How can I type this into a calculator?
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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
add a comment |
up vote
0
down vote
favorite
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
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 at 23:05
Theexp
function is a function, not the number $e$.
– Sean Roberson
Nov 15 at 23:05
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
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
edited Nov 20 at 19:27
Robert Howard
1,804722
1,804722
asked Nov 15 at 23:02
Doug Fir
1656
1656
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 at 23:05
Theexp
function is a function, not the number $e$.
– Sean Roberson
Nov 15 at 23:05
add a comment |
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 at 23:05
Theexp
function is a function, not the number $e$.
– Sean Roberson
Nov 15 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 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 at 23:05
The
exp
function is a function, not the number $e$.– Sean Roberson
Nov 15 at 23:05
The
exp
function is a function, not the number $e$.– Sean Roberson
Nov 15 at 23:05
add a comment |
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
You should use exp as function: 0.5*log(exp(22))
Ah, thank you so much. Accepting this once the time limit comes off
– Doug Fir
Nov 15 at 23:07
1
+1. As a Python user, I can confirm that's the syntax.
– J.G.
Nov 15 at 23:08
add a comment |
up vote
3
down vote
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 $frac{1}{2} ln(e^{22}) = 11.$
My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
– MPW
Nov 15 at 23:10
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
You should use exp as function: 0.5*log(exp(22))
Ah, thank you so much. Accepting this once the time limit comes off
– Doug Fir
Nov 15 at 23:07
1
+1. As a Python user, I can confirm that's the syntax.
– J.G.
Nov 15 at 23:08
add a comment |
up vote
4
down vote
accepted
You should use exp as function: 0.5*log(exp(22))
Ah, thank you so much. Accepting this once the time limit comes off
– Doug Fir
Nov 15 at 23:07
1
+1. As a Python user, I can confirm that's the syntax.
– J.G.
Nov 15 at 23:08
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
You should use exp as function: 0.5*log(exp(22))
You should use exp as function: 0.5*log(exp(22))
edited Nov 15 at 23:07
J.G.
19.2k21932
19.2k21932
answered Nov 15 at 23:05
gammatester
16.4k21631
16.4k21631
Ah, thank you so much. Accepting this once the time limit comes off
– Doug Fir
Nov 15 at 23:07
1
+1. As a Python user, I can confirm that's the syntax.
– J.G.
Nov 15 at 23:08
add a comment |
Ah, thank you so much. Accepting this once the time limit comes off
– Doug Fir
Nov 15 at 23:07
1
+1. As a Python user, I can confirm that's the syntax.
– J.G.
Nov 15 at 23:08
Ah, thank you so much. Accepting this once the time limit comes off
– Doug Fir
Nov 15 at 23:07
Ah, thank you so much. Accepting this once the time limit comes off
– Doug Fir
Nov 15 at 23:07
1
1
+1. As a Python user, I can confirm that's the syntax.
– J.G.
Nov 15 at 23:08
+1. As a Python user, I can confirm that's the syntax.
– J.G.
Nov 15 at 23:08
add a comment |
up vote
3
down vote
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 $frac{1}{2} ln(e^{22}) = 11.$
My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
– MPW
Nov 15 at 23:10
add a comment |
up vote
3
down vote
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 $frac{1}{2} ln(e^{22}) = 11.$
My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
– MPW
Nov 15 at 23:10
add a comment |
up vote
3
down vote
up vote
3
down vote
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 $frac{1}{2} ln(e^{22}) = 11.$
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 $frac{1}{2} ln(e^{22}) = 11.$
answered Nov 15 at 23:08
Sean Roberson
6,31131327
6,31131327
My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
– MPW
Nov 15 at 23:10
add a comment |
My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
– MPW
Nov 15 at 23:10
My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
– MPW
Nov 15 at 23:10
My first reaction as well—but note that this does NOT answer OP’s question in the slightest.
– MPW
Nov 15 at 23:10
add a comment |
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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 at 23:05
The
exp
function is a function, not the number $e$.– Sean Roberson
Nov 15 at 23:05