Solving an equation with radicals in the exponent The Next CEO of Stack OverflowA Math...

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Solving an equation with radicals in the exponent



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1












$begingroup$



Given the following equation:$$2^{x^{frac{1}{12}}}+2^{x^{frac{1}{4}}}=2cdot2^{x^{frac{1}{6}}}$$




I have to solve for $x$, I tried working a bit with logarithms but the plus sign is quite annoying so I didn't think that was the correct way, I saw the '$2cdot$' on the RHS and that suggested me that maybe I should bring it to the other side and the equation will turn out to be a perfect square, but that wasn't the case... I didn't get much further (the problem doesn't specify so I assumed that $x$ is a real number)






contest-math







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you tried a change of variable?
    $endgroup$
    – Ertxiem
    Mar 16 at 19:10










  • $begingroup$
    @Ertxiem Yes, but I'm not sure if I'm missing out on non-trivial solutions
    $endgroup$
    – Spasoje Durovic
    Mar 16 at 19:26
















1












$begingroup$



Given the following equation:$$2^{x^{frac{1}{12}}}+2^{x^{frac{1}{4}}}=2cdot2^{x^{frac{1}{6}}}$$




I have to solve for $x$, I tried working a bit with logarithms but the plus sign is quite annoying so I didn't think that was the correct way, I saw the '$2cdot$' on the RHS and that suggested me that maybe I should bring it to the other side and the equation will turn out to be a perfect square, but that wasn't the case... I didn't get much further (the problem doesn't specify so I assumed that $x$ is a real number)






contest-math







share|cite|improve this question











$endgroup$












  • $begingroup$
    Have you tried a change of variable?
    $endgroup$
    – Ertxiem
    Mar 16 at 19:10










  • $begingroup$
    @Ertxiem Yes, but I'm not sure if I'm missing out on non-trivial solutions
    $endgroup$
    – Spasoje Durovic
    Mar 16 at 19:26














1












1








1





$begingroup$



Given the following equation:$$2^{x^{frac{1}{12}}}+2^{x^{frac{1}{4}}}=2cdot2^{x^{frac{1}{6}}}$$




I have to solve for $x$, I tried working a bit with logarithms but the plus sign is quite annoying so I didn't think that was the correct way, I saw the '$2cdot$' on the RHS and that suggested me that maybe I should bring it to the other side and the equation will turn out to be a perfect square, but that wasn't the case... I didn't get much further (the problem doesn't specify so I assumed that $x$ is a real number)






contest-math







share|cite|improve this question











$endgroup$





Given the following equation:$$2^{x^{frac{1}{12}}}+2^{x^{frac{1}{4}}}=2cdot2^{x^{frac{1}{6}}}$$




I have to solve for $x$, I tried working a bit with logarithms but the plus sign is quite annoying so I didn't think that was the correct way, I saw the '$2cdot$' on the RHS and that suggested me that maybe I should bring it to the other side and the equation will turn out to be a perfect square, but that wasn't the case... I didn't get much further (the problem doesn't specify so I assumed that $x$ is a real number)






contest-math




contest-math






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 16 at 19:32









Maria Mazur

48.9k1260122




48.9k1260122










asked Mar 16 at 19:02









Spasoje DurovicSpasoje Durovic

40811




40811












  • $begingroup$
    Have you tried a change of variable?
    $endgroup$
    – Ertxiem
    Mar 16 at 19:10










  • $begingroup$
    @Ertxiem Yes, but I'm not sure if I'm missing out on non-trivial solutions
    $endgroup$
    – Spasoje Durovic
    Mar 16 at 19:26


















  • $begingroup$
    Have you tried a change of variable?
    $endgroup$
    – Ertxiem
    Mar 16 at 19:10










  • $begingroup$
    @Ertxiem Yes, but I'm not sure if I'm missing out on non-trivial solutions
    $endgroup$
    – Spasoje Durovic
    Mar 16 at 19:26
















$begingroup$
Have you tried a change of variable?
$endgroup$
– Ertxiem
Mar 16 at 19:10




$begingroup$
Have you tried a change of variable?
$endgroup$
– Ertxiem
Mar 16 at 19:10












$begingroup$
@Ertxiem Yes, but I'm not sure if I'm missing out on non-trivial solutions
$endgroup$
– Spasoje Durovic
Mar 16 at 19:26




$begingroup$
@Ertxiem Yes, but I'm not sure if I'm missing out on non-trivial solutions
$endgroup$
– Spasoje Durovic
Mar 16 at 19:26










1 Answer
1






active

oldest

votes


















4












$begingroup$

By Am Gm inequality (we use it twice) we have $$2cdot2^{x^{frac{1}{6}}} =2^{x^{frac{1}{12}}}+2^{x^{frac{1}{4}}}geq 2sqrt{cdot2^{x^{frac{1}{12}}+x^{frac{1}{4}}}}geq 2cdot sqrt{2^{2{sqrt{x^{frac{1}{12}+frac{1}{4}}}}}}=2cdot2^{{x^{frac{1}{6}}}}$$



So we have equality case which is achieved iff $$ x^{frac{1}{12}}=x^{frac{1}{4}}implies x^3=ximplies xin{0,1,-1}$$



Since radicand $x$ must be nonegative we have $x=0$ or $x=1$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It always surprises me how clever manipulations of the simple Am-Gm inequality can lead to these results... How did you come to this idea?
    $endgroup$
    – Dr. Mathva
    Mar 16 at 19:47












