Mixing
$(x*(log_{2}(x))^2)/2 = x^{3/2}$ how to solve it?
$begingroup$
Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.
logarithms
edited Jan 2 at 10:53
Bernard
119k639112
asked Jan 2 at 10:51
HmmmanHmmman
13
$endgroup$
add a comment |
$begingroup$
Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.
logarithms
edited Jan 2 at 10:53
Bernard
119k639112
asked Jan 2 at 10:51
HmmmanHmmman
13
$endgroup$
$begingroup$
Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
$endgroup$
– Matti P.
Jan 2 at 10:55
$begingroup$
You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
$endgroup$
– Matti P.
Jan 2 at 10:58
$begingroup$
Are you looking for only an approximate solution?
$endgroup$
– Matti P.
Jan 2 at 10:59
$begingroup$
@MattiP. thanks, your idea with $x = 2^y$ was great!
$endgroup$
– Hmmman
Jan 2 at 11:09
$begingroup$
see herehttps://planetmath.org/approximationofthelogfunction
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:37
add a comment |
$begingroup$
Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.
logarithms
edited Jan 2 at 10:53
Bernard
119k639112
asked Jan 2 at 10:51
HmmmanHmmman
13
$endgroup$
Is there a manual solution for this equation? Or I should use Wolfram:
result from Wolfram.
logarithms
logarithms
edited Jan 2 at 10:53
Bernard
119k639112
asked Jan 2 at 10:51
HmmmanHmmman
13
edited Jan 2 at 10:53
Bernard
119k639112
asked Jan 2 at 10:51
HmmmanHmmman
13
edited Jan 2 at 10:53
Bernard
119k639112
edited Jan 2 at 10:53
Bernard
119k639112
edited Jan 2 at 10:53
Bernard
119k639112
119k639112
asked Jan 2 at 10:51
HmmmanHmmman
13
asked Jan 2 at 10:51
HmmmanHmmman
13
asked Jan 2 at 10:51
HmmmanHmmman
13
13
$begingroup$
Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
$endgroup$
– Matti P.
Jan 2 at 10:55
$begingroup$
You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
$endgroup$
– Matti P.
Jan 2 at 10:58
$begingroup$
Are you looking for only an approximate solution?
$endgroup$
– Matti P.
Jan 2 at 10:59
$begingroup$
@MattiP. thanks, your idea with $x = 2^y$ was great!
$endgroup$
– Hmmman
Jan 2 at 11:09
$begingroup$
see herehttps://planetmath.org/approximationofthelogfunction
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:37
add a comment |
$begingroup$
Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
$endgroup$
– Matti P.
Jan 2 at 10:55
$begingroup$
You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
$endgroup$
– Matti P.
Jan 2 at 10:58
$begingroup$
Are you looking for only an approximate solution?
$endgroup$
– Matti P.
Jan 2 at 10:59
$begingroup$
@MattiP. thanks, your idea with $x = 2^y$ was great!
$endgroup$
– Hmmman
Jan 2 at 11:09
$begingroup$
see herehttps://planetmath.org/approximationofthelogfunction
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:37
$begingroup$
Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
$endgroup$
– Matti P.
Jan 2 at 10:55
$begingroup$
Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
$endgroup$
– Matti P.
Jan 2 at 10:55
$begingroup$
You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
$endgroup$
– Matti P.
Jan 2 at 10:58
$begingroup$
You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
$endgroup$
– Matti P.
Jan 2 at 10:58
$begingroup$
Are you looking for only an approximate solution?
$endgroup$
– Matti P.
Jan 2 at 10:59
$begingroup$
Are you looking for only an approximate solution?
$endgroup$
– Matti P.
Jan 2 at 10:59
$begingroup$
@MattiP. thanks, your idea with $x = 2^y$ was great!
$endgroup$
– Hmmman
Jan 2 at 11:09
$begingroup$
@MattiP. thanks, your idea with $x = 2^y$ was great!
$endgroup$
– Hmmman
Jan 2 at 11:09
$begingroup$
see herehttps://planetmath.org/approximationofthelogfunction
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:37
$begingroup$
see herehttps://planetmath.org/approximationofthelogfunction
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:37
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:
The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.
Then:
begin{eqnarray}
frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
end{eqnarray}
Then with Lambert function:
begin{eqnarray}
-frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
end{eqnarray}
Then the solutions are:
begin{eqnarray}
x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.448369089606ldots \
x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.652053581ldots-i101297.962454054ldots \
x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.459946970862ldots \
end{eqnarray}
The function is numeric too, but it is very elegant.
