Algebraic expression to the algebraic expressionth power equation
$begingroup$
How can you solve problems like $ x^{x-1}=7 $? More generally, how can you solve equations like $(ax+b)^{cx+d}=e$ , where $a,b,c,d,e$ are given?$($Give all the roots, including complex ones$)$
exponential-function
edited Jan 4 at 20:37
mrtaurho
5,68551540
asked Jan 4 at 20:29
Math LoverMath Lover
15910
$endgroup$
add a comment |
$begingroup$
How can you solve problems like $ x^{x-1}=7 $? More generally, how can you solve equations like $(ax+b)^{cx+d}=e$ , where $a,b,c,d,e$ are given?$($Give all the roots, including complex ones$)$
exponential-function
edited Jan 4 at 20:37
mrtaurho
5,68551540
asked Jan 4 at 20:29
Math LoverMath Lover
15910
$endgroup$
1
$begingroup$
I think you are looking for the Lambert W-Function. Concerning your given equation it seems like even the Lambert W-Function does not help in order to produce a closed-form hence WolframAlpha only gives two numerical solutions.
$endgroup$
– mrtaurho
Jan 4 at 20:34
$begingroup$
I think WolframAlpha is only giving real solutions, but I want the complex solutions,too.
$endgroup$
– Math Lover
Jan 4 at 20:51
$begingroup$
I can understand you doubts but take a look at this similiar request which indeed produces complex values too $($take for example the $W_{-1}$ solution for $n=0$ which is a complex number, to be precise look here$)$.
$endgroup$
– mrtaurho
Jan 4 at 20:56
add a comment |
$begingroup$
How can you solve problems like $ x^{x-1}=7 $? More generally, how can you solve equations like $(ax+b)^{cx+d}=e$ , where $a,b,c,d,e$ are given?$($Give all the roots, including complex ones$)$
exponential-function
edited Jan 4 at 20:37
mrtaurho
5,68551540
asked Jan 4 at 20:29
Math LoverMath Lover
15910
$endgroup$
How can you solve problems like $ x^{x-1}=7 $? More generally, how can you solve equations like $(ax+b)^{cx+d}=e$ , where $a,b,c,d,e$ are given?$($Give all the roots, including complex ones$)$
exponential-function
exponential-function
edited Jan 4 at 20:37
mrtaurho
5,68551540
asked Jan 4 at 20:29
Math LoverMath Lover
15910
edited Jan 4 at 20:37
mrtaurho
5,68551540
asked Jan 4 at 20:29
Math LoverMath Lover
15910
edited Jan 4 at 20:37
mrtaurho
5,68551540
edited Jan 4 at 20:37
mrtaurho
5,68551540
edited Jan 4 at 20:37
mrtaurho
5,68551540
5,68551540
asked Jan 4 at 20:29
Math LoverMath Lover
15910
asked Jan 4 at 20:29
Math LoverMath Lover
15910
asked Jan 4 at 20:29
Math LoverMath Lover
15910
15910
1
$begingroup$
I think you are looking for the Lambert W-Function. Concerning your given equation it seems like even the Lambert W-Function does not help in order to produce a closed-form hence WolframAlpha only gives two numerical solutions.
$endgroup$
– mrtaurho
Jan 4 at 20:34
$begingroup$
I think WolframAlpha is only giving real solutions, but I want the complex solutions,too.
$endgroup$
– Math Lover
Jan 4 at 20:51
$begingroup$
I can understand you doubts but take a look at this similiar request which indeed produces complex values too $($take for example the $W_{-1}$ solution for $n=0$ which is a complex number, to be precise look here$)$.
$endgroup$
– mrtaurho
Jan 4 at 20:56
add a comment |
1
$begingroup$
I think you are looking for the Lambert W-Function. Concerning your given equation it seems like even the Lambert W-Function does not help in order to produce a closed-form hence WolframAlpha only gives two numerical solutions.
$endgroup$
– mrtaurho
Jan 4 at 20:34
$begingroup$
I think WolframAlpha is only giving real solutions, but I want the complex solutions,too.
