Equation with several exponents
5
$begingroup$
Is there a way to get the result for this value of x?
x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2])
I've tried Solve, FindInstance and I did not get a solution
I tried this:
Solve[x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2]),{x}]
equation-solving
asked 13 hours ago
LCarvalhoLCarvalho
5,78642986
$endgroup$
add a comment |
5
$begingroup$
Is there a way to get the result for this value of x?
x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2])
I've tried Solve, FindInstance and I did not get a solution
I tried this:
Solve[x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2]),{x}]
equation-solving
asked 13 hours ago
LCarvalhoLCarvalho
5,78642986
$endgroup$
add a comment |
5
5
5
$begingroup$
Is there a way to get the result for this value of x?
x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2])
I've tried Solve, FindInstance and I did not get a solution
I tried this:
Solve[x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2]),{x}]
equation-solving
asked 13 hours ago
LCarvalhoLCarvalho
5,78642986
$endgroup$
Is there a way to get the result for this value of x?
x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2])
I've tried Solve, FindInstance and I did not get a solution
I tried this:
Solve[x^x^(x^(1/6)/6)==Sqrt[2^9]^(2*Sqrt[2])^(2/3+2*Sqrt[2]),{x}]
equation-solving
equation-solving
asked 13 hours ago
LCarvalhoLCarvalho
5,78642986
asked 13 hours ago
LCarvalhoLCarvalho
5,78642986
asked 13 hours ago
LCarvalhoLCarvalho
5,78642986
asked 13 hours ago
LCarvalhoLCarvalho
5,78642986
asked 13 hours ago
LCarvalhoLCarvalho
5,78642986
5,78642986
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
6
$begingroup$
Try
NSolve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]
(*{x->512}*)
answered 13 hours ago
Ulrich NeumannUlrich Neumann
9,181516
$endgroup$
$begingroup$
Little detail that makes all the difference. Thanks a lot!!!
$endgroup$
– LCarvalho
13 hours ago
4
$begingroup$
Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]also works!
$endgroup$
– Ulrich Neumann
13 hours ago
add a comment |
0
$begingroup$
This is essentially the same as Ulrich's solution but provides the motivation for restricting the domain of Solve
eqn = x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]) //
Simplify
(* x^x^(x^(1/6)/6) == 8^(3 8^Sqrt[2]) *)
The RHS of eqn is real
Element[eqn[[-1]], Reals]
(* True *)
Consequently, finding the FunctionDomain for the LHS of eqn
const = FunctionDomain[eqn[[1]], x] // Simplify
(* x > 0 *)
Then,
Solve[eqn && const, x][[1]]
(* {x -> 512} *)
answered 6 hours ago
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
6
$begingroup$
Try
NSolve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]
(*{x->512}*)
answered 13 hours ago
Ulrich NeumannUlrich Neumann
9,181516
$endgroup$
$begingroup$
Little detail that makes all the difference. Thanks a lot!!!
$endgroup$
– LCarvalho
13 hours ago
4
$begingroup$
Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]also works!
$endgroup$
– Ulrich Neumann
13 hours ago
add a comment |
6
$begingroup$
Try
NSolve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]
(*{x->512}*)
answered 13 hours ago
Ulrich NeumannUlrich Neumann
9,181516
$endgroup$
$begingroup$
Little detail that makes all the difference. Thanks a lot!!!
$endgroup$
– LCarvalho
13 hours ago
4
$begingroup$
Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]also works!
$endgroup$
– Ulrich Neumann
13 hours ago
add a comment |
6
6
6
$begingroup$
Try
NSolve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]
(*{x->512}*)
answered 13 hours ago
Ulrich NeumannUlrich Neumann
9,181516
$endgroup$
Try
NSolve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]
(*{x->512}*)
answered 13 hours ago
Ulrich NeumannUlrich Neumann
9,181516
answered 13 hours ago
Ulrich NeumannUlrich Neumann
9,181516
answered 13 hours ago
Ulrich NeumannUlrich Neumann
9,181516
answered 13 hours ago
Ulrich NeumannUlrich Neumann
9,181516
9,181516
$begingroup$
Little detail that makes all the difference. Thanks a lot!!!
