How to parse $2^{1+2^3}$?












3












$begingroup$


I'm fairly comfortable with arithmetic and high school math overall, but I'm having trouble being 100% sure that $2^{large {1+2^3}}$ should be parsed as 2^(1+(2^3)) and not as 2^((1+2)^3), because I've heard that




One must evaluate the base of an exponent before it is raised to the exponent.




As I understand that is actually true and the only way to support the second alternative is to claim that the power is $1+2^3$ (instead of $2^3$ alone) and that its base is $1+2$.



Can anyone confirm that the correct interpretation is 2^(1+(2^3)) and that it yields $512$?






algebra-precalculus arithmetic exponentiation







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $$2^{1+2^3}=2^{1+8}=2^9=512$$
    $endgroup$
    – Dr. Mathva
    Jan 22 at 19:34






  • 2




    $begingroup$
    Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
    $endgroup$
    – Shubham Johri
    Jan 22 at 19:36






  • 2




    $begingroup$
    You are correct. Exponents take precedence over addition.
    $endgroup$
    – Wojowu
    Jan 22 at 20:05






  • 1




    $begingroup$
    This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
    $endgroup$
    – Git Gud
    Jan 23 at 9:09










  • $begingroup$
    Thank you everyone for all your comments.
    $endgroup$
    – Sahil Sunny
    Jan 25 at 6:57
















3












$begingroup$


I'm fairly comfortable with arithmetic and high school math overall, but I'm having trouble being 100% sure that $2^{large {1+2^3}}$ should be parsed as 2^(1+(2^3)) and not as 2^((1+2)^3), because I've heard that




One must evaluate the base of an exponent before it is raised to the exponent.




As I understand that is actually true and the only way to support the second alternative is to claim that the power is $1+2^3$ (instead of $2^3$ alone) and that its base is $1+2$.



Can anyone confirm that the correct interpretation is 2^(1+(2^3)) and that it yields $512$?






algebra-precalculus arithmetic exponentiation







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    $$2^{1+2^3}=2^{1+8}=2^9=512$$
    $endgroup$
    – Dr. Mathva
    Jan 22 at 19:34






  • 2




    $begingroup$
    Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
    $endgroup$
    – Shubham Johri
    Jan 22 at 19:36






  • 2




    $begingroup$
    You are correct. Exponents take precedence over addition.
    $endgroup$
    – Wojowu
    Jan 22 at 20:05






  • 1




    $begingroup$
    This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
    $endgroup$
    – Git Gud
    Jan 23 at 9:09










  • $begingroup$
    Thank you everyone for all your comments.
    $endgroup$
    – Sahil Sunny
    Jan 25 at 6:57














3












3








3





$begingroup$


I'm fairly comfortable with arithmetic and high school math overall, but I'm having trouble being 100% sure that $2^{large {1+2^3}}$ should be parsed as 2^(1+(2^3)) and not as 2^((1+2)^3), because I've heard that




One must evaluate the base of an exponent before it is raised to the exponent.




As I understand that is actually true and the only way to support the second alternative is to claim that the power is $1+2^3$ (instead of $2^3$ alone) and that its base is $1+2$.



Can anyone confirm that the correct interpretation is 2^(1+(2^3)) and that it yields $512$?






algebra-precalculus arithmetic exponentiation







share|cite|improve this question









$endgroup$




I'm fairly comfortable with arithmetic and high school math overall, but I'm having trouble being 100% sure that $2^{large {1+2^3}}$ should be parsed as 2^(1+(2^3)) and not as 2^((1+2)^3), because I've heard that




One must evaluate the base of an exponent before it is raised to the exponent.




As I understand that is actually true and the only way to support the second alternative is to claim that the power is $1+2^3$ (instead of $2^3$ alone) and that its base is $1+2$.



Can anyone confirm that the correct interpretation is 2^(1+(2^3)) and that it yields $512$?






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










asked Jan 22 at 19:32









Sahil SunnySahil Sunny

183




183








  • 2




    $begingroup$
    $$2^{1+2^3}=2^{1+8}=2^9=512$$
    $endgroup$
    – Dr. Mathva
    Jan 22 at 19:34






  • 2




    $begingroup$
    Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
    $endgroup$
    – Shubham Johri
    Jan 22 at 19:36






  • 2




    $begingroup$
    You are correct. Exponents take precedence over addition.
    $endgroup$
    – Wojowu
    Jan 22 at 20:05






  • 1




    $begingroup$
    This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
    $endgroup$
    – Git Gud
    Jan 23 at 9:09










  • $begingroup$
    Thank you everyone for all your comments.
    $endgroup$
    – Sahil Sunny
    Jan 25 at 6:57














  • 2




    $begingroup$
    $$2^{1+2^3}=2^{1+8}=2^9=512$$
    $endgroup$
    – Dr. Mathva
    Jan 22 at 19:34






  • 2




    $begingroup$
    Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
    $endgroup$
    – Shubham Johri
    Jan 22 at 19:36






  • 2




    $begingroup$
    You are correct. Exponents take precedence over addition.
    $endgroup$
    – Wojowu
    Jan 22 at 20:05