$endgroup$
– Math Lover
Jan 4 at 20:51
$begingroup$
I can understand you doubts but take a look at this similiar request which indeed produces complex values too $($take for example the $W_{-1}$ solution for $n=0$ which is a complex number, to be precise look here$)$.
$endgroup$
– mrtaurho
Jan 4 at 20:56
1
1
$begingroup$
I think you are looking for the Lambert W-Function. Concerning your given equation it seems like even the Lambert W-Function does not help in order to produce a closed-form hence WolframAlpha only gives two numerical solutions.
$endgroup$
– mrtaurho
Jan 4 at 20:34
$begingroup$
I think you are looking for the Lambert W-Function. Concerning your given equation it seems like even the Lambert W-Function does not help in order to produce a closed-form hence WolframAlpha only gives two numerical solutions.
$endgroup$
– mrtaurho
Jan 4 at 20:34
$begingroup$
I think WolframAlpha is only giving real solutions, but I want the complex solutions,too.
$endgroup$
– Math Lover
Jan 4 at 20:51
$begingroup$
I think WolframAlpha is only giving real solutions, but I want the complex solutions,too.
$endgroup$
– Math Lover
Jan 4 at 20:51
$begingroup$
I can understand you doubts but take a look at this similiar request which indeed produces complex values too $($take for example the $W_{-1}$ solution for $n=0$ which is a complex number, to be precise look here$)$.
$endgroup$
– mrtaurho
Jan 4 at 20:56
$begingroup$
I can understand you doubts but take a look at this similiar request which indeed produces complex values too $($take for example the $W_{-1}$ solution for $n=0$ which is a complex number, to be precise look here$)$.
$endgroup$
– mrtaurho
Jan 4 at 20:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Interesting enough for me to post an answer is the fact that the problem for your first equation only lies in the fact that the exponent is given by $x-1$ and not by $x$. I will demnonstrate how to solve this equation for the latter case. First of all rewrite the $x^x$ term in terms of the exponential and do not forget assuming a complex valued logarithm to get
$$begin{align*}
x^x&=7\
e^{xlog(x)+2pi i n}&=7\
xlog(x)+2pi i n&=log(7)\
log(x)e^{log(x)}&=log(7)-2pi i n\
log(x)&=W(log(7)-2pi i n)
end{align*}$$
$$therefore~x=e^{W(log(7)-2pi i n)}~~~ninmathbb Z$$
I have doubts that on can deduce a general formula for arbitrary $a,b,c,d,e$ $($just take $a=c=1$,$b=0$,$d=-1$ and $e=7$ to reproduce your first equation$)$. Anyway considering that $a=c$ and $b=d$ it is indeed possible since this is basically the same as $x^x$ and can be solved using the Lamber W-Function but note that you have to consider the different branches of this function with regard to the values of $a,b$ and $e$.
From hereon I have to admit that I have not enough experience with the Lambert W-Function to give an detailed outline of the different branches and why they are important.
answered Jan 4 at 20:56
mrtaurhomrtaurho
5,68551540
$endgroup$
$begingroup$
Can you explain a bitthe Lambert W-Function, and how you solved $x^x=7$? I don’t understand your steps. Thanks! Also, will it be helpful to multiply both sides of the equation by x, to get $x^x=7x$?
$endgroup$
– Math Lover
Jan 4 at 21:15
$begingroup$
The Lambet W-Function is defined as the inverse function $f(z)=ze^z$, i.e. $W(z)=f^{-1}(z)$. Thus, by having an equation of this structure on can apply this definition, namely $$ze^z=rRightarrow z=W(r)$$ By performing the steps to get this structure with your given equation leaves us with $$log(x)e^{log(x)}=log(7)-2pi i nRightarrow log(x)=W(log(7)-2pi i n)$$Concerning the extra $2pi i n$: this happens since the logarithm as a complex-valued function is far away from being injective. In fact the logarithm inherits a additive period from the exponential since $e^{2pi in}=1$.
$endgroup$
– mrtaurho
Jan 4 at 21:34
$begingroup$
Can you tell me what is $pi$in?