There is an article in wikipedia for this Lambert W function.
answered Jan 2 at 12:42
El PastaEl Pasta
34515
$endgroup$
1
$begingroup$
Thanks for your exhaustive explanation!!!
$endgroup$
– Hmmman
Jan 2 at 12:48
1
$begingroup$
This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
$endgroup$
– Claude Leibovici
Jan 3 at 5:29
add a comment |
$begingroup$
Welcome to the world of Lambert function !
Using Matti P.'s suggestion $x=2^y$, the equation becomes
$$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
$$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
$$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
$$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$
It have been faster to use $x=2^{e^t}$ for the same result.
answered Jan 2 at 11:49
Claude LeiboviciClaude Leibovici
119k1157132
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add a comment |
$begingroup$
This equation has three real solutions:
$$xapprox 0.4483690898$$
$$x=4$$
$$xapprox 6380.459941$$
This can be obtained by the Newton Raphson method.
answered Jan 2 at 11:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
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How did you find them? I asked how to solve, I know answers by myself!
$endgroup$
– Hmmman
Jan 2 at 11:22
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see here sosmath.com/calculus/diff/der07/der07.html
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:25
$begingroup$
Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
$endgroup$
– Hmmman
Jan 2 at 11:31
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:
The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.
Then:
begin{eqnarray}
frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
end{eqnarray}
Then with Lambert function:
begin{eqnarray}
-frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
end{eqnarray}
Then the solutions are:
begin{eqnarray}
x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.448369089606ldots \
x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.652053581ldots-i101297.962454054ldots \
x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.459946970862ldots \
end{eqnarray}
The function is numeric too, but it is very elegant.
There is an article in wikipedia for this Lambert W function.
answered Jan 2 at 12:42
El PastaEl Pasta
34515
$endgroup$
1
$begingroup$
Thanks for your exhaustive explanation!!!
$endgroup$
– Hmmman
Jan 2 at 12:48
1
$begingroup$
This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
$endgroup$
– Claude Leibovici
Jan 3 at 5:29
add a comment |
$begingroup$
There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:
The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.
Then:
begin{eqnarray}
frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
end{eqnarray}
Then with Lambert function:
begin{eqnarray}
-frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
end{eqnarray}
Then the solutions are:
begin{eqnarray}
x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.448369089606ldots \
x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.652053581ldots-i101297.962454054ldots \
x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.459946970862ldots \
end{eqnarray}
The function is numeric too, but it is very elegant.
There is an article in wikipedia for this Lambert W function.
answered Jan 2 at 12:42
El PastaEl Pasta
34515
$endgroup$
1
$begingroup$
Thanks for your exhaustive explanation!!!
$endgroup$
– Hmmman
Jan 2 at 12:48
1
$begingroup$
This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
$endgroup$
– Claude Leibovici
Jan 3 at 5:29
add a comment |
$begingroup$
There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:
The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.
Then:
begin{eqnarray}
frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
end{eqnarray}
Then with Lambert function:
begin{eqnarray}
-frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
end{eqnarray}
Then the solutions are:
begin{eqnarray}
x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.448369089606ldots \
x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.652053581ldots-i101297.962454054ldots \
x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.459946970862ldots \
end{eqnarray}
The function is numeric too, but it is very elegant.
There is an article in wikipedia for this Lambert W function.
answered Jan 2 at 12:42
El PastaEl Pasta
34515
$endgroup$
There is no algebraic solution, but you can use numeric methods like Newton-Raphson method or the Lambert $W$ function:
The Lambert $W$ function is defined by $z=W(ze^z)$ with the domain as $mathbb{C}$ and there are two possible functions $W_0$ with $textrm{Re}(W_0(x))geq -1$ and $W_{-1}$ with $textrm{Re}(W_{-1}) < -1$.
