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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$begingroup$
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
edited Mar 16 at 19:32
Maria Mazur
48.9k1260122
asked Mar 16 at 19:02
Spasoje DurovicSpasoje Durovic
40811
$endgroup$
add a comment |
$begingroup$
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
edited Mar 16 at 19:32
Maria Mazur
48.9k1260122
asked Mar 16 at 19:02
Spasoje DurovicSpasoje Durovic
40811
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Mar 16 at 19:32
Maria Mazur
48.9k1260122
asked Mar 16 at 19:02
Spasoje DurovicSpasoje Durovic
40811
$endgroup$
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
edited Mar 16 at 19:32
Maria Mazur
48.9k1260122
asked Mar 16 at 19:02
Spasoje DurovicSpasoje Durovic
40811
edited Mar 16 at 19:32
Maria Mazur
48.9k1260122
asked Mar 16 at 19:02
Spasoje DurovicSpasoje Durovic
40811
edited Mar 16 at 19:32
Maria Mazur
48.9k1260122
edited Mar 16 at 19:32
Maria Mazur
48.9k1260122
edited Mar 16 at 19:32
Maria Mazur
48.9k1260122
48.9k1260122
asked Mar 16 at 19:02
Spasoje DurovicSpasoje Durovic
40811
asked Mar 16 at 19:02
Spasoje DurovicSpasoje Durovic
40811
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $x=1$
answered Mar 16 at 19:16
Maria MazurMaria Mazur
48.9k1260122
$endgroup$
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
add a comment |
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1 Answer
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active
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votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $x=1$
answered Mar 16 at 19:16
Maria MazurMaria Mazur
48.9k1260122
$endgroup$
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
add a comment |
$begingroup$
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 $x=1$
answered Mar 16 at 19:16
Maria MazurMaria Mazur
48.9k1260122
$endgroup$
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
add a comment |
$begingroup$
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 $x=1$
answered Mar 16 at 19:16
Maria MazurMaria Mazur
48.9k1260122
$endgroup$
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 $x=1$
answered Mar 16 at 19:16
Maria MazurMaria Mazur
48.9k1260122
edited Mar 16 at 19:29
answered Mar 16 at 19:16
Maria MazurMaria Mazur
48.9k1260122
answered Mar 16 at 19:16
Maria MazurMaria Mazur
48.9k1260122
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
add a comment |
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
add a comment |
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$begingroup$
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