Find $x$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$












7












$begingroup$


Find all $x in (-1, +infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.



What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:



$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y > 1$, which can then be rewritten as:
$$ left( frac{y^2 + 1}{2y} right)^{y - 2} - left( frac{y^2 - 1}{2y} right)^{y - 2} = 1 $$
Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.



Do you have any suggestion on how I can continue?






algebra-precalculus exponential-function







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$endgroup$












  • $begingroup$
    $y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
    $endgroup$
    – Viktor Glombik
    Dec 26 '18 at 13:01


















7












$begingroup$


Find all $x in (-1, +infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.



What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:



$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y > 1$, which can then be rewritten as:
$$ left( frac{y^2 + 1}{2y} right)^{y - 2} - left( frac{y^2 - 1}{2y} right)^{y - 2} = 1 $$
Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.



Do you have any suggestion on how I can continue?






algebra-precalculus exponential-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    $y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
    $endgroup$
    – Viktor Glombik
    Dec 26 '18 at 13:01
















7












7








7


0



$begingroup$


Find all $x in (-1, +infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.



What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:



$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y > 1$, which can then be rewritten as:
$$ left( frac{y^2 + 1}{2y} right)^{y - 2} - left( frac{y^2 - 1}{2y} right)^{y - 2} = 1 $$
Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.



Do you have any suggestion on how I can continue?






algebra-precalculus exponential-function







share|cite|improve this question











$endgroup$




Find all $x in (-1, +infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.



What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:



$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y > 1$, which can then be rewritten as:
$$ left( frac{y^2 + 1}{2y} right)^{y - 2} - left( frac{y^2 - 1}{2y} right)^{y - 2} = 1 $$
Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.



Do you have any suggestion on how I can continue?






algebra-precalculus exponential-function




algebra-precalculus exponential-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 27 '18 at 6:57









Dylan

13.2k31027




13.2k31027










asked Dec 26 '18 at 12:54









SandelSandel

1816




1816












  • $begingroup$
    $y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
    $endgroup$
    – Viktor Glombik
    Dec 26 '18 at 13:01




















  • $begingroup$
    $y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
    $endgroup$
    – Viktor Glombik
    Dec 26 '18 at 13:01


















$begingroup$
$y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
$endgroup$
– Viktor Glombik
Dec 26 '18 at 13:01






$begingroup$
$y = 4$ is in fact the only solution to the last equation, and $x = 2$ is the only solution to the given equation.
$endgroup$
– Viktor Glombik
Dec 26 '18 at 13:01












1 Answer
1






active

oldest

votes


















6












$begingroup$

Hint:



$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$



$$iff1=left(dfrac{2x+4}{x^2+4x+5}right)^2+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^2$$



We have
$$left(dfrac{2x+4}{x^2+4x+5}right)^x+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^x=1$$



WLOG $dfrac{2x+4}{x^2+4x+5}=cos t,dfrac{x^2+4x+3}{x^2+4x+5}=sin t$ with $0<t<dfracpi2$ for $-1<x<infty$



Clearly, $$(cos t)^x+(sin t)^x$$ is a decreasing function to forbid multiple solutions



Generalization : $$left(dfrac{2x+4}{x^2+4x+5}right)^y+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^y=1$$



$implies y=2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:37










  • $begingroup$
    @Sandel, For $0le yle1, y^x$ is decreasing right?
    $endgroup$
    – lab bhattacharjee
    Dec 26 '18 at 13:48










  • $begingroup$
    Yes, I agree with that part.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:53











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1 Answer
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active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

Hint:



$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$



$$iff1=left(dfrac{2x+4}{x^2+4x+5}right)^2+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^2$$



We have
$$left(dfrac{2x+4}{x^2+4x+5}right)^x+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^x=1$$



WLOG $dfrac{2x+4}{x^2+4x+5}=cos t,dfrac{x^2+4x+3}{x^2+4x+5}=sin t$ with $0<t<dfracpi2$ for $-1<x<infty$



Clearly, $$(cos t)^x+(sin t)^x$$ is a decreasing function to forbid multiple solutions



Generalization : $$left(dfrac{2x+4}{x^2+4x+5}right)^y+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^y=1$$



$implies y=2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:37










  • $begingroup$
    @Sandel, For $0le yle1, y^x$ is decreasing right?
    $endgroup$
    – lab bhattacharjee
    Dec 26 '18 at 13:48