$endgroup$
– Math Lover
Jan 4 at 21:46
$begingroup$
Since you were asking for the complex roots of your equation too I assume you are used to complex numbers in general. Thus, recall Euler's Formula $e^{it}=cos(t)+isin(t)$ We know that the sine function aswell as the cosine function are both periodic with period $2pi$. So plugging in $t+2pi$ does not change the value $$e^{i(t+2pi)}=e^{it}$$ And this can be taken further so that we can conclude $$e^{i(t+2pi n)}=e^{it}~~~ninmathbb Z$$ This is what this extra term is all about: ensuring that we get all possible values of the exponential, and the logarithm repectively, at not just one.
$endgroup$
– mrtaurho
Jan 4 at 21:56
$begingroup$
Shouldn’t the right side of your third line of your solution be ln(7),instead of log(7)?
$endgroup$
– Math Lover
Jan 4 at 22:57
|
show 13 more comments
$begingroup$
$$x^{x-1}=e^{(x-1)log x}implies x^{x-1}=7iff e^{(x-1)log x}=e^{log 7}implies(x-1)log x=log7$$
and you'll probably need some trascendental function like Lambert function or stuff.
answered Jan 4 at 20:34
DonAntonioDonAntonio
179k1494232
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Interesting enough for me to post an answer is the fact that the problem for your first equation only lies in the fact that the exponent is given by $x-1$ and not by $x$. I will demnonstrate how to solve this equation for the latter case. First of all rewrite the $x^x$ term in terms of the exponential and do not forget assuming a complex valued logarithm to get
$$begin{align*}
x^x&=7\
e^{xlog(x)+2pi i n}&=7\
xlog(x)+2pi i n&=log(7)\
log(x)e^{log(x)}&=log(7)-2pi i n\
log(x)&=W(log(7)-2pi i n)
end{align*}$$
$$therefore~x=e^{W(log(7)-2pi i n)}~~~ninmathbb Z$$
I have doubts that on can deduce a general formula for arbitrary $a,b,c,d,e$ $($just take $a=c=1$,$b=0$,$d=-1$ and $e=7$ to reproduce your first equation$)$. Anyway considering that $a=c$ and $b=d$ it is indeed possible since this is basically the same as $x^x$ and can be solved using the Lamber W-Function but note that you have to consider the different branches of this function with regard to the values of $a,b$ and $e$.
From hereon I have to admit that I have not enough experience with the Lambert W-Function to give an detailed outline of the different branches and why they are important.
answered Jan 4 at 20:56
mrtaurhomrtaurho
5,68551540
$endgroup$
$begingroup$
Can you explain a bitthe Lambert W-Function, and how you solved $x^x=7$? I don’t understand your steps. Thanks! Also, will it be helpful to multiply both sides of the equation by x, to get $x^x=7x$?
$endgroup$
– Math Lover
Jan 4 at 21:15
$begingroup$
The Lambet W-Function is defined as the inverse function $f(z)=ze^z$, i.e. $W(z)=f^{-1}(z)$. Thus, by having an equation of this structure on can apply this definition, namely $$ze^z=rRightarrow z=W(r)$$ By performing the steps to get this structure with your given equation leaves us with $$log(x)e^{log(x)}=log(7)-2pi i nRightarrow log(x)=W(log(7)-2pi i n)$$Concerning the extra $2pi i n$: this happens since the logarithm as a complex-valued function is far away from being injective. In fact the logarithm inherits a additive period from the exponential since $e^{2pi in}=1$.
$endgroup$
– mrtaurho
Jan 4 at 21:34
$begingroup$
Can you tell me what is $pi$in?
$endgroup$
– Math Lover
Jan 4 at 21:46
$begingroup$
Since you were asking for the complex roots of your equation too I assume you are used to complex numbers in general. Thus, recall Euler's Formula $e^{it}=cos(t)+isin(t)$ We know that the sine function aswell as the cosine function are both periodic with period $2pi$. So plugging in $t+2pi$ does not change the value $$e^{i(t+2pi)}=e^{it}$$ And this can be taken further so that we can conclude $$e^{i(t+2pi n)}=e^{it}~~~ninmathbb Z$$ This is what this extra term is all about: ensuring that we get all possible values of the exponential, and the logarithm repectively, at not just one.
