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Help solving log equation for x

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 "{๐ฅโฃ๐ฅโโโงโ๐ฅ2log(2)+๐(๐ฅ+1)1log(2)=0}โ{๐ฅโฃ๐ฅโโโง๐ฅ2log(2)=0}"

The answer I am looking for is "x = 1+3^(1/2)."

 2 None slelievre 17194 ●21 ●154 ●341 http://carva.org/samue...

Help solving log equation for x

Hi, I am trying to solve the following equation for x, 1+2log(x+1, 4)==2log(x,2)x:

1 + 2 * log(x + 1, 4) == 2 * log(x, 2)


I entered in the input "(1+2log(x+1, 4)==2log(x,2)).solve(x,algorithm='sympy', domain='all')"

(1 + 2 * log(x + 1, 4) == 2 * log(x, 2)).solve(x, algorithm='sympy', domain='all')


and the Output output was "{๐ฅโฃ๐ฅโโโงโ๐ฅ2log(2)+๐(๐ฅ+1)1log(2)=0}โ{๐ฅโฃ๐ฅโโโง๐ฅ2log(2)=0}"

{๐ฅโฃ๐ฅโโโงโ๐ฅ2log(2)+๐(๐ฅ+1)1log(2)=0}โ{๐ฅโฃ๐ฅโโโง๐ฅ2log(2)=0}

The answer I am looking for is "x is

x = 1+3^(1/2)."1+3^(1/2)