Solve $(x^2-5x+5)^x^2-7x+12= 1$.

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What are all the possible values of $x$ if $(x^2-5x+5)^x^2-7x+12= 1$. I have already found three answers: 4, 3, and 1, however, apparently there are more possibilities. I don't know how to figure this out so it would be extremely appreciated if someone found the other possibilities and showed me how to do it.










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  • 1




    Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
    – Blue
    Aug 31 at 1:18







  • 1




    I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
    – vadim123
    Aug 31 at 1:29










  • $x^2-7x+12 = (x-3)(x-4)$
    – amsmath
    Aug 31 at 1:31










  • For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
    – amsmath
    Aug 31 at 1:38











  • No, wait. For $x^2-5x+5 < 0$ for $x=3$.
    – amsmath
    Aug 31 at 1:52














up vote
3
down vote

favorite












What are all the possible values of $x$ if $(x^2-5x+5)^x^2-7x+12= 1$. I have already found three answers: 4, 3, and 1, however, apparently there are more possibilities. I don't know how to figure this out so it would be extremely appreciated if someone found the other possibilities and showed me how to do it.










share|cite|improve this question



















  • 1




    Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
    – Blue
    Aug 31 at 1:18







  • 1




    I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
    – vadim123
    Aug 31 at 1:29










  • $x^2-7x+12 = (x-3)(x-4)$
    – amsmath
    Aug 31 at 1:31










  • For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
    – amsmath
    Aug 31 at 1:38











  • No, wait. For $x^2-5x+5 < 0$ for $x=3$.
    – amsmath
    Aug 31 at 1:52












up vote
3
down vote

favorite









up vote
3
down vote

favorite











What are all the possible values of $x$ if $(x^2-5x+5)^x^2-7x+12= 1$. I have already found three answers: 4, 3, and 1, however, apparently there are more possibilities. I don't know how to figure this out so it would be extremely appreciated if someone found the other possibilities and showed me how to do it.










share|cite|improve this question















What are all the possible values of $x$ if $(x^2-5x+5)^x^2-7x+12= 1$. I have already found three answers: 4, 3, and 1, however, apparently there are more possibilities. I don't know how to figure this out so it would be extremely appreciated if someone found the other possibilities and showed me how to do it.







quadratics problem-solving integers






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edited Aug 31 at 8:10









tarit goswami

1,158219




1,158219










asked Aug 31 at 1:11









Julie

364




364







  • 1




    Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
    – Blue
    Aug 31 at 1:18







  • 1




    I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
    – vadim123
    Aug 31 at 1:29










  • $x^2-7x+12 = (x-3)(x-4)$
    – amsmath
    Aug 31 at 1:31










  • For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
    – amsmath
    Aug 31 at 1:38











  • No, wait. For $x^2-5x+5 < 0$ for $x=3$.
    – amsmath
    Aug 31 at 1:52












  • 1




    Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
    – Blue
    Aug 31 at 1:18







  • 1




    I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
    – vadim123
    Aug 31 at 1:29










  • $x^2-7x+12 = (x-3)(x-4)$
    – amsmath
    Aug 31 at 1:31










  • For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
    – amsmath
    Aug 31 at 1:38











  • No, wait. For $x^2-5x+5 < 0$ for $x=3$.
    – amsmath
    Aug 31 at 1:52







1




1




Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
– Blue
Aug 31 at 1:18





Please check the equation. The parentheses don't balance. (While you're editing, put the equation into the title of your question.)
– Blue
Aug 31 at 1:18





1




1




I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
– vadim123
Aug 31 at 1:29




I have adjusted the problem's typesetting to be consistent with the $4,3,1$ solutions given.
– vadim123
Aug 31 at 1:29












$x^2-7x+12 = (x-3)(x-4)$
– amsmath
Aug 31 at 1:31




$x^2-7x+12 = (x-3)(x-4)$
– amsmath
Aug 31 at 1:31












For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
– amsmath
Aug 31 at 1:38





For positive $x^2-5x+5$ your solutions are the only ones. The question is whether negative $x^2-5x+5$ is allowed.
– amsmath
Aug 31 at 1:38













No, wait. For $x^2-5x+5 < 0$ for $x=3$.
– amsmath
Aug 31 at 1:52




No, wait. For $x^2-5x+5 < 0$ for $x=3$.
– amsmath
Aug 31 at 1:52










1 Answer
1






active

oldest

votes

















up vote
5
down vote



accepted










Hint:




$(-1)^k=1$ (if $k$ is an even integer).



In other words, you forgot the case that $x^2-5x+5=-1.$




Solving this, we get $x^2-5x+6=0rightarrow x=2,3.$



We check that the exponent is even.



If $x=3,$ then $(x-3)(x-4)=0.$



If $x=2,$ then $(-1)(-2)=2.$



Either way, it is even, so both solutions are valid.



All solutions are $ 1,2,3,4$.






share|cite|improve this answer






















  • So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
    – Julie
    Aug 31 at 1:33











  • Solve for $x^2-5x+5=-1$ and check the exponent is even.
    – Jason Kim
    Aug 31 at 1:34











  • In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
    – Jason Kim
    Aug 31 at 1:37






  • 1




    @Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
    – amsmath
    Aug 31 at 1:59







  • 1




    @JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain : How we are sure there are no other solutions?
    – tarit goswami
    Aug 31 at 4:57











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
5
down vote



accepted










Hint:




$(-1)^k=1$ (if $k$ is an even integer).