Then:
begin{eqnarray}
frac{1}{2}x left(log_2(x)right)^2 &=& x^{3/2} \
frac{1}{2}x^{-1/2} left(log_2(x)right)^2 &=& 1 \
frac{1}{sqrt{2}}x^{-1/4}log_2(x) &=& pm 1 \
x^{-1/4}frac{ln(x)}{ln(2)} &=& pm sqrt{2} \
x^{-1/4}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}ln(x) &=& pm sqrt{2}ln(2) \
e^{(-1/4)ln(x)}left(-frac{1}{4}ln(x)right) &=& pm frac{1}{4}sqrt{2}ln(2) \
end{eqnarray}
Then with Lambert function:
begin{eqnarray}
-frac{1}{4}ln(x) &=& Wleft(pmfrac{1}{4}sqrt{2}ln(2)right) \
x &=& exp leftlbrace-4Wleft(pmfrac{1}{4}sqrt{2}ln(2)right)rightrbrace \
end{eqnarray}
Then the solutions are:
begin{eqnarray}
x_1 &=& exp leftlbrace-4W_0left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace = 0.448369089606ldots \
x_2 &=& exp leftlbrace-4W_0left(-frac{1}{4}sqrt{2}ln(2)right)rightrbrace=4 \
x_3 &=& exp leftlbrace-4W_{-1}left(frac{1}{4}sqrt{2}ln(2)right)rightrbrace=-153792.652053581ldots-i101297.962454054ldots \
x_4 &=& exp leftlbrace-4W_{-1}left(-frac{1}{4}sqrt{2}ln(2) right)rightrbrace=6380.459946970862ldots \
end{eqnarray}
The function is numeric too, but it is very elegant.
There is an article in wikipedia for this Lambert W function.
answered Jan 2 at 12:42
El PastaEl Pasta
34515
edited Jan 6 at 23:47
answered Jan 2 at 12:42
El PastaEl Pasta
34515
answered Jan 2 at 12:42
El PastaEl Pasta
34515
answered Jan 2 at 12:42
El PastaEl Pasta
34515
34515
1
$begingroup$
Thanks for your exhaustive explanation!!!
$endgroup$
– Hmmman
Jan 2 at 12:48
1
$begingroup$
This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
$endgroup$
– Claude Leibovici
Jan 3 at 5:29
add a comment |
1
$begingroup$
Thanks for your exhaustive explanation!!!
$endgroup$
– Hmmman
Jan 2 at 12:48
1
$begingroup$
This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
$endgroup$
– Claude Leibovici
Jan 3 at 5:29
1
1
$begingroup$
Thanks for your exhaustive explanation!!!
$endgroup$
– Hmmman
Jan 2 at 12:48
$begingroup$
Thanks for your exhaustive explanation!!!
$endgroup$
– Hmmman
Jan 2 at 12:48
1
1
$begingroup$
This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
$endgroup$
– Claude Leibovici
Jan 3 at 5:29
$begingroup$
This is a nicely detailed and illustrated solution ! May I confess that I missed the complex root ? $to +1$ for sure.
$endgroup$
– Claude Leibovici
Jan 3 at 5:29
add a comment |
$begingroup$
Welcome to the world of Lambert function !
Using Matti P.'s suggestion $x=2^y$, the equation becomes
$$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
$$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
$$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
$$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$
It have been faster to use $x=2^{e^t}$ for the same result.
answered Jan 2 at 11:49
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
add a comment |
$begingroup$
Welcome to the world of Lambert function !
Using Matti P.'s suggestion $x=2^y$, the equation becomes
$$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
$$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
$$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
$$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$
It have been faster to use $x=2^{e^t}$ for the same result.
answered Jan 2 at 11:49
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
add a comment |
$begingroup$
Welcome to the world of Lambert function !
Using Matti P.'s suggestion $x=2^y$, the equation becomes
$$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
$$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
$$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
$$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$
It have been faster to use $x=2^{e^t}$ for the same result.
answered Jan 2 at 11:49
Claude LeiboviciClaude Leibovici
119k1157132
$endgroup$
Welcome to the world of Lambert function !
Using Matti P.'s suggestion $x=2^y$, the equation becomes
$$2^{y-1} y^2=2^{frac{3y}{2}}$$ that is to say
$$y^2=2^{frac{y}{2}+1}=2 times2^{frac{y}{2}}$$ Now, let $2^{frac{y}{2}}=t$, that is to say $y=frac{2 t}{log (2)}$ to make the equation
$$e^{2 t} left(frac{2 t^2}{log ^2(2)}-e^tright)=0implies e^t=frac{2 t^2}{log ^2(2)}$$ and then the three roots
$$t_1=-2 W_0left(frac{log (2)}{2 sqrt{2}}right)qquad t_2=-2 W_0left(-frac{log (2)}{2 sqrt{2}}right)qquad t_3=-2 W_{-1}left(-frac{log (2)}{2 sqrt{2}}right)$$
It have been faster to use $x=2^{e^t}$ for the same result.
