How can I type this into a calculator?












0














In a textbook I am asked to calculate an expression using a calculator: $$frac{1}{2}lnleft(e^{22}right)$$



I tried using:



0.5*log(exp**22)


This returned an error:



File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/decorators.py", line 91, in __sympifyit_wrapper return func(a, b) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/decorators.py", line 132, in binary_op_wrapper return func(self, other) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/expr.py", line 172, in __rpow__ return Pow(other, self) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/cache.py", line 95, in wrapper retval = func(*args, **kwargs) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/power.py", line 286, in __new__ obj = b._eval_power(e) TypeError: unbound method _eval_power() must be called with exp instance as first argument (got Integer instance instead)


I tried replacing $exp$ with just $e$ and received a similar error.



From my solutions page I can see that the answer is $11$. How can I type the expression in order to get $11$ back?






python







share|cite|improve this question
























  • I believe it's $.5*log(exp(1)**22)$, but anyway you should ask these questions on a stack exchange more relevant to python errors.
    – TrostAft
    Nov 15 '18 at 23:05










  • The exp function is a function, not the number $e$.
    – Sean Roberson
    Nov 15 '18 at 23:05
















0














In a textbook I am asked to calculate an expression using a calculator: $$frac{1}{2}lnleft(e^{22}right)$$



I tried using:



0.5*log(exp**22)


This returned an error:



File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/decorators.py", line 91, in __sympifyit_wrapper return func(a, b) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/decorators.py", line 132, in binary_op_wrapper return func(self, other) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/expr.py", line 172, in __rpow__ return Pow(other, self) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/cache.py", line 95, in wrapper retval = func(*args, **kwargs) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/power.py", line 286, in __new__ obj = b._eval_power(e) TypeError: unbound method _eval_power() must be called with exp instance as first argument (got Integer instance instead)


I tried replacing $exp$ with just $e$ and received a similar error.



From my solutions page I can see that the answer is $11$. How can I type the expression in order to get $11$ back?






python







share|cite|improve this question
























  • I believe it's $.5*log(exp(1)**22)$, but anyway you should ask these questions on a stack exchange more relevant to python errors.
    – TrostAft
    Nov 15 '18 at 23:05










  • The exp function is a function, not the number $e$.
    – Sean Roberson
    Nov 15 '18 at 23:05














0












0








0







In a textbook I am asked to calculate an expression using a calculator: $$frac{1}{2}lnleft(e^{22}right)$$



I tried using:



0.5*log(exp**22)


This returned an error:



File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/decorators.py", line 91, in __sympifyit_wrapper return func(a, b) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/decorators.py", line 132, in binary_op_wrapper return func(self, other) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/expr.py", line 172, in __rpow__ return Pow(other, self) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/cache.py", line 95, in wrapper retval = func(*args, **kwargs) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/power.py", line 286, in __new__ obj = b._eval_power(e) TypeError: unbound method _eval_power() must be called with exp instance as first argument (got Integer instance instead)


I tried replacing $exp$ with just $e$ and received a similar error.



From my solutions page I can see that the answer is $11$. How can I type the expression in order to get $11$ back?






python







share|cite|improve this question















In a textbook I am asked to calculate an expression using a calculator: $$frac{1}{2}lnleft(e^{22}right)$$



I tried using:



0.5*log(exp**22)


This returned an error:



File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/decorators.py", line 91, in __sympifyit_wrapper return func(a, b) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/decorators.py", line 132, in binary_op_wrapper return func(self, other) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/expr.py", line 172, in __rpow__ return Pow(other, self) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/cache.py", line 95, in wrapper retval = func(*args, **kwargs) File "/base/data/home/apps/s~sympy-live-hrd/56.412575950366508090/sympy/sympy/core/power.py", line 286, in __new__ obj = b._eval_power(e) TypeError: unbound method _eval_power() must be called with exp instance as first argument (got Integer instance instead)


I tried replacing $exp$ with just $e$ and received a similar error.



From my solutions page I can see that the answer is $11$. How can I type the expression in order to get $11$ back?






python




python






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 '18 at 19:27









Robert Howard

1,9161822




1,9161822










asked Nov 15 '18 at 23:02









Doug Fir

2177




2177












  • I believe it's $.5*log(exp(1)**22)$, but anyway you should ask these questions on a stack exchange more relevant to python errors.
    – TrostAft
    Nov 15 '18 at 23:05










  • The exp function is a function, not the number $e$.
    – Sean Roberson
    Nov 15 '18 at 23:05


















  • I believe it's $.5*log(exp(1)**22)$, but anyway you should ask these questions on a stack exchange more relevant to python errors.
    – TrostAft
    Nov 15 '18 at 23:05










  • The exp function is a function, not the number $e$.
    – Sean Roberson
    Nov 15 '18 at 23:05
















