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Solve in $mathbb Rquad 5^sqrtx - 5^x-7 = 100$
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up vote
7
down vote
favorite
Solve in $mathbb R$
$$5^sqrtx - 5^x-7 = 100$$
$mathbf My Attempt$
I converted the eq. to this form
$$5^(sqrtx-3)(sqrtx+3)-5.5^sqrtx-3+4=0$$
It's apparent that $;mathbf x=9; $ is a solution, but I can't find the reasoning for this mathematically.
Any hint?
$mathbf Edit$
I'll post my solution tomorrow.
algebra-precalculus
asked Jul 14 at 20:27
Wolfdale
24919
add a comment |Â
up vote
7
down vote
favorite
Solve in $mathbb R$
$$5^sqrtx - 5^x-7 = 100$$
$mathbf My Attempt$
I converted the eq. to this form
$$5^(sqrtx-3)(sqrtx+3)-5.5^sqrtx-3+4=0$$
It's apparent that $;mathbf x=9; $ is a solution, but I can't find the reasoning for this mathematically.
Any hint?
$mathbf Edit$
I'll post my solution tomorrow.
algebra-precalculus
asked Jul 14 at 20:27
Wolfdale
24919
1
I wouldn't expect a very pleasant analytic solution...trial and error (as you have done) looks sensible.
– lulu
Jul 14 at 20:32
@lulu, thx for the hint maybe some inequality or number analysis can help.
– Wolfdale
Jul 14 at 20:34
You might be able to show whether there is any other solution, but an analytic way to arrive at $9$ is highly unlikely.
– Thomas Andrews
Jul 14 at 20:35
Oh, numerical methods work fine, though of course you'll just get an approximation that way. (which of course might lead you to guess the exact answer)
– lulu
Jul 14 at 20:37
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Solve in $mathbb R$
$$5^sqrtx - 5^x-7 = 100$$
$mathbf My Attempt$
I converted the eq. to this form
$$5^(sqrtx-3)(sqrtx+3)-5.5^sqrtx-3+4=0$$
It's apparent that $;mathbf x=9; $ is a solution, but I can't find the reasoning for this mathematically.
Any hint?
$mathbf Edit$
I'll post my solution tomorrow.
algebra-precalculus
asked Jul 14 at 20:27
Wolfdale
24919
Solve in $mathbb R$
$$5^sqrtx - 5^x-7 = 100$$
$mathbf My Attempt$
I converted the eq. to this form
$$5^(sqrtx-3)(sqrtx+3)-5.5^sqrtx-3+4=0$$
It's apparent that $;mathbf x=9; $ is a solution, but I can't find the reasoning for this mathematically.
Any hint?
$mathbf Edit$
I'll post my solution tomorrow.
algebra-precalculus
asked Jul 14 at 20:27
Wolfdale
24919
edited Jul 15 at 3:55
asked Jul 14 at 20:27
Wolfdale
24919
asked Jul 14 at 20:27
Wolfdale
24919
asked Jul 14 at 20:27
Wolfdale
24919
24919
1
I wouldn't expect a very pleasant analytic solution...trial and error (as you have done) looks sensible.
– lulu
Jul 14 at 20:32
@lulu, thx for the hint maybe some inequality or number analysis can help.
– Wolfdale
Jul 14 at 20:34
You might be able to show whether there is any other solution, but an analytic way to arrive at $9$ is highly unlikely.
– Thomas Andrews
Jul 14 at 20:35
Oh, numerical methods work fine, though of course you'll just get an approximation that way. (which of course might lead you to guess the exact answer)
– lulu
Jul 14 at 20:37
add a comment |Â
1
I wouldn't expect a very pleasant analytic solution...trial and error (as you have done) looks sensible.
– lulu
Jul 14 at 20:32
@lulu, thx for the hint maybe some inequality or number analysis can help.
– Wolfdale
Jul 14 at 20:34
You might be able to show whether there is any other solution, but an analytic way to arrive at $9$ is highly unlikely.
– Thomas Andrews
Jul 14 at 20:35
Oh, numerical methods work fine, though of course you'll just get an approximation that way. (which of course might lead you to guess the exact answer)
– lulu
Jul 14 at 20:37
1
1
I wouldn't expect a very pleasant analytic solution...trial and error (as you have done) looks sensible.
– lulu
Jul 14 at 20:32
I wouldn't expect a very pleasant analytic solution...trial and error (as you have done) looks sensible.
