Find the value of $x$, where $sqrt{(2+sqrt{3})^x} + sqrt{(2-sqrt{3})^x} = 2^x$?
$$sqrt{(2+sqrt{3})^x} + sqrt{(2-sqrt{3})^x} = 2^x$$
I am not able to get to the result logically.
What I tried and got to this result
$$
LHS
= [(sqrt{3}+1)/sqrt{2}]^x +[(sqrt{3}-1)/sqrt{2}]^x
$$
algebra-precalculus
edited Nov 28 at 17:55
Brahadeesh
6,10242360
asked Nov 28 at 17:51
user349915
244
|
show 1 more comment
$$sqrt{(2+sqrt{3})^x} + sqrt{(2-sqrt{3})^x} = 2^x$$
I am not able to get to the result logically.
What I tried and got to this result
$$
LHS
= [(sqrt{3}+1)/sqrt{2}]^x +[(sqrt{3}-1)/sqrt{2}]^x
$$
algebra-precalculus
edited Nov 28 at 17:55
Brahadeesh
6,10242360
asked Nov 28 at 17:51
user349915
244
what makes you believe it has a nice form?
– Jorge Fernández
Nov 28 at 18:01
The answer is $2$.
– Jan
Nov 28 at 18:01
2 is ONE solution. But that doesn't mean it is the only one.
– Alejandro Nasif Salum
Nov 28 at 18:09
Please prove your answer.
– user349915
Nov 28 at 18:12
if you square both parts, you get $(2-sqrt3)^x+(2+sqrt3)^x=2^{2x}-2$. It seems $2$ is the only solution
– Vasya
Nov 28 at 18:19
|
show 1 more comment
$$sqrt{(2+sqrt{3})^x} + sqrt{(2-sqrt{3})^x} = 2^x$$
I am not able to get to the result logically.
What I tried and got to this result
$$
LHS
= [(sqrt{3}+1)/sqrt{2}]^x +[(sqrt{3}-1)/sqrt{2}]^x
$$
algebra-precalculus
edited Nov 28 at 17:55
Brahadeesh
6,10242360
asked Nov 28 at 17:51
user349915
244
$$sqrt{(2+sqrt{3})^x} + sqrt{(2-sqrt{3})^x} = 2^x$$
I am not able to get to the result logically.
What I tried and got to this result
$$
LHS
= [(sqrt{3}+1)/sqrt{2}]^x +[(sqrt{3}-1)/sqrt{2}]^x
$$
algebra-precalculus
algebra-precalculus
edited Nov 28 at 17:55
Brahadeesh
6,10242360
asked Nov 28 at 17:51
user349915
244
edited Nov 28 at 17:55
Brahadeesh
6,10242360
asked Nov 28 at 17:51
user349915
244
edited Nov 28 at 17:55
Brahadeesh
6,10242360
edited Nov 28 at 17:55
Brahadeesh
6,10242360
edited Nov 28 at 17:55
Brahadeesh
6,10242360
6,10242360
asked Nov 28 at 17:51
user349915
244
asked Nov 28 at 17:51
user349915
244
asked Nov 28 at 17:51
user349915
244
244
what makes you believe it has a nice form?
– Jorge Fernández
Nov 28 at 18:01
The answer is $2$.
– Jan
Nov 28 at 18:01
2 is ONE solution. But that doesn't mean it is the only one.
– Alejandro Nasif Salum
Nov 28 at 18:09
Please prove your answer.
– user349915
Nov 28 at 18:12
if you square both parts, you get $(2-sqrt3)^x+(2+sqrt3)^x=2^{2x}-2$. It seems $2$ is the only solution
– Vasya
Nov 28 at 18:19
|
show 1 more comment
what makes you believe it has a nice form?
– Jorge Fernández
Nov 28 at 18:01
The answer is $2$.
– Jan
Nov 28 at 18:01
2 is ONE solution. But that doesn't mean it is the only one.
– Alejandro Nasif Salum
Nov 28 at 18:09
Please prove your answer.