I believe it's $.5*log(exp(1)**22)$, but anyway you should ask these questions on a stack exchange more relevant to python errors.
– TrostAft
Nov 15 '18 at 23:05




I believe it's $.5*log(exp(1)**22)$, but anyway you should ask these questions on a stack exchange more relevant to python errors.
– TrostAft
Nov 15 '18 at 23:05












The exp function is a function, not the number $e$.
– Sean Roberson
Nov 15 '18 at 23:05




The exp function is a function, not the number $e$.
– Sean Roberson
Nov 15 '18 at 23:05










2 Answers
2






active

oldest

votes


















5














You should use exp as function: 0.5*log(exp(22))






share|cite|improve this answer























  • Ah, thank you so much. Accepting this once the time limit comes off
    – Doug Fir
    Nov 15 '18 at 23:07






  • 1




    +1. As a Python user, I can confirm that's the syntax.
    – J.G.
    Nov 15 '18 at 23:08



















3














You don't even need a calculator.



Note that $ln(e) = 1$ as $e^1 = e$. Recall also that $log_b (x^n) = n log_b (x).$ Thus $frac{1}{2} ln(e^{22}) = 11.$






share|cite|improve this answer





















  • My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
    – MPW
    Nov 15 '18 at 23:10











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









5














You should use exp as function: 0.5*log(exp(22))






share|cite|improve this answer























  • Ah, thank you so much. Accepting this once the time limit comes off
    – Doug Fir
    Nov 15 '18 at 23:07






  • 1




    +1. As a Python user, I can confirm that's the syntax.
    – J.G.
    Nov 15 '18 at 23:08
















5














You should use exp as function: 0.5*log(exp(22))






share|cite|improve this answer























  • Ah, thank you so much. Accepting this once the time limit comes off
    – Doug Fir
    Nov 15 '18 at 23:07






  • 1




    +1. As a Python user, I can confirm that's the syntax.
    – J.G.
    Nov 15 '18 at 23:08














5












5








5






You should use exp as function: 0.5*log(exp(22))






share|cite|improve this answer














You should use exp as function: 0.5*log(exp(22))







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 15 '18 at 23:07









J.G.

23k22137




23k22137










answered Nov 15 '18 at 23:05









gammatester

16.7k21632




16.7k21632












  • Ah, thank you so much. Accepting this once the time limit comes off
    – Doug Fir
    Nov 15 '18 at 23:07






  • 1




    +1. As a Python user, I can confirm that's the syntax.
    – J.G.
    Nov 15 '18 at 23:08


















  • Ah, thank you so much. Accepting this once the time limit comes off
    – Doug Fir
    Nov 15 '18 at 23:07






  • 1




    +1. As a Python user, I can confirm that's the syntax.
    – J.G.
    Nov 15 '18 at 23:08
















Ah, thank you so much. Accepting this once the time limit comes off
– Doug Fir
Nov 15 '18 at 23:07




Ah, thank you so much. Accepting this once the time limit comes off
– Doug Fir
Nov 15 '18 at 23:07




1




1




+1. As a Python user, I can confirm that's the syntax.
– J.G.
Nov 15 '18 at 23:08




+1. As a Python user, I can confirm that's the syntax.
– J.G.
Nov 15 '18 at 23:08











3














You don't even need a calculator.



Note that $ln(e) = 1$ as $e^1 = e$. Recall also that $log_b (x^n) = n log_b (x).$ Thus $frac{1}{2} ln(e^{22}) = 11.$






share|cite|improve this answer





















  • My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
    – MPW
    Nov 15 '18 at 23:10
















3














You don't even need a calculator.



Note that $ln(e) = 1$ as $e^1 = e$. Recall also that $log_b (x^n) = n log_b (x).$ Thus $frac{1}{2} ln(e^{22}) = 11.$






share|cite|improve this answer





















  • My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
    – MPW
    Nov 15 '18 at 23:10














3












3








3






You don't even need a calculator.



Note that $ln(e) = 1$ as $e^1 = e$. Recall also that $log_b (x^n) = n log_b (x).$ Thus $frac{1}{2} ln(e^{22}) = 11.$






share|cite|improve this answer












You don't even need a calculator.



Note that $ln(e) = 1$ as $e^1 = e$. Recall also that $log_b (x^n) = n log_b (x).$ Thus $frac{1}{2} ln(e^{22}) = 11.$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 15 '18 at 23:08









Sean Roberson

6,40531327




6,40531327












  • My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
    – MPW
    Nov 15 '18 at 23:10


















  • My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
    – MPW
    Nov 15 '18 at 23:10
















My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
– MPW
Nov 15 '18 at 23:10




My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
– MPW
Nov 15 '18 at 23:10


















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