  • 1




    $begingroup$
    This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
    $endgroup$
    – Git Gud
    Jan 23 at 9:09










  • $begingroup$
    Thank you everyone for all your comments.
    $endgroup$
    – Sahil Sunny
    Jan 25 at 6:57








2




2




$begingroup$
$$2^{1+2^3}=2^{1+8}=2^9=512$$
$endgroup$
– Dr. Mathva
Jan 22 at 19:34




$begingroup$
$$2^{1+2^3}=2^{1+8}=2^9=512$$
$endgroup$
– Dr. Mathva
Jan 22 at 19:34




2




2




$begingroup$
Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
$endgroup$
– Shubham Johri
Jan 22 at 19:36




$begingroup$
Yes, the answer is $512$. The base of the exponent in $1+2^3$ is just $2$. Had there been brackets as $(1+2)^3$, it would have meant $3^3$
$endgroup$
– Shubham Johri
Jan 22 at 19:36




2




2




$begingroup$
You are correct. Exponents take precedence over addition.
$endgroup$
– Wojowu
Jan 22 at 20:05




$begingroup$
You are correct. Exponents take precedence over addition.
$endgroup$
– Wojowu
Jan 22 at 20:05




1




1




$begingroup$
This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
$endgroup$
– Git Gud
Jan 23 at 9:09




$begingroup$
This is a well established convention. Powers (or exponents) are evaluated first and indeed, in $1+2^3$ the only power is $2^3$, there's no way to argue that the base is $1+2$.
$endgroup$
– Git Gud
Jan 23 at 9:09












$begingroup$
Thank you everyone for all your comments.
$endgroup$
– Sahil Sunny
Jan 25 at 6:57




$begingroup$
Thank you everyone for all your comments.
$endgroup$
– Sahil Sunny
Jan 25 at 6:57










1 Answer
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$begingroup$

Yes, your work is indeed correct.



Note also: $2^{large(1+2^3)} = 2^1cdot 2^{large(2^3)} = 2 cdot 2^8=2cdot 256 = 512$, just as an exercise in working with exponents.



This can be useful, e.g., if you are working with $3^{large(2+2^2)} = 3^{2 + 4}= 3^2cdot 3^4 = 9cdot 81= 729.$



Which ever you find easiest to do in various such examples, if one has $a^{b+c}$, one can add $b+c = d$, then calculate $a^d$, or, equivalently, one can multiply $a^bcdot a^c$.






share|cite|improve this answer











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    $begingroup$

    Yes, your work is indeed correct.



    Note also: $2^{large(1+2^3)} = 2^1cdot 2^{large(2^3)} = 2 cdot 2^8=2cdot 256 = 512$, just as an exercise in working with exponents.



    This can be useful, e.g., if you are working with $3^{large(2+2^2)} = 3^{2 + 4}= 3^2cdot 3^4 = 9cdot 81= 729.$



    Which ever you find easiest to do in various such examples, if one has $a^{b+c}$, one can add $b+c = d$, then calculate $a^d$, or, equivalently, one can multiply $a^bcdot a^c$.






    share|cite|improve this answer











    $endgroup$


















      4












      $begingroup$

      Yes, your work is indeed correct.



      Note also: $2^{large(1+2^3)} = 2^1cdot 2^{large(2^3)} = 2 cdot 2^8=2cdot 256 = 512$, just as an exercise in working with exponents.



      This can be useful, e.g., if you are working with $3^{large(2+2^2)} = 3^{2 + 4}= 3^2cdot 3^4 = 9cdot 81= 729.$



      Which ever you find easiest to do in various such examples, if one has $a^{b+c}$, one can add $b+c = d$, then calculate $a^d$, or, equivalently, one can multiply $a^bcdot a^c$.






      share|cite|improve this answer











      $endgroup$
















        4












        4








        4





        $begingroup$

        Yes, your work is indeed correct.



        Note also: $2^{large(1+2^3)} = 2^1cdot 2^{large(2^3)} = 2 cdot 2^8=2cdot 256 = 512$, just as an exercise in working with exponents.



        This can be useful, e.g., if you are working with $3^{large(2+2^2)} = 3^{2 + 4}= 3^2cdot 3^4 = 9cdot 81= 729.$



        Which ever you find easiest to do in various such examples, if one has $a^{b+c}$, one can add $b+c = d$, then calculate $a^d$, or, equivalently, one can multiply $a^bcdot a^c$.






        share|cite|improve this answer











        $endgroup$



        Yes, your work is indeed correct.



        Note also: $2^{large(1+2^3)} = 2^1cdot 2^{large(2^3)} = 2 cdot 2^8=2cdot 256 = 512$, just as an exercise in working with exponents.



        This can be useful, e.g., if you are working with $3^{large(2+2^2)} = 3^{2 + 4}= 3^2cdot 3^4 = 9cdot 81= 729.$



        Which ever you find easiest to do in various such examples, if one has $a^{b+c}$, one can add $b+c = d$, then calculate $a^d$, or, equivalently, one can multiply $a^bcdot a^c$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 22 at 22:56

























        answered Jan 22 at 19:47









        jordan_glenjordan_glen

        1




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