– user349915
Nov 28 at 18:12
if you square both parts, you get $(2-sqrt3)^x+(2+sqrt3)^x=2^{2x}-2$. It seems $2$ is the only solution
– Vasya
Nov 28 at 18:19
what makes you believe it has a nice form?
– Jorge Fernández
Nov 28 at 18:01
what makes you believe it has a nice form?
– Jorge Fernández
Nov 28 at 18:01
The answer is $2$.
– Jan
Nov 28 at 18:01
The answer is $2$.
– Jan
Nov 28 at 18:01
2 is ONE solution. But that doesn't mean it is the only one.
– Alejandro Nasif Salum
Nov 28 at 18:09
2 is ONE solution. But that doesn't mean it is the only one.
– Alejandro Nasif Salum
Nov 28 at 18:09
Please prove your answer.
– user349915
Nov 28 at 18:12
Please prove your answer.
– user349915
Nov 28 at 18:12
if you square both parts, you get $(2-sqrt3)^x+(2+sqrt3)^x=2^{2x}-2$. It seems $2$ is the only solution
– Vasya
Nov 28 at 18:19
if you square both parts, you get $(2-sqrt3)^x+(2+sqrt3)^x=2^{2x}-2$. It seems $2$ is the only solution
– Vasya
Nov 28 at 18:19
|
show 1 more comment
1 Answer
1
active
oldest
votes
$$dfrac{2+sqrt3}4=left(dfrac{sqrt3+1}{2sqrt2}right)^2$$
$cos(45^circ-30^circ)=?$
$sin(45-30)=?$
Finally $sin^mA+cos^mA$ is a decreasing function for $0le Aledfracpi2$
answered Nov 28 at 18:32
lab bhattacharjee
222k15155273
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$$dfrac{2+sqrt3}4=left(dfrac{sqrt3+1}{2sqrt2}right)^2$$
$cos(45^circ-30^circ)=?$
$sin(45-30)=?$
Finally $sin^mA+cos^mA$ is a decreasing function for $0le Aledfracpi2$
answered Nov 28 at 18:32
lab bhattacharjee
222k15155273
add a comment |
$$dfrac{2+sqrt3}4=left(dfrac{sqrt3+1}{2sqrt2}right)^2$$
$cos(45^circ-30^circ)=?$
$sin(45-30)=?$
Finally $sin^mA+cos^mA$ is a decreasing function for $0le Aledfracpi2$
answered Nov 28 at 18:32
lab bhattacharjee
222k15155273
add a comment |
$$dfrac{2+sqrt3}4=left(dfrac{sqrt3+1}{2sqrt2}right)^2$$
$cos(45^circ-30^circ)=?$
$sin(45-30)=?$
Finally $sin^mA+cos^mA$ is a decreasing function for $0le Aledfracpi2$
answered Nov 28 at 18:32
lab bhattacharjee
222k15155273
$$dfrac{2+sqrt3}4=left(dfrac{sqrt3+1}{2sqrt2}right)^2$$
$cos(45^circ-30^circ)=?$
$sin(45-30)=?$
Finally $sin^mA+cos^mA$ is a decreasing function for $0le Aledfracpi2$
answered Nov 28 at 18:32
lab bhattacharjee
222k15155273
answered Nov 28 at 18:32
lab bhattacharjee
222k15155273
answered Nov 28 at 18:32
lab bhattacharjee
222k15155273
answered Nov 28 at 18:32
lab bhattacharjee
222k15155273
222k15155273
add a comment |
add a comment |
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what makes you believe it has a nice form?
– Jorge Fernández
Nov 28 at 18:01
The answer is $2$.
– Jan
Nov 28 at 18:01
2 is ONE solution. But that doesn't mean it is the only one.
– Alejandro Nasif Salum
Nov 28 at 18:09
Please prove your answer.
– user349915
Nov 28 at 18:12
if you square both parts, you get $(2-sqrt3)^x+(2+sqrt3)^x=2^{2x}-2$. It seems $2$ is the only solution
– Vasya
Nov 28 at 18:19