can anyone prove it ??
x to the power x = 2
answer ASAP
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xˣ = 2 given. To find x.
I dont understand what is to be proved here.
x Log₂ x = Log₂ 2 = 1
x = 1/Log₂ x or, Log₂ x = 1/x
=> x = 2^(1/x)
for x = 0.01, xˣ ≈ 1
1 1
2 4
So xˣ = 2, for 1 < x < 2, xˣ increases continuously for x > 1
x = 1.5 , xˣ = 1.83
1.6 xˣ = 2.12 more than 2.
So value of x can be found successive iterations as follows:
let x = 1.5, the find next iteration value of x = 2^(1/x)
for x = 1.5, next x = 2^(1/1.5) = 1.5874
= 1.5874, 1.5475
= 1.5475 1.565
1.565 1.5572
1.5572 1.5606
1.5606 1.5591
1.5591 1.5598
1.5598 1.5595
1.5595 1.5596
1.55961 1.55961
So answer is close to 1.55961 -- it is a non terminating irrational number.
There are iterative methods to solve such equations. We need to do as many iterations as possible to get the answer to the required accuracy.
I dont understand what is to be proved here.
x Log₂ x = Log₂ 2 = 1
x = 1/Log₂ x or, Log₂ x = 1/x
=> x = 2^(1/x)
for x = 0.01, xˣ ≈ 1
1 1
2 4
So xˣ = 2, for 1 < x < 2, xˣ increases continuously for x > 1
x = 1.5 , xˣ = 1.83
1.6 xˣ = 2.12 more than 2.
So value of x can be found successive iterations as follows:
let x = 1.5, the find next iteration value of x = 2^(1/x)
for x = 1.5, next x = 2^(1/1.5) = 1.5874
= 1.5874, 1.5475
= 1.5475 1.565
1.565 1.5572
1.5572 1.5606
1.5606 1.5591
1.5591 1.5598
1.5598 1.5595
1.5595 1.5596
1.55961 1.55961
So answer is close to 1.55961 -- it is a non terminating irrational number.
There are iterative methods to solve such equations. We need to do as many iterations as possible to get the answer to the required accuracy.
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kvnmurty:
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this is answer you want
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