$endgroup$
– mrtaurho
Jan 4 at 21:56
$begingroup$
Shouldn’t the right side of your third line of your solution be ln(7),instead of log(7)?
$endgroup$
– Math Lover
Jan 4 at 22:57
|
show 13 more comments
$begingroup$
Interesting enough for me to post an answer is the fact that the problem for your first equation only lies in the fact that the exponent is given by $x-1$ and not by $x$. I will demnonstrate how to solve this equation for the latter case. First of all rewrite the $x^x$ term in terms of the exponential and do not forget assuming a complex valued logarithm to get
$$begin{align*}
x^x&=7\
e^{xlog(x)+2pi i n}&=7\
xlog(x)+2pi i n&=log(7)\
log(x)e^{log(x)}&=log(7)-2pi i n\
log(x)&=W(log(7)-2pi i n)
end{align*}$$
$$therefore~x=e^{W(log(7)-2pi i n)}~~~ninmathbb Z$$
I have doubts that on can deduce a general formula for arbitrary $a,b,c,d,e$ $($just take $a=c=1$,$b=0$,$d=-1$ and $e=7$ to reproduce your first equation$)$. Anyway considering that $a=c$ and $b=d$ it is indeed possible since this is basically the same as $x^x$ and can be solved using the Lamber W-Function but note that you have to consider the different branches of this function with regard to the values of $a,b$ and $e$.
From hereon I have to admit that I have not enough experience with the Lambert W-Function to give an detailed outline of the different branches and why they are important.
answered Jan 4 at 20:56
mrtaurhomrtaurho
5,68551540
$endgroup$
$begingroup$
Can you explain a bitthe Lambert W-Function, and how you solved $x^x=7$? I don’t understand your steps. Thanks! Also, will it be helpful to multiply both sides of the equation by x, to get $x^x=7x$?
$endgroup$
– Math Lover
Jan 4 at 21:15
$begingroup$
The Lambet W-Function is defined as the inverse function $f(z)=ze^z$, i.e. $W(z)=f^{-1}(z)$. Thus, by having an equation of this structure on can apply this definition, namely $$ze^z=rRightarrow z=W(r)$$ By performing the steps to get this structure with your given equation leaves us with $$log(x)e^{log(x)}=log(7)-2pi i nRightarrow log(x)=W(log(7)-2pi i n)$$Concerning the extra $2pi i n$: this happens since the logarithm as a complex-valued function is far away from being injective. In fact the logarithm inherits a additive period from the exponential since $e^{2pi in}=1$.
$endgroup$
– mrtaurho
Jan 4 at 21:34
$begingroup$
Can you tell me what is $pi$in?
$endgroup$
– Math Lover
Jan 4 at 21:46
$begingroup$
Since you were asking for the complex roots of your equation too I assume you are used to complex numbers in general. Thus, recall Euler's Formula $e^{it}=cos(t)+isin(t)$ We know that the sine function aswell as the cosine function are both periodic with period $2pi$. So plugging in $t+2pi$ does not change the value $$e^{i(t+2pi)}=e^{it}$$ And this can be taken further so that we can conclude $$e^{i(t+2pi n)}=e^{it}~~~ninmathbb Z$$ This is what this extra term is all about: ensuring that we get all possible values of the exponential, and the logarithm repectively, at not just one.
$endgroup$
– mrtaurho
Jan 4 at 21:56
$begingroup$
Shouldn’t the right side of your third line of your solution be ln(7),instead of log(7)?
