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Pourquoi faire simple quand on peut faire compliqué ?x>0log_3( x^( log_3( x ) ) = log_3( 81 ) ( log_3( x) )² = log_3( 3⁴) ( log_3( x) )² = 4 | log_3( x) | = 2 log_3( x) = 2 ou log_3( x) = -2 x = 3² ou x = 3^(-2) x = 9 ou x = 1/9
This is the simplest and most rational way: to prolog both parts by base 3.
@@Sergey_Moskvichev Thanks
X^(Log[3,X])=81 X=0.1 recurring=1/9 X=9
log(3)x * log(3)x=log(3)81 , log(3)81=4 , log(3)x * log(3)x = 4 , --> log(3)4 * log(3)4 = 2 * 2 , log(3)x=2 , --> , x=3^2 , x =9 , test , 9^log(3)9=9^2 , 9^2=81 , OK ,
Please Can you solve this équation sqrt (x-1) +4 sqrt(y-4) = (x+y)/2
We suppose a= sqrt(x-1) , b= sqrt(y-4) finnaly WE get (a-1)^2 +(b-4)^2 = 12 ......
Thanks i solved it
I just guessworked this lol
Don't guess . Put in some works!🥰
x^{Log[3](x)} = 81x^{Ln(x) / Ln(3)} = 81Ln[x^{Ln(x) / Ln(3)}] = Ln(81)[Ln(x) / Ln(3)].Ln(x) = Ln(81) → let: a = Ln(x)[a / Ln(3)].a = Ln(81)a² / Ln(3) = Ln(81)a² = Ln(3) * Ln(81)a² = Ln(3) * Ln(3⁴)a² = Ln(3) * 4.Ln(3)a² = 4 * [Ln(3)]²a² = [2.Ln(3)]²a = ± 2.Ln(3)a = ± Ln(9) → recall: a = Ln(x)Ln(x) = ± Ln(9)First case: Ln(x) = Ln(9)x = 9Second case: Ln(x) = - Ln(9)Ln(x) = - Ln(9)Ln(x) = Ln(9)^(- 1)x = 9^(- 1)x = 1/9
Pourquoi faire simple quand on peut faire compliqué ?
x>0
log_3( x^( log_3( x ) ) = log_3( 81 )
( log_3( x) )² = log_3( 3⁴)
( log_3( x) )² = 4
| log_3( x) | = 2
log_3( x) = 2 ou log_3( x) = -2
x = 3² ou x = 3^(-2)
x = 9 ou x = 1/9
This is the simplest and most rational way: to prolog both parts by base 3.
@@Sergey_Moskvichev Thanks
X^(Log[3,X])=81 X=0.1 recurring=1/9 X=9
log(3)x * log(3)x=log(3)81 , log(3)81=4 , log(3)x * log(3)x = 4 , --> log(3)4 * log(3)4 = 2 * 2 ,
log(3)x=2 , --> , x=3^2 , x =9 , test , 9^log(3)9=9^2 , 9^2=81 , OK ,
Please Can you solve this équation
sqrt (x-1) +4 sqrt(y-4) = (x+y)/2
We suppose a= sqrt(x-1) , b= sqrt(y-4) finnaly WE get (a-1)^2 +(b-4)^2 = 12 ......
Thanks i solved it
I just guessworked this lol
Don't guess . Put in some works!🥰
x^{Log[3](x)} = 81
x^{Ln(x) / Ln(3)} = 81
Ln[x^{Ln(x) / Ln(3)}] = Ln(81)
[Ln(x) / Ln(3)].Ln(x) = Ln(81) → let: a = Ln(x)
[a / Ln(3)].a = Ln(81)
a² / Ln(3) = Ln(81)
a² = Ln(3) * Ln(81)
a² = Ln(3) * Ln(3⁴)
a² = Ln(3) * 4.Ln(3)
a² = 4 * [Ln(3)]²
a² = [2.Ln(3)]²
a = ± 2.Ln(3)
a = ± Ln(9) → recall: a = Ln(x)
Ln(x) = ± Ln(9)
First case: Ln(x) = Ln(9)
x = 9
Second case: Ln(x) = - Ln(9)
Ln(x) = - Ln(9)
Ln(x) = Ln(9)^(- 1)
x = 9^(- 1)
x = 1/9