In other words, you forgot the case that $x^2-5x+5=-1.$




Solving this, we get $x^2-5x+6=0rightarrow x=2,3.$



We check that the exponent is even.



If $x=3,$ then $(x-3)(x-4)=0.$



If $x=2,$ then $(-1)(-2)=2.$



Either way, it is even, so both solutions are valid.



All solutions are $ 1,2,3,4$.






share|cite|improve this answer






















  • So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
    – Julie
    Aug 31 at 1:33











  • Solve for $x^2-5x+5=-1$ and check the exponent is even.
    – Jason Kim
    Aug 31 at 1:34











  • In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
    – Jason Kim
    Aug 31 at 1:37






  • 1




    @Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
    – amsmath
    Aug 31 at 1:59







  • 1




    @JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain : How we are sure there are no other solutions?
    – tarit goswami
    Aug 31 at 4:57















up vote
5
down vote



accepted










Hint:




$(-1)^k=1$ (if $k$ is an even integer).



In other words, you forgot the case that $x^2-5x+5=-1.$




Solving this, we get $x^2-5x+6=0rightarrow x=2,3.$



We check that the exponent is even.



If $x=3,$ then $(x-3)(x-4)=0.$



If $x=2,$ then $(-1)(-2)=2.$



Either way, it is even, so both solutions are valid.



All solutions are $ 1,2,3,4$.






share|cite|improve this answer






















  • So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
    – Julie
    Aug 31 at 1:33











  • Solve for $x^2-5x+5=-1$ and check the exponent is even.
    – Jason Kim
    Aug 31 at 1:34











  • In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
    – Jason Kim
    Aug 31 at 1:37






  • 1




    @Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
    – amsmath
    Aug 31 at 1:59







  • 1




    @JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain : How we are sure there are no other solutions?
    – tarit goswami
    Aug 31 at 4:57













up vote
5
down vote



accepted







up vote
5
down vote



accepted






Hint:




$(-1)^k=1$ (if $k$ is an even integer).



In other words, you forgot the case that $x^2-5x+5=-1.$




Solving this, we get $x^2-5x+6=0rightarrow x=2,3.$



We check that the exponent is even.



If $x=3,$ then $(x-3)(x-4)=0.$



If $x=2,$ then $(-1)(-2)=2.$



Either way, it is even, so both solutions are valid.



All solutions are $ 1,2,3,4$.






share|cite|improve this answer














Hint:




$(-1)^k=1$ (if $k$ is an even integer).



In other words, you forgot the case that $x^2-5x+5=-1.$




Solving this, we get $x^2-5x+6=0rightarrow x=2,3.$



We check that the exponent is even.



If $x=3,$ then $(x-3)(x-4)=0.$



If $x=2,$ then $(-1)(-2)=2.$



Either way, it is even, so both solutions are valid.



All solutions are $ 1,2,3,4$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 31 at 8:43









tarit goswami

1,158219




1,158219










answered Aug 31 at 1:31









Jason Kim

53016




53016











  • So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
    – Julie
    Aug 31 at 1:33











  • Solve for $x^2-5x+5=-1$ and check the exponent is even.
    – Jason Kim
    Aug 31 at 1:34











  • In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
    – Jason Kim
    Aug 31 at 1:37






  • 1




    @Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
    – amsmath
    Aug 31 at 1:59







  • 1




    @JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain : How we are sure there are no other solutions?
    – tarit goswami
    Aug 31 at 4:57

















  • So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
    – Julie
    Aug 31 at 1:33











  • Solve for $x^2-5x+5=-1$ and check the exponent is even.
    – Jason Kim
    Aug 31 at 1:34











  • In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
    – Jason Kim
    Aug 31 at 1:37






  • 1




    @Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
    – amsmath
    Aug 31 at 1:59







  • 1




    @JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain : How we are sure there are no other solutions?
    – tarit goswami
    Aug 31 at 4:57
















So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
– Julie
Aug 31 at 1:33





So how would you use this to get the other possibilities? Sorry, I'm still quite confused.
– Julie
Aug 31 at 1:33













Solve for $x^2-5x+5=-1$ and check the exponent is even.
– Jason Kim
Aug 31 at 1:34





Solve for $x^2-5x+5=-1$ and check the exponent is even.
– Jason Kim
Aug 31 at 1:34













In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
– Jason Kim
Aug 31 at 1:37




In particular ~ there are three possibilities: The exponent is $0,$ the base is $1/-1.$ $-1$ is the case that was not included in the solution.
– Jason Kim
Aug 31 at 1:37




1




1




@Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
– amsmath
Aug 31 at 1:59





@Julie You should actually answer this question yourself. Do you just have this equation or any more information? For example, is the question to find all integer values of $x$ solving the equation?
– amsmath
Aug 31 at 1:59





1




1




@JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain : How we are sure there are no other solutions?
– tarit goswami
Aug 31 at 4:57





@JasonKim Recently in an entrance exam it was asked to explain why these are all solutions and there is no other(in first step asked to find the solutions). I have found all, but was not able to explain properly(I have explained by taking $log$ both side) why these are all solutions and there are no other. Can you explain : How we are sure there are no other solutions?
– tarit goswami
Aug 31 at 4:57


















 

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