– lulu
Jul 14 at 20:32
@lulu, thx for the hint maybe some inequality or number analysis can help.
– Wolfdale
Jul 14 at 20:34
@lulu, thx for the hint maybe some inequality or number analysis can help.
– Wolfdale
Jul 14 at 20:34
You might be able to show whether there is any other solution, but an analytic way to arrive at $9$ is highly unlikely.
– Thomas Andrews
Jul 14 at 20:35
You might be able to show whether there is any other solution, but an analytic way to arrive at $9$ is highly unlikely.
– Thomas Andrews
Jul 14 at 20:35
Oh, numerical methods work fine, though of course you'll just get an approximation that way. (which of course might lead you to guess the exact answer)
– lulu
Jul 14 at 20:37
Oh, numerical methods work fine, though of course you'll just get an approximation that way. (which of course might lead you to guess the exact answer)
– lulu
Jul 14 at 20:37
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
2
down vote
My sugestion. Proceed as suggested by Dr. Sonnhard Graubner. Set the function $F:(0,10)to mathbbR$ by
$$
F(x)=5^x-frac15^75^x^2-100.
$$
In the absence of a method of finite steps to solve an equation $F(x)=0$ there is a powerful method of resolution by interaction. The Kantorovich's theorem on Newton's interactions. The method sometimes (very rarely) results in a finite step method and thus exact solution.
Use the Kantorovich's theorem on Newton's
method in its classical formulation.
Let $Isubseteq mathbbR$
and $F:Ito mathbbR$ a continuous function, continuously
differentiable on $mathrmint(I)$. Take $x_0in mathrmint(I)$,
$L,, b>0$ and suppose that
.1 $F '(x_0)$ is non-singular,
.2 $ | F'(x_0)^-1left[ F'(y)-F'(x)right]
| leq L|x-y|
;;$ for any $x,yin I$,
.3$ |F'(x_0)^-1F(x_0)|leq b$,
.4 $2bLleq 1$.
Define
beginequation
t_*:=frac1-sqrt1-2bLL,qquad
t_**:=frac1+sqrt1-2bLL.
endequation
If
$
[x_0-t_*,x_0+t_*]subset I,
$
then the sequences $x_k$ generated by Newton's Method for
solving $F(x)=0$ with starting point $x_0$,
beginequation labelns.KT
x_k+1 =x_k-F'(x_k) ^-1F(x_k), qquad k=0,1,cdots,
endequation
is well defined, is contained in $(x_0-t_*,x_0+t_*)$, converges to a
point $x_*in [x_0-t_*,x_0+t_*]$ which is the unique zero of $F$ in
$[x_0-t_*,x_0+t_*]$ and
beginequation
labeleq:q.conv.x
|x_*-x_k+1|leq frac12 |x_*-x_k |, qquad
k=0,1,,cdots.
endequation
Moreover, if assumption .4 holds as an strict inequality, i.e.
$2bL<1$, then
beginequation
|x_*-x_k+1|leqfrac1-theta^2^k1+theta^2^k
frac L2sqrt1-2bL|x_*-x_k|^2leq
frac L2sqrt1-2bL|x_*-x_k|^2, quad k=0,1,cdots,
endequation
where $theta:=t_*/t_**<1$, and $x_*$ is the
unique zero of $F$ in $[x_0-t_*,x_0+t_*]$ for any $rho$ such that
$ t_*leqrho<t_**,qquad [x_0-rho,x_0+rho]subset I.$
answered Jul 14 at 20:51
MathOverview
7,95242962
add a comment |Â
up vote
1
down vote
Note: The right-hand side $100$ has a nice representation by powers of $5$.
We have
beginalign*
5^colorbluesqrtx-5^x-7=100=5^colorblue3-5^2
endalign*
which indicates a trial via
beginalign*
sqrtx=3 qquadtextandqquad x-7=2
endalign*
giving the solution $x=9$.