$endgroup$
– Math Lover
Jan 4 at 22:57
|
show 13 more comments
$begingroup$
Interesting enough for me to post an answer is the fact that the problem for your first equation only lies in the fact that the exponent is given by $x-1$ and not by $x$. I will demnonstrate how to solve this equation for the latter case. First of all rewrite the $x^x$ term in terms of the exponential and do not forget assuming a complex valued logarithm to get
$$begin{align*}
x^x&=7\
e^{xlog(x)+2pi i n}&=7\
xlog(x)+2pi i n&=log(7)\
log(x)e^{log(x)}&=log(7)-2pi i n\
log(x)&=W(log(7)-2pi i n)
end{align*}$$
$$therefore~x=e^{W(log(7)-2pi i n)}~~~ninmathbb Z$$
I have doubts that on can deduce a general formula for arbitrary $a,b,c,d,e$ $($just take $a=c=1$,$b=0$,$d=-1$ and $e=7$ to reproduce your first equation$)$. Anyway considering that $a=c$ and $b=d$ it is indeed possible since this is basically the same as $x^x$ and can be solved using the Lamber W-Function but note that you have to consider the different branches of this function with regard to the values of $a,b$ and $e$.
From hereon I have to admit that I have not enough experience with the Lambert W-Function to give an detailed outline of the different branches and why they are important.
answered Jan 4 at 20:56
mrtaurhomrtaurho
5,68551540
$endgroup$
Interesting enough for me to post an answer is the fact that the problem for your first equation only lies in the fact that the exponent is given by $x-1$ and not by $x$. I will demnonstrate how to solve this equation for the latter case. First of all rewrite the $x^x$ term in terms of the exponential and do not forget assuming a complex valued logarithm to get
$$begin{align*}
x^x&=7\
e^{xlog(x)+2pi i n}&=7\
xlog(x)+2pi i n&=log(7)\
log(x)e^{log(x)}&=log(7)-2pi i n\
log(x)&=W(log(7)-2pi i n)
end{align*}$$
$$therefore~x=e^{W(log(7)-2pi i n)}~~~ninmathbb Z$$
I have doubts that on can deduce a general formula for arbitrary $a,b,c,d,e$ $($just take $a=c=1$,$b=0$,$d=-1$ and $e=7$ to reproduce your first equation$)$. Anyway considering that $a=c$ and $b=d$ it is indeed possible since this is basically the same as $x^x$ and can be solved using the Lamber W-Function but note that you have to consider the different branches of this function with regard to the values of $a,b$ and $e$.
From hereon I have to admit that I have not enough experience with the Lambert W-Function to give an detailed outline of the different branches and why they are important.
answered Jan 4 at 20:56
mrtaurhomrtaurho
5,68551540
edited Jan 4 at 21:43
answered Jan 4 at 20:56
mrtaurhomrtaurho
5,68551540
answered Jan 4 at 20:56
mrtaurhomrtaurho
5,68551540
answered Jan 4 at 20:56
mrtaurhomrtaurho
5,68551540
5,68551540
$begingroup$
Can you explain a bitthe Lambert W-Function, and how you solved $x^x=7$? I don’t understand your steps. Thanks! Also, will it be helpful to multiply both sides of the equation by x, to get $x^x=7x$?
$endgroup$
– Math Lover
Jan 4 at 21:15
$begingroup$
The Lambet W-Function is defined as the inverse function $f(z)=ze^z$, i.e. $W(z)=f^{-1}(z)$. Thus, by having an equation of this structure on can apply this definition, namely $$ze^z=rRightarrow z=W(r)$$ By performing the steps to get this structure with your given equation leaves us with $$log(x)e^{log(x)}=log(7)-2pi i nRightarrow log(x)=W(log(7)-2pi i n)$$Concerning the extra $2pi i n$: this happens since the logarithm as a complex-valued function is far away from being injective. In fact the logarithm inherits a additive period from the exponential since $e^{2pi in}=1$.
$endgroup$
– mrtaurho
Jan 4 at 21:34
$begingroup$
Can you tell me what is $pi$in?
$endgroup$
– Math Lover
Jan 4 at 21:46
$begingroup$
Since you were asking for the complex roots of your equation too I assume you are used to complex numbers in general. Thus, recall Euler's Formula $e^{it}=cos(t)+isin(t)$ We know that the sine function aswell as the cosine function are both periodic with period $2pi$. So plugging in $t+2pi$ does not change the value $$e^{i(t+2pi)}=e^{it}$$ And this can be taken further so that we can conclude $$e^{i(t+2pi n)}=e^{it}~~~ninmathbb Z$$ This is what this extra term is all about: ensuring that we get all possible values of the exponential, and the logarithm repectively, at not just one.
