Algebraic expression to the algebraic expressionth power equation












0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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


















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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
















0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 20:37









mrtaurho

5,68551540




5,68551540










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
















  • 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












2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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





















0












$begingroup$

$$x^{x-1}=e^{(x-1)log x}implies x^{x-1}=7iff e^{(x-1)log x}=e^{log 7}implies(x-1)log x=log7$$



and you'll probably need some trascendental function like Lambert function or stuff.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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


















    1












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    1












    1








    1





    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 4 at 21:43

























    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




















    • $begingroup$
      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 $e^{2pi in}=1$.
    $endgroup$
    – mrtaurho
    Jan 4 at 21:34














    $begingroup$
    Can you tell me what is $pi$in?
    $endgroup$
    – Math Lover
    Jan 4 at 21:46




    $begingroup$
    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






    $begingroup$
    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






    $begingroup$
    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













    0












    $begingroup$

    $$x^{x-1}=e^{(x-1)log x}implies x^{x-1}=7iff e^{(x-1)log x}=e^{log 7}implies(x-1)log x=log7$$



    and you'll probably need some trascendental function like Lambert function or stuff.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      $$x^{x-1}=e^{(x-1)log x}implies x^{x-1}=7iff e^{(x-1)log x}=e^{log 7}implies(x-1)log x=log7$$



      and you'll probably need some trascendental function like Lambert function or stuff.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        $$x^{x-1}=e^{(x-1)log x}implies x^{x-1}=7iff e^{(x-1)log x}=e^{log 7}implies(x-1)log x=log7$$



        and you'll probably need some trascendental function like Lambert function or stuff.






        share|cite|improve this answer









        $endgroup$



        $$x^{x-1}=e^{(x-1)log x}implies x^{x-1}=7iff e^{(x-1)log x}=e^{log 7}implies(x-1)log x=log7$$



        and you'll probably need some trascendental function like Lambert function or stuff.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 20:34









        DonAntonioDonAntonio

        179k1494232




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