I think the challenging aspect is if there is an algebraic/analytic method besides some iterative approach which gives us the second solution close at hand.
answered Jul 15 at 14:58
Markus Scheuer
56.2k450135
add a comment |Â
up vote
0
down vote
accepted
This is the method I used to justify my found solution
Let $; t=5^sqrtx-3;$, we arrive at
$$t^sqrtx+3-5 t+4=0$$
$$t^sqrtx+3 - t^2 + t^2 -5 t+4=0$$
$$t^2(t^sqrtx+1 -1)+(t-1)(t-4)=0$$
Assume $;sqrtx in mathbb N$
$$t^2(t-1)(t^sqrtx+^sqrtx-1+cdots +1)+(t-1)(t-4)=0$$
$$(t-1)(t^sqrtx+2+^sqrtx+1+cdots +t^2+t-4)=0$$
$$t-1=0 quad Rightarrowquad 5^sqrtx-3=1 quad Rightarrowquad sqrtx=3 in mathbb N$$
$x=9quad $ is a valid solution.
answered Jul 15 at 13:26
Wolfdale
24919
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
My sugestion. Proceed as suggested by Dr. Sonnhard Graubner. Set the function $F:(0,10)to mathbbR$ by
$$
F(x)=5^x-frac15^75^x^2-100.
$$
In the absence of a method of finite steps to solve an equation $F(x)=0$ there is a powerful method of resolution by interaction. The Kantorovich's theorem on Newton's interactions. The method sometimes (very rarely) results in a finite step method and thus exact solution.
Use the Kantorovich's theorem on Newton's
method in its classical formulation.
Let $Isubseteq mathbbR$
and $F:Ito mathbbR$ a continuous function, continuously
differentiable on $mathrmint(I)$. Take $x_0in mathrmint(I)$,
$L,, b>0$ and suppose that
.1 $F '(x_0)$ is non-singular,
.2 $ | F'(x_0)^-1left[ F'(y)-F'(x)right]
| leq L|x-y|
;;$ for any $x,yin I$,
.3$ |F'(x_0)^-1F(x_0)|leq b$,
.4 $2bLleq 1$.
Define
beginequation
t_*:=frac1-sqrt1-2bLL,qquad
t_**:=frac1+sqrt1-2bLL.
endequation
If
$
[x_0-t_*,x_0+t_*]subset I,
$
then the sequences $x_k$ generated by Newton's Method for
solving $F(x)=0$ with starting point $x_0$,
beginequation labelns.KT
x_k+1 =x_k-F'(x_k) ^-1F(x_k), qquad k=0,1,cdots,
endequation
is well defined, is contained in $(x_0-t_*,x_0+t_*)$, converges to a
point $x_*in [x_0-t_*,x_0+t_*]$ which is the unique zero of $F$ in
$[x_0-t_*,x_0+t_*]$ and
beginequation
labeleq:q.conv.x
|x_*-x_k+1|leq frac12 |x_*-x_k |, qquad
k=0,1,,cdots.
endequation
Moreover, if assumption .4 holds as an strict inequality, i.e.
$2bL<1$, then
beginequation
|x_*-x_k+1|leqfrac1-theta^2^k1+theta^2^k
frac L2sqrt1-2bL|x_*-x_k|^2leq
frac L2sqrt1-2bL|x_*-x_k|^2, quad k=0,1,cdots,
endequation
where $theta:=t_*/t_**<1$, and $x_*$ is the
unique zero of $F$ in $[x_0-t_*,x_0+t_*]$ for any $rho$ such that
$ t_*leqrho<t_**,qquad [x_0-rho,x_0+rho]subset I.$
answered Jul 14 at 20:51
MathOverview
7,95242962
add a comment |Â
up vote
2
down vote
My sugestion. Proceed as suggested by Dr. Sonnhard Graubner. Set the function $F:(0,10)to mathbbR$ by
$$
F(x)=5^x-frac15^75^x^2-100.
$$
In the absence of a method of finite steps to solve an equation $F(x)=0$ there is a powerful method of resolution by interaction. The Kantorovich's theorem on Newton's interactions. The method sometimes (very rarely) results in a finite step method and thus exact solution.
Use the Kantorovich's theorem on Newton's
method in its classical formulation.
Let $Isubseteq mathbbR$
and $F:Ito mathbbR$ a continuous function, continuously
differentiable on $mathrmint(I)$. Take $x_0in mathrmint(I)$,
$L,, b>0$ and suppose that
.1 $F '(x_0)$ is non-singular,
.2 $ | F'(x_0)^-1left[ F'(y)-F'(x)right]
| leq L|x-y|
;;$ for any $x,yin I$,
.3$ |F'(x_0)^-1F(x_0)|leq b$,
.4 $2bLleq 1$.
