BTW :
sage: solve(1+2*log(x+1, 4)==2*log(x,2), x, to_poly_solve="force")
[x == -sqrt(3) + 1, x == sqrt(3) + 1]
HTH,
β
Le samedi 9 avril 2022 Γ 11:34:10 UTC+2, wdjo...@gmail.com a Γ©crit :
> On Sat, Apr 9, 2022 at 5:18 AM Paolo Robillos <paolo.r...@gmail.com>
> wrote:
> >
> > Hi,
> >
> > I am trying to solve the following equation for x, 1+2log(x+1,
> 4)==2log(x,2)
> >
> > I entered in the input "(1+2log(x+1,
> 4)==2log(x,2)).solve(x,algorithm='sympy', domain='all')"
> >
> > and the Output was
> "{π₯β£π₯βββ§βπ₯2log(2)+π(π₯+1)1log(2)=0}β{π₯β£π₯βββ§π₯2log(2)=0}"
> >
> > The answer I am looking for is "x = 1+3^(1/2)."
> >
>
> You're right, but here is a comment to show SymPy is on the right track:
>
> The first equation is x^(2/log(2)) == e*(x + 1)^(1/log(2)). If you
> first raise both sides ro the log(2) power, SageMath can solve it:
>
> sage: solve(x^(2) == e^(log(2))*(x + 1), x)
> [x == -sqrt(3) + 1, x == sqrt(3) + 1]
>
>
> >
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