$endgroup$
– mrtaurho
Jan 4 at 21:56
$begingroup$
Shouldn’t the right side of your third line of your solution be ln(7),instead of log(7)?
$endgroup$
– Math Lover
Jan 4 at 22:57
|
show 13 more comments
$begingroup$
Can you explain a bitthe Lambert W-Function, and how you solved $x^x=7$? I don’t understand your steps. Thanks! Also, will it be helpful to multiply both sides of the equation by x, to get $x^x=7x$?
$endgroup$
– Math Lover
Jan 4 at 21:15
$begingroup$
The Lambet W-Function is defined as the inverse function $f(z)=ze^z$, i.e. $W(z)=f^{-1}(z)$. Thus, by having an equation of this structure on can apply this definition, namely $$ze^z=rRightarrow z=W(r)$$ By performing the steps to get this structure with your given equation leaves us with $$log(x)e^{log(x)}=log(7)-2pi i nRightarrow log(x)=W(log(7)-2pi i n)$$Concerning the extra $2pi i n$: this happens since the logarithm as a complex-valued function is far away from being injective. In fact the logarithm inherits a additive period from the exponential since $e^{2pi in}=1$.
$endgroup$
– mrtaurho
Jan 4 at 21:34
$begingroup$
Can you tell me what is $pi$in?
$endgroup$
– Math Lover
Jan 4 at 21:46
$begingroup$
Since you were asking for the complex roots of your equation too I assume you are used to complex numbers in general. Thus, recall Euler's Formula $e^{it}=cos(t)+isin(t)$ We know that the sine function aswell as the cosine function are both periodic with period $2pi$. So plugging in $t+2pi$ does not change the value $$e^{i(t+2pi)}=e^{it}$$ And this can be taken further so that we can conclude $$e^{i(t+2pi n)}=e^{it}~~~ninmathbb Z$$ This is what this extra term is all about: ensuring that we get all possible values of the exponential, and the logarithm repectively, at not just one.
$endgroup$
– mrtaurho
Jan 4 at 21:56
$begingroup$
Shouldn’t the right side of your third line of your solution be ln(7),instead of log(7)?
$endgroup$
– Math Lover
Jan 4 at 22:57
$begingroup$
Can you explain a bitthe Lambert W-Function, and how you solved $x^x=7$? I don’t understand your steps. Thanks! Also, will it be helpful to multiply both sides of the equation by x, to get $x^x=7x$?
$endgroup$
– Math Lover
Jan 4 at 21:15
$begingroup$
Can you explain a bitthe Lambert W-Function, and how you solved $x^x=7$? I don’t understand your steps. Thanks! Also, will it be helpful to multiply both sides of the equation by x, to get $x^x=7x$?
$endgroup$
– Math Lover
Jan 4 at 21:15
$begingroup$
The Lambet W-Function is defined as the inverse function $f(z)=ze^z$, i.e. $W(z)=f^{-1}(z)$. Thus, by having an equation of this structure on can apply this definition, namely $$ze^z=rRightarrow z=W(r)$$ By performing the steps to get this structure with your given equation leaves us with $$log(x)e^{log(x)}=log(7)-2pi i nRightarrow log(x)=W(log(7)-2pi i n)$$Concerning the extra $2pi i n$: this happens since the logarithm as a complex-valued function is far away from being injective. In fact the logarithm inherits a additive period from the exponential since $e^{2pi in}=1$.
$endgroup$
– mrtaurho
Jan 4 at 21:34
$begingroup$
The Lambet W-Function is defined as the inverse function $f(z)=ze^z$, i.e. $W(z)=f^{-1}(z)$. Thus, by having an equation of this structure on can apply this definition, namely $$ze^z=rRightarrow z=W(r)$$ By performing the steps to get this structure with your given equation leaves us with $$log(x)e^{log(x)}=log(7)-2pi i nRightarrow log(x)=W(log(7)-2pi i n)$$Concerning the extra $2pi i n$: this happens since the logarithm as a complex-valued function is far away from being injective. In fact the logarithm inherits a additive period from the exponential since $e^{2pi in}=1$.