Define
beginequation
t_*:=frac1-sqrt1-2bLL,qquad
t_**:=frac1+sqrt1-2bLL.
endequation
If
$
[x_0-t_*,x_0+t_*]subset I,
$
then the sequences $x_k$ generated by Newton's Method for
solving $F(x)=0$ with starting point $x_0$,
beginequation labelns.KT
x_k+1 =x_k-F'(x_k) ^-1F(x_k), qquad k=0,1,cdots,
endequation
is well defined, is contained in $(x_0-t_*,x_0+t_*)$, converges to a
point $x_*in [x_0-t_*,x_0+t_*]$ which is the unique zero of $F$ in
$[x_0-t_*,x_0+t_*]$ and
beginequation
labeleq:q.conv.x
|x_*-x_k+1|leq frac12 |x_*-x_k |, qquad
k=0,1,,cdots.
endequation
Moreover, if assumption .4 holds as an strict inequality, i.e.
$2bL<1$, then
beginequation
|x_*-x_k+1|leqfrac1-theta^2^k1+theta^2^k
frac L2sqrt1-2bL|x_*-x_k|^2leq
frac L2sqrt1-2bL|x_*-x_k|^2, quad k=0,1,cdots,
endequation
where $theta:=t_*/t_**<1$, and $x_*$ is the
unique zero of $F$ in $[x_0-t_*,x_0+t_*]$ for any $rho$ such that
$ t_*leqrho<t_**,qquad [x_0-rho,x_0+rho]subset I.$
answered Jul 14 at 20:51
MathOverview
7,95242962
add a comment |Â
up vote
2
down vote
up vote
2
down vote
My sugestion. Proceed as suggested by Dr. Sonnhard Graubner. Set the function $F:(0,10)to mathbbR$ by
$$
F(x)=5^x-frac15^75^x^2-100.
$$
In the absence of a method of finite steps to solve an equation $F(x)=0$ there is a powerful method of resolution by interaction. The Kantorovich's theorem on Newton's interactions. The method sometimes (very rarely) results in a finite step method and thus exact solution.
Use the Kantorovich's theorem on Newton's
method in its classical formulation.
Let $Isubseteq mathbbR$
and $F:Ito mathbbR$ a continuous function, continuously
differentiable on $mathrmint(I)$. Take $x_0in mathrmint(I)$,
$L,, b>0$ and suppose that
.1 $F '(x_0)$ is non-singular,
.2 $ | F'(x_0)^-1left[ F'(y)-F'(x)right]
| leq L|x-y|
;;$ for any $x,yin I$,
.3$ |F'(x_0)^-1F(x_0)|leq b$,
.4 $2bLleq 1$.
Define
beginequation
t_*:=frac1-sqrt1-2bLL,qquad
t_**:=frac1+sqrt1-2bLL.
endequation
If
$
[x_0-t_*,x_0+t_*]subset I,
$
then the sequences $x_k$ generated by Newton's Method for
solving $F(x)=0$ with starting point $x_0$,
beginequation labelns.KT
x_k+1 =x_k-F'(x_k) ^-1F(x_k), qquad k=0,1,cdots,
endequation
is well defined, is contained in $(x_0-t_*,x_0+t_*)$, converges to a
point $x_*in [x_0-t_*,x_0+t_*]$ which is the unique zero of $F$ in
$[x_0-t_*,x_0+t_*]$ and
beginequation
labeleq:q.conv.x
|x_*-x_k+1|leq frac12 |x_*-x_k |, qquad
k=0,1,,cdots.
endequation
Moreover, if assumption .4 holds as an strict inequality, i.e.
$2bL<1$, then
beginequation
|x_*-x_k+1|leqfrac1-theta^2^k1+theta^2^k
frac L2sqrt1-2bL|x_*-x_k|^2leq
frac L2sqrt1-2bL|x_*-x_k|^2, quad k=0,1,cdots,
endequation
where $theta:=t_*/t_**<1$, and $x_*$ is the
unique zero of $F$ in $[x_0-t_*,x_0+t_*]$ for any $rho$ such that
$ t_*leqrho<t_**,qquad [x_0-rho,x_0+rho]subset I.$
answered Jul 14 at 20:51
MathOverview
7,95242962
My sugestion. Proceed as suggested by Dr. Sonnhard Graubner. Set the function $F:(0,10)to mathbbR$ by
$$
F(x)=5^x-frac15^75^x^2-100.
$$
In the absence of a method of finite steps to solve an equation $F(x)=0$ there is a powerful method of resolution by interaction. The Kantorovich's theorem on Newton's interactions. The method sometimes (very rarely) results in a finite step method and thus exact solution.
Use the Kantorovich's theorem on Newton's
method in its classical formulation.