$endgroup$
– LCarvalho
13 hours ago
4
$begingroup$
Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]also works!
$endgroup$
– Ulrich Neumann
13 hours ago
add a comment |
$begingroup$
Little detail that makes all the difference. Thanks a lot!!!
$endgroup$
– LCarvalho
13 hours ago
4
$begingroup$
Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals]also works!
$endgroup$
– Ulrich Neumann
13 hours ago
$begingroup$
Little detail that makes all the difference. Thanks a lot!!!
$endgroup$
– LCarvalho
13 hours ago
$begingroup$
Little detail that makes all the difference. Thanks a lot!!!
$endgroup$
– LCarvalho
13 hours ago
4
4
$begingroup$
Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals] also works!$endgroup$
– Ulrich Neumann
13 hours ago
$begingroup$
Solve[x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]), x, Reals] also works!$endgroup$
– Ulrich Neumann
13 hours ago
add a comment |
0
$begingroup$
This is essentially the same as Ulrich's solution but provides the motivation for restricting the domain of Solve
eqn = x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]) //
Simplify
(* x^x^(x^(1/6)/6) == 8^(3 8^Sqrt[2]) *)
The RHS of eqn is real
Element[eqn[[-1]], Reals]
(* True *)
Consequently, finding the FunctionDomain for the LHS of eqn
const = FunctionDomain[eqn[[1]], x] // Simplify
(* x > 0 *)
Then,
Solve[eqn && const, x][[1]]
(* {x -> 512} *)
answered 6 hours ago
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
add a comment |
0
$begingroup$
This is essentially the same as Ulrich's solution but provides the motivation for restricting the domain of Solve
eqn = x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]) //
Simplify
(* x^x^(x^(1/6)/6) == 8^(3 8^Sqrt[2]) *)
The RHS of eqn is real
Element[eqn[[-1]], Reals]
(* True *)
Consequently, finding the FunctionDomain for the LHS of eqn
const = FunctionDomain[eqn[[1]], x] // Simplify
(* x > 0 *)
Then,
Solve[eqn && const, x][[1]]
(* {x -> 512} *)
answered 6 hours ago
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
add a comment |
0
0
0
$begingroup$
This is essentially the same as Ulrich's solution but provides the motivation for restricting the domain of Solve
eqn = x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]) //
Simplify
(* x^x^(x^(1/6)/6) == 8^(3 8^Sqrt[2]) *)
The RHS of eqn is real
Element[eqn[[-1]], Reals]
(* True *)
Consequently, finding the FunctionDomain for the LHS of eqn
const = FunctionDomain[eqn[[1]], x] // Simplify
(* x > 0 *)
Then,
Solve[eqn && const, x][[1]]
(* {x -> 512} *)
answered 6 hours ago
Bob HanlonBob Hanlon
60.6k33597
$endgroup$
This is essentially the same as Ulrich's solution but provides the motivation for restricting the domain of Solve
eqn = x^x^(x^(1/6)/6) == Sqrt[2^9]^(2*Sqrt[2])^(2/3 + 2*Sqrt[2]) //
Simplify
(* x^x^(x^(1/6)/6) == 8^(3 8^Sqrt[2]) *)
The RHS of eqn is real
Element[eqn[[-1]], Reals]
(* True *)
Consequently, finding the FunctionDomain for the LHS of eqn
const = FunctionDomain[eqn[[1]], x] // Simplify
(* x > 0 *)
Then,
Solve[eqn && const, x][[1]]
(* {x -> 512} *)
answered 6 hours ago
Bob HanlonBob Hanlon
60.6k33597
answered 6 hours ago
Bob HanlonBob Hanlon
60.6k33597
answered 6 hours ago
Bob HanlonBob Hanlon
60.6k33597
answered 6 hours ago
Bob HanlonBob Hanlon
60.6k33597
60.6k33597
add a comment |
add a comment |
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