  • $begingroup$
    I solved some similar problems before.
    $endgroup$
    – Maria Mazur
    Mar 16 at 19:48












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1 Answer
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active

oldest

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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









4












$begingroup$

By Am Gm inequality (we use it twice) we have $$2cdot2^{x^{frac{1}{6}}} =2^{x^{frac{1}{12}}}+2^{x^{frac{1}{4}}}geq 2sqrt{cdot2^{x^{frac{1}{12}}+x^{frac{1}{4}}}}geq 2cdot sqrt{2^{2{sqrt{x^{frac{1}{12}+frac{1}{4}}}}}}=2cdot2^{{x^{frac{1}{6}}}}$$



So we have equality case which is achieved iff $$ x^{frac{1}{12}}=x^{frac{1}{4}}implies x^3=ximplies xin{0,1,-1}$$



Since radicand $x$ must be nonegative we have $x=0$ or $x=1$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It always surprises me how clever manipulations of the simple Am-Gm inequality can lead to these results... How did you come to this idea?
    $endgroup$
    – Dr. Mathva
    Mar 16 at 19:47












  • $begingroup$
    I solved some similar problems before.
    $endgroup$
    – Maria Mazur
    Mar 16 at 19:48
















4












$begingroup$

By Am Gm inequality (we use it twice) we have $$2cdot2^{x^{frac{1}{6}}} =2^{x^{frac{1}{12}}}+2^{x^{frac{1}{4}}}geq 2sqrt{cdot2^{x^{frac{1}{12}}+x^{frac{1}{4}}}}geq 2cdot sqrt{2^{2{sqrt{x^{frac{1}{12}+frac{1}{4}}}}}}=2cdot2^{{x^{frac{1}{6}}}}$$



So we have equality case which is achieved iff $$ x^{frac{1}{12}}=x^{frac{1}{4}}implies x^3=ximplies xin{0,1,-1}$$



Since radicand $x$ must be nonegative we have $x=0$ or $x=1$






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    It always surprises me how clever manipulations of the simple Am-Gm inequality can lead to these results... How did you come to this idea?
    $endgroup$
    – Dr. Mathva
    Mar 16 at 19:47












  • $begingroup$
    I solved some similar problems before.
    $endgroup$
    – Maria Mazur
    Mar 16 at 19:48














4












4








4





$begingroup$

By Am Gm inequality (we use it twice) we have $$2cdot2^{x^{frac{1}{6}}} =2^{x^{frac{1}{12}}}+2^{x^{frac{1}{4}}}geq 2sqrt{cdot2^{x^{frac{1}{12}}+x^{frac{1}{4}}}}geq 2cdot sqrt{2^{2{sqrt{x^{frac{1}{12}+frac{1}{4}}}}}}=2cdot2^{{x^{frac{1}{6}}}}$$



So we have equality case which is achieved iff $$ x^{frac{1}{12}}=x^{frac{1}{4}}implies x^3=ximplies xin{0,1,-1}$$



Since radicand $x$ must be nonegative we have $x=0$ or $x=1$






share|cite|improve this answer











$endgroup$



By Am Gm inequality (we use it twice) we have $$2cdot2^{x^{frac{1}{6}}} =2^{x^{frac{1}{12}}}+2^{x^{frac{1}{4}}}geq 2sqrt{cdot2^{x^{frac{1}{12}}+x^{frac{1}{4}}}}geq 2cdot sqrt{2^{2{sqrt{x^{frac{1}{12}+frac{1}{4}}}}}}=2cdot2^{{x^{frac{1}{6}}}}$$



So we have equality case which is achieved iff $$ x^{frac{1}{12}}=x^{frac{1}{4}}implies x^3=ximplies xin{0,1,-1}$$



Since radicand $x$ must be nonegative we have $x=0$ or $x=1$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Mar 16 at 19:29

























answered Mar 16 at 19:16









Maria MazurMaria Mazur

48.9k1260122




48.9k1260122








  • 1




    $begingroup$
    It always surprises me how clever manipulations of the simple Am-Gm inequality can lead to these results... How did you come to this idea?
    $endgroup$
    – Dr. Mathva
    Mar 16 at 19:47












  • $begingroup$
    I solved some similar problems before.
    $endgroup$
    – Maria Mazur
    Mar 16 at 19:48














  • 1




    $begingroup$
    It always surprises me how clever manipulations of the simple Am-Gm inequality can lead to these results... How did you come to this idea?
    $endgroup$
    – Dr. Mathva
    Mar 16 at 19:47












  • $begingroup$
    I solved some similar problems before.
    $endgroup$
    – Maria Mazur
    Mar 16 at 19:48








1




1




$begingroup$
It always surprises me how clever manipulations of the simple Am-Gm inequality can lead to these results... How did you come to this idea?
$endgroup$
– Dr. Mathva
Mar 16 at 19:47






$begingroup$
It always surprises me how clever manipulations of the simple Am-Gm inequality can lead to these results... How did you come to this idea?
$endgroup$
– Dr. Mathva
Mar 16 at 19:47














$begingroup$
I solved some similar problems before.
$endgroup$
– Maria Mazur
Mar 16 at 19:48




$begingroup$
I solved some similar problems before.
$endgroup$
– Maria Mazur
Mar 16 at 19:48


















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