Let $Isubseteq mathbbR$
and $F:Ito mathbbR$ a continuous function, continuously
differentiable on $mathrmint(I)$. Take $x_0in mathrmint(I)$,
$L,, b>0$ and suppose that
.1 $F '(x_0)$ is non-singular,
.2 $ | F'(x_0)^-1left[ F'(y)-F'(x)right]
| leq L|x-y|
;;$ for any $x,yin I$,
.3$ |F'(x_0)^-1F(x_0)|leq b$,
.4 $2bLleq 1$.
Define
beginequation
t_*:=frac1-sqrt1-2bLL,qquad
t_**:=frac1+sqrt1-2bLL.
endequation
If
$
[x_0-t_*,x_0+t_*]subset I,
$
then the sequences $x_k$ generated by Newton's Method for
solving $F(x)=0$ with starting point $x_0$,
beginequation labelns.KT
x_k+1 =x_k-F'(x_k) ^-1F(x_k), qquad k=0,1,cdots,
endequation
is well defined, is contained in $(x_0-t_*,x_0+t_*)$, converges to a
point $x_*in [x_0-t_*,x_0+t_*]$ which is the unique zero of $F$ in
$[x_0-t_*,x_0+t_*]$ and
beginequation
labeleq:q.conv.x
|x_*-x_k+1|leq frac12 |x_*-x_k |, qquad
k=0,1,,cdots.
endequation
Moreover, if assumption .4 holds as an strict inequality, i.e.
$2bL<1$, then
beginequation
|x_*-x_k+1|leqfrac1-theta^2^k1+theta^2^k
frac L2sqrt1-2bL|x_*-x_k|^2leq
frac L2sqrt1-2bL|x_*-x_k|^2, quad k=0,1,cdots,
endequation
where $theta:=t_*/t_**<1$, and $x_*$ is the
unique zero of $F$ in $[x_0-t_*,x_0+t_*]$ for any $rho$ such that
$ t_*leqrho<t_**,qquad [x_0-rho,x_0+rho]subset I.$
answered Jul 14 at 20:51
MathOverview
7,95242962
edited Jul 14 at 20:57
answered Jul 14 at 20:51
MathOverview
7,95242962
answered Jul 14 at 20:51
MathOverview
7,95242962
answered Jul 14 at 20:51
MathOverview
7,95242962
7,95242962
add a comment |Â
add a comment |Â
up vote
1
down vote
Note: The right-hand side $100$ has a nice representation by powers of $5$.
We have
beginalign*
5^colorbluesqrtx-5^x-7=100=5^colorblue3-5^2
endalign*
which indicates a trial via
beginalign*
sqrtx=3 qquadtextandqquad x-7=2
endalign*
giving the solution $x=9$.
I think the challenging aspect is if there is an algebraic/analytic method besides some iterative approach which gives us the second solution close at hand.
answered Jul 15 at 14:58
Markus Scheuer
56.2k450135
add a comment |Â
up vote
1
down vote
Note: The right-hand side $100$ has a nice representation by powers of $5$.
We have
beginalign*
5^colorbluesqrtx-5^x-7=100=5^colorblue3-5^2
endalign*
which indicates a trial via
beginalign*
sqrtx=3 qquadtextandqquad x-7=2
endalign*
giving the solution $x=9$.
I think the challenging aspect is if there is an algebraic/analytic method besides some iterative approach which gives us the second solution close at hand.
answered Jul 15 at 14:58
Markus Scheuer
56.2k450135
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note: The right-hand side $100$ has a nice representation by powers of $5$.
We have
beginalign*
5^colorbluesqrtx-5^x-7=100=5^colorblue3-5^2
endalign*
which indicates a trial via
beginalign*
sqrtx=3 qquadtextandqquad x-7=2
endalign*
giving the solution $x=9$.
I think the challenging aspect is if there is an algebraic/analytic method besides some iterative approach which gives us the second solution close at hand.
answered Jul 15 at 14:58
Markus Scheuer
56.2k450135
Note: The right-hand side $100$ has a nice representation by powers of $5$.
We have
beginalign*
5^colorbluesqrtx-5^x-7=100=5^colorblue3-5^2
endalign*
which indicates a trial via
beginalign*
sqrtx=3 qquadtextandqquad x-7=2
endalign*
giving the solution $x=9$.