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– mrtaurho
Jan 4 at 21:34
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Can you tell me what is $pi$in?
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– Math Lover
Jan 4 at 21:46
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Can you tell me what is $pi$in?
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– Math Lover
Jan 4 at 21:46
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Since you were asking for the complex roots of your equation too I assume you are used to complex numbers in general. Thus, recall Euler's Formula $e^{it}=cos(t)+isin(t)$ We know that the sine function aswell as the cosine function are both periodic with period $2pi$. So plugging in $t+2pi$ does not change the value $$e^{i(t+2pi)}=e^{it}$$ And this can be taken further so that we can conclude $$e^{i(t+2pi n)}=e^{it}~~~ninmathbb Z$$ This is what this extra term is all about: ensuring that we get all possible values of the exponential, and the logarithm repectively, at not just one.
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– mrtaurho
Jan 4 at 21:56
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Since you were asking for the complex roots of your equation too I assume you are used to complex numbers in general. Thus, recall Euler's Formula $e^{it}=cos(t)+isin(t)$ We know that the sine function aswell as the cosine function are both periodic with period $2pi$. So plugging in $t+2pi$ does not change the value $$e^{i(t+2pi)}=e^{it}$$ And this can be taken further so that we can conclude $$e^{i(t+2pi n)}=e^{it}~~~ninmathbb Z$$ This is what this extra term is all about: ensuring that we get all possible values of the exponential, and the logarithm repectively, at not just one.
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– mrtaurho
Jan 4 at 21:56
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Shouldn’t the right side of your third line of your solution be ln(7),instead of log(7)?
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– Math Lover
Jan 4 at 22:57
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Shouldn’t the right side of your third line of your solution be ln(7),instead of log(7)?
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– Math Lover
Jan 4 at 22:57
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show 13 more comments
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$$x^{x-1}=e^{(x-1)log x}implies x^{x-1}=7iff e^{(x-1)log x}=e^{log 7}implies(x-1)log x=log7$$
and you'll probably need some trascendental function like Lambert function or stuff.
answered Jan 4 at 20:34
DonAntonioDonAntonio
179k1494232
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add a comment |
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$$x^{x-1}=e^{(x-1)log x}implies x^{x-1}=7iff e^{(x-1)log x}=e^{log 7}implies(x-1)log x=log7$$
and you'll probably need some trascendental function like Lambert function or stuff.
answered Jan 4 at 20:34
DonAntonioDonAntonio
179k1494232
$endgroup$
add a comment |
$begingroup$
$$x^{x-1}=e^{(x-1)log x}implies x^{x-1}=7iff e^{(x-1)log x}=e^{log 7}implies(x-1)log x=log7$$
and you'll probably need some trascendental function like Lambert function or stuff.
answered Jan 4 at 20:34
DonAntonioDonAntonio
179k1494232
$endgroup$
$$x^{x-1}=e^{(x-1)log x}implies x^{x-1}=7iff e^{(x-1)log x}=e^{log 7}implies(x-1)log x=log7$$
and you'll probably need some trascendental function like Lambert function or stuff.
answered Jan 4 at 20:34
DonAntonioDonAntonio
179k1494232
answered Jan 4 at 20:34
DonAntonioDonAntonio
179k1494232
answered Jan 4 at 20:34
DonAntonioDonAntonio
179k1494232
answered Jan 4 at 20:34
DonAntonioDonAntonio
179k1494232
179k1494232
add a comment |
add a comment |
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I think you are looking for the Lambert W-Function. Concerning your given equation it seems like even the Lambert W-Function does not help in order to produce a closed-form hence WolframAlpha only gives two numerical solutions.
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– mrtaurho
Jan 4 at 20:34
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I think WolframAlpha is only giving real solutions, but I want the complex solutions,too.
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– Math Lover
Jan 4 at 20:51
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I can understand you doubts but take a look at this similiar request which indeed produces complex values too $($take for example the $W_{-1}$ solution for $n=0$ which is a complex number, to be precise look here$)$.
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– mrtaurho
Jan 4 at 20:56