  • $begingroup$
    Yes, I agree with that part.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:53
















6












$begingroup$

Hint:



$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$



$$iff1=left(dfrac{2x+4}{x^2+4x+5}right)^2+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^2$$



We have
$$left(dfrac{2x+4}{x^2+4x+5}right)^x+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^x=1$$



WLOG $dfrac{2x+4}{x^2+4x+5}=cos t,dfrac{x^2+4x+3}{x^2+4x+5}=sin t$ with $0<t<dfracpi2$ for $-1<x<infty$



Clearly, $$(cos t)^x+(sin t)^x$$ is a decreasing function to forbid multiple solutions



Generalization : $$left(dfrac{2x+4}{x^2+4x+5}right)^y+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^y=1$$



$implies y=2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:37










  • $begingroup$
    @Sandel, For $0le yle1, y^x$ is decreasing right?
    $endgroup$
    – lab bhattacharjee
    Dec 26 '18 at 13:48










  • $begingroup$
    Yes, I agree with that part.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:53














6












6








6





$begingroup$

Hint:



$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$



$$iff1=left(dfrac{2x+4}{x^2+4x+5}right)^2+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^2$$



We have
$$left(dfrac{2x+4}{x^2+4x+5}right)^x+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^x=1$$



WLOG $dfrac{2x+4}{x^2+4x+5}=cos t,dfrac{x^2+4x+3}{x^2+4x+5}=sin t$ with $0<t<dfracpi2$ for $-1<x<infty$



Clearly, $$(cos t)^x+(sin t)^x$$ is a decreasing function to forbid multiple solutions



Generalization : $$left(dfrac{2x+4}{x^2+4x+5}right)^y+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^y=1$$



$implies y=2$






share|cite|improve this answer











$endgroup$



Hint:



$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$



$$iff1=left(dfrac{2x+4}{x^2+4x+5}right)^2+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^2$$



We have
$$left(dfrac{2x+4}{x^2+4x+5}right)^x+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^x=1$$



WLOG $dfrac{2x+4}{x^2+4x+5}=cos t,dfrac{x^2+4x+3}{x^2+4x+5}=sin t$ with $0<t<dfracpi2$ for $-1<x<infty$



Clearly, $$(cos t)^x+(sin t)^x$$ is a decreasing function to forbid multiple solutions



Generalization : $$left(dfrac{2x+4}{x^2+4x+5}right)^y+left(dfrac{x^2+4x+3}{x^2+4x+5}right)^y=1$$



$implies y=2$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 26 '18 at 13:06

























answered Dec 26 '18 at 13:01









lab bhattacharjeelab bhattacharjee

226k15157275




226k15157275












  • $begingroup$
    The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:37










  • $begingroup$
    @Sandel, For $0le yle1, y^x$ is decreasing right?
    $endgroup$
    – lab bhattacharjee
    Dec 26 '18 at 13:48










  • $begingroup$
    Yes, I agree with that part.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:53


















  • $begingroup$
    The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:37










  • $begingroup$
    @Sandel, For $0le yle1, y^x$ is decreasing right?
    $endgroup$
    – lab bhattacharjee
    Dec 26 '18 at 13:48










  • $begingroup$
    Yes, I agree with that part.
    $endgroup$
    – Sandel
    Dec 26 '18 at 13:53
















$begingroup$
The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
$endgroup$
– Sandel
Dec 26 '18 at 13:37




$begingroup$
The only thing unclear to me is when you say that the function $g(x) = (cos t)^x + (sin t)^x$ is a decreasing function, as $t$ also depends on $x$, so you cannot treat $t$ as a constant.
$endgroup$
– Sandel
Dec 26 '18 at 13:37












$begingroup$
@Sandel, For $0le yle1, y^x$ is decreasing right?
$endgroup$
– lab bhattacharjee
Dec 26 '18 at 13:48




$begingroup$
@Sandel, For $0le yle1, y^x$ is decreasing right?
$endgroup$
– lab bhattacharjee
Dec 26 '18 at 13:48












$begingroup$
Yes, I agree with that part.
$endgroup$
– Sandel
Dec 26 '18 at 13:53




$begingroup$
Yes, I agree with that part.
$endgroup$
– Sandel
Dec 26 '18 at 13:53


















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