On Sat, Apr 9, 2022 at 5:18 AM Paolo Robillos <paolo.robil...@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 "{๐ฅโฃ๐ฅโโโงโ๐ฅ2log(2)+๐(๐ฅ+1)1log(2)=0}โ{๐ฅโฃ๐ฅโโโง๐ฅ2log(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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