I think the challenging aspect is if there is an algebraic/analytic method besides some iterative approach which gives us the second solution close at hand.
answered Jul 15 at 14:58
Markus Scheuer
56.2k450135
edited Jul 15 at 15:07
answered Jul 15 at 14:58
Markus Scheuer
56.2k450135
answered Jul 15 at 14:58
Markus Scheuer
56.2k450135
answered Jul 15 at 14:58
Markus Scheuer
56.2k450135
56.2k450135
add a comment |Â
add a comment |Â
up vote
0
down vote
accepted
This is the method I used to justify my found solution
Let $; t=5^sqrtx-3;$, we arrive at
$$t^sqrtx+3-5 t+4=0$$
$$t^sqrtx+3 - t^2 + t^2 -5 t+4=0$$
$$t^2(t^sqrtx+1 -1)+(t-1)(t-4)=0$$
Assume $;sqrtx in mathbb N$
$$t^2(t-1)(t^sqrtx+^sqrtx-1+cdots +1)+(t-1)(t-4)=0$$
$$(t-1)(t^sqrtx+2+^sqrtx+1+cdots +t^2+t-4)=0$$
$$t-1=0 quad Rightarrowquad 5^sqrtx-3=1 quad Rightarrowquad sqrtx=3 in mathbb N$$
$x=9quad $ is a valid solution.
answered Jul 15 at 13:26
Wolfdale
24919
add a comment |Â
up vote
0
down vote
accepted
This is the method I used to justify my found solution
Let $; t=5^sqrtx-3;$, we arrive at
$$t^sqrtx+3-5 t+4=0$$
$$t^sqrtx+3 - t^2 + t^2 -5 t+4=0$$
$$t^2(t^sqrtx+1 -1)+(t-1)(t-4)=0$$
Assume $;sqrtx in mathbb N$
$$t^2(t-1)(t^sqrtx+^sqrtx-1+cdots +1)+(t-1)(t-4)=0$$
$$(t-1)(t^sqrtx+2+^sqrtx+1+cdots +t^2+t-4)=0$$
$$t-1=0 quad Rightarrowquad 5^sqrtx-3=1 quad Rightarrowquad sqrtx=3 in mathbb N$$
$x=9quad $ is a valid solution.
answered Jul 15 at 13:26
Wolfdale
24919
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
This is the method I used to justify my found solution
Let $; t=5^sqrtx-3;$, we arrive at
$$t^sqrtx+3-5 t+4=0$$
$$t^sqrtx+3 - t^2 + t^2 -5 t+4=0$$
$$t^2(t^sqrtx+1 -1)+(t-1)(t-4)=0$$
Assume $;sqrtx in mathbb N$
$$t^2(t-1)(t^sqrtx+^sqrtx-1+cdots +1)+(t-1)(t-4)=0$$
$$(t-1)(t^sqrtx+2+^sqrtx+1+cdots +t^2+t-4)=0$$
$$t-1=0 quad Rightarrowquad 5^sqrtx-3=1 quad Rightarrowquad sqrtx=3 in mathbb N$$
$x=9quad $ is a valid solution.
answered Jul 15 at 13:26
Wolfdale
24919
This is the method I used to justify my found solution
Let $; t=5^sqrtx-3;$, we arrive at
$$t^sqrtx+3-5 t+4=0$$
$$t^sqrtx+3 - t^2 + t^2 -5 t+4=0$$
$$t^2(t^sqrtx+1 -1)+(t-1)(t-4)=0$$
Assume $;sqrtx in mathbb N$
$$t^2(t-1)(t^sqrtx+^sqrtx-1+cdots +1)+(t-1)(t-4)=0$$
$$(t-1)(t^sqrtx+2+^sqrtx+1+cdots +t^2+t-4)=0$$
$$t-1=0 quad Rightarrowquad 5^sqrtx-3=1 quad Rightarrowquad sqrtx=3 in mathbb N$$
$x=9quad $ is a valid solution.
answered Jul 15 at 13:26
Wolfdale
24919
answered Jul 15 at 13:26
Wolfdale
24919
answered Jul 15 at 13:26
Wolfdale
24919
answered Jul 15 at 13:26
Wolfdale
24919
24919
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1
I wouldn't expect a very pleasant analytic solution...trial and error (as you have done) looks sensible.
– lulu
Jul 14 at 20:32
@lulu, thx for the hint maybe some inequality or number analysis can help.
– Wolfdale
Jul 14 at 20:34
You might be able to show whether there is any other solution, but an analytic way to arrive at $9$ is highly unlikely.
– Thomas Andrews
Jul 14 at 20:35
Oh, numerical methods work fine, though of course you'll just get an approximation that way. (which of course might lead you to guess the exact answer)
– lulu
Jul 14 at 20:37