answered Jan 2 at 11:49
Claude LeiboviciClaude Leibovici
119k1157132
edited Jan 3 at 5:30
answered Jan 2 at 11:49
Claude LeiboviciClaude Leibovici
119k1157132
answered Jan 2 at 11:49
Claude LeiboviciClaude Leibovici
119k1157132
answered Jan 2 at 11:49
Claude LeiboviciClaude Leibovici
119k1157132
119k1157132
add a comment |
add a comment |
$begingroup$
This equation has three real solutions:
$$xapprox 0.4483690898$$
$$x=4$$
$$xapprox 6380.459941$$
This can be obtained by the Newton Raphson method.
answered Jan 2 at 11:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
$begingroup$
How did you find them? I asked how to solve, I know answers by myself!
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– Hmmman
Jan 2 at 11:22
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see here sosmath.com/calculus/diff/der07/der07.html
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– Dr. Sonnhard Graubner
Jan 2 at 11:25
$begingroup$
Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
$endgroup$
– Hmmman
Jan 2 at 11:31
add a comment |
$begingroup$
This equation has three real solutions:
$$xapprox 0.4483690898$$
$$x=4$$
$$xapprox 6380.459941$$
This can be obtained by the Newton Raphson method.
answered Jan 2 at 11:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
$begingroup$
How did you find them? I asked how to solve, I know answers by myself!
$endgroup$
– Hmmman
Jan 2 at 11:22
$begingroup$
see here sosmath.com/calculus/diff/der07/der07.html
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:25
$begingroup$
Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
$endgroup$
– Hmmman
Jan 2 at 11:31
add a comment |
$begingroup$
This equation has three real solutions:
$$xapprox 0.4483690898$$
$$x=4$$
$$xapprox 6380.459941$$
This can be obtained by the Newton Raphson method.
answered Jan 2 at 11:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
$endgroup$
This equation has three real solutions:
$$xapprox 0.4483690898$$
$$x=4$$
$$xapprox 6380.459941$$
This can be obtained by the Newton Raphson method.
answered Jan 2 at 11:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
edited Jan 2 at 11:23
answered Jan 2 at 11:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 2 at 11:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
answered Jan 2 at 11:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
73.6k42864
73.6k42864
$begingroup$
How did you find them? I asked how to solve, I know answers by myself!
$endgroup$
– Hmmman
Jan 2 at 11:22
$begingroup$
see here sosmath.com/calculus/diff/der07/der07.html
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:25
$begingroup$
Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
$endgroup$
– Hmmman
Jan 2 at 11:31
add a comment |
$begingroup$
How did you find them? I asked how to solve, I know answers by myself!
$endgroup$
– Hmmman
Jan 2 at 11:22
$begingroup$
see here sosmath.com/calculus/diff/der07/der07.html
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:25
$begingroup$
Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
$endgroup$
– Hmmman
Jan 2 at 11:31
$begingroup$
How did you find them? I asked how to solve, I know answers by myself!
$endgroup$
– Hmmman
Jan 2 at 11:22
$begingroup$
How did you find them? I asked how to solve, I know answers by myself!
$endgroup$
– Hmmman
Jan 2 at 11:22
$begingroup$
see here sosmath.com/calculus/diff/der07/der07.html
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:25
$begingroup$
see here sosmath.com/calculus/diff/der07/der07.html
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:25
$begingroup$
Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
$endgroup$
– Hmmman
Jan 2 at 11:31
$begingroup$
Thanks, but I was just trying to solve it without computer. Maybe using some approximation formulas for $(log_{2}(x))^2$, with rough approximation. Seems like it s impossible!
$endgroup$
– Hmmman
Jan 2 at 11:31
add a comment |
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$begingroup$
Well, as you can see, the Wolfram Alpha results at a solution involving some special functions (namely the $W$ function). So it seems like there is likely no algebraic solution.
$endgroup$
– Matti P.
Jan 2 at 10:55
$begingroup$
You could try by approaching this with a substitution like $x=2^y$, $e^y$ or something similar.
$endgroup$
– Matti P.
Jan 2 at 10:58
$begingroup$
Are you looking for only an approximate solution?
$endgroup$
– Matti P.
Jan 2 at 10:59
$begingroup$
@MattiP. thanks, your idea with $x = 2^y$ was great!
$endgroup$
– Hmmman
Jan 2 at 11:09
$begingroup$
see herehttps://planetmath.org/approximationofthelogfunction
$endgroup$
– Dr. Sonnhard Graubner
Jan 2 at 11:37