33}$. The only set of parantheses needed is the outermost one, to imply that you do the addition before you do the exponentiation.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 25 at 23:46









          NicNic8

          4,63831123




          4,63831123










          answered Jan 25 at 23:43









          MikeMike

          4,386412




          4,386412








          • 1




            $begingroup$
            And even the outer ones aren't really needed since they're implied by the fact the whole exponent is one superscript!
            $endgroup$
            – timtfj
            Jan 26 at 1:07














          • 1




            $begingroup$
            And even the outer ones aren't really needed since they're implied by the fact the whole exponent is one superscript!
            $endgroup$
            – timtfj
            Jan 26 at 1:07








          1




          1




          $begingroup$
          And even the outer ones aren't really needed since they're implied by the fact the whole exponent is one superscript!
          $endgroup$
          – timtfj
          Jan 26 at 1:07




          $begingroup$
          And even the outer ones aren't really needed since they're implied by the fact the whole exponent is one superscript!
          $endgroup$
          – timtfj
          Jan 26 at 1:07











          2












          $begingroup$

          $3+(4+(5+(6+(7+(8))))) = 3+4+5+6+7+8 = 33$



          Therefore, $2^{(3+(4+(5+(6+(7+(8))))))} = 2^{33}$.






          share|cite|improve this answer











          $endgroup$


















            2












            $begingroup$

            $3+(4+(5+(6+(7+(8))))) = 3+4+5+6+7+8 = 33$



            Therefore, $2^{(3+(4+(5+(6+(7+(8))))))} = 2^{33}$.






            share|cite|improve this answer











            $endgroup$
















              2












              2








              2





              $begingroup$

              $3+(4+(5+(6+(7+(8))))) = 3+4+5+6+7+8 = 33$



              Therefore, $2^{(3+(4+(5+(6+(7+(8))))))} = 2^{33}$.






              share|cite|improve this answer











              $endgroup$



              $3+(4+(5+(6+(7+(8))))) = 3+4+5+6+7+8 = 33$



              Therefore, $2^{(3+(4+(5+(6+(7+(8))))))} = 2^{33}$.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited Jan 26 at 0:10

























              answered Jan 25 at 23:44









              NicNic8NicNic8

              4,63831123




              4,63831123























                  2












                  $begingroup$

                  Addition is associative so in your case, we have:



                  $$2^{(3+(4+(5+(6))))}=2^{(3+4+5+6)}$$



                  No need for an inordinate number of parentheses here.






                  share|cite|improve this answer











                  $endgroup$


















                    2












                    $begingroup$

                    Addition is associative so in your case, we have:



                    $$2^{(3+(4+(5+(6))))}=2^{(3+4+5+6)}$$



                    No need for an inordinate number of parentheses here.






                    share|cite|improve this answer











                    $endgroup$
















                      2












                      2








                      2





                      $begingroup$

                      Addition is associative so in your case, we have:



                      $$2^{(3+(4+(5+(6))))}=2^{(3+4+5+6)}$$



                      No need for an inordinate number of parentheses here.






                      share|cite|improve this answer











                      $endgroup$



                      Addition is associative so in your case, we have:



                      $$2^{(3+(4+(5+(6))))}=2^{(3+4+5+6)}$$



                      No need for an inordinate number of parentheses here.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Jan 26 at 0:55

























                      answered Jan 25 at 23:45









                      GnumbertesterGnumbertester

                      672114




                      672114























                          1












                          $begingroup$

                          This would be identical to simply simplifying the exponent.



                          So what would an equivalent expression be for the exponent, i.e. what does



                          $$(3+(4+(5+(6+(7+(8)))))))$$



                          equal? (Or the sum with the numbers up to $6$, you have two different expressions in the question.) Obviously it would just be the sum of these numbers, right? Thus, you can just substitute the sum into the exponent. (This applies to either one.)






                          share|cite|improve this answer









                          $endgroup$


















                            1












                            $begingroup$

                            This would be identical to simply simplifying the exponent.



                            So what would an equivalent expression be for the exponent, i.e. what does



                            $$(3+(4+(5+(6+(7+(8)))))))$$



                            equal? (Or the sum with the numbers up to $6$, you have two different expressions in the question.) Obviously it would just be the sum of these numbers, right? Thus, you can just substitute the sum into the exponent. (This applies to either one.)






                            share|cite|improve this answer









                            $endgroup$
















                              1












                              1








                              1





                              $begingroup$

                              This would be identical to simply simplifying the exponent.



                              So what would an equivalent expression be for the exponent, i.e. what does



                              $$(3+(4+(5+(6+(7+(8)))))))$$



                              equal? (Or the sum with the numbers up to $6$, you have two different expressions in the question.) Obviously it would just be the sum of these numbers, right? Thus, you can just substitute the sum into the exponent. (This applies to either one.)






                              share|cite|improve this answer









                              $endgroup$



                              This would be identical to simply simplifying the exponent.



                              So what would an equivalent expression be for the exponent, i.e. what does



                              $$(3+(4+(5+(6+(7+(8)))))))$$



                              equal? (Or the sum with the numbers up to $6$, you have two different expressions in the question.) Obviously it would just be the sum of these numbers, right? Thus, you can just substitute the sum into the exponent. (This applies to either one.)







                              share|cite|improve this answer












                              share|cite|improve this answer



                              share|cite|improve this answer










                              answered Jan 25 at 23:44









                              Eevee TrainerEevee Trainer

                              7,82421339




                              7,82421339























                                  1












                                  $begingroup$

                                  To combat this kind of problem when typesetting documents, we often either just evaluate the exponent (like Mike suggested), or use dots, like $2(3+(4+(5+cdots+20(21))))$. Alternatively, I've seen people use $)cdots)$, but that's exceedingly rare. If working in a context when you have commutativity and associativity though, the first option is probably the way to go.






                                  share|cite|improve this answer









                                  $endgroup$


















                                    1












                                    $begingroup$

                                    To combat this kind of problem when typesetting documents, we often either just evaluate the exponent (like Mike suggested), or use dots, like $2(3+(4+(5+cdots+20(21))))$. Alternatively, I've seen people use $)cdots)$, but that's exceedingly rare. If working in a context when you have commutativity and associativity though, the first option is probably the way to go.






                                    share|cite|improve this answer









                                    $endgroup$
















                                      1












                                      1








                                      1





                                      $begingroup$

                                      To combat this kind of problem when typesetting documents, we often either just evaluate the exponent (like Mike suggested), or use dots, like $2(3+(4+(5+cdots+20(21))))$. Alternatively, I've seen people use $)cdots)$, but that's exceedingly rare. If working in a context when you have commutativity and associativity though, the first option is probably the way to go.






                                      share|cite|improve this answer









                                      $endgroup$



                                      To combat this kind of problem when typesetting documents, we often either just evaluate the exponent (like Mike suggested), or use dots, like $2(3+(4+(5+cdots+20(21))))$. Alternatively, I've seen people use $)cdots)$, but that's exceedingly rare. If working in a context when you have commutativity and associativity though, the first option is probably the way to go.







                                      share|cite|improve this answer












                                      share|cite|improve this answer



                                      share|cite|improve this answer










                                      answered Jan 25 at 23:46









                                      YiFanYiFan

                                      4,7111727




                                      4,7111727















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