Hello again... The 'usual' transformations lead to the cubic equation x^3-3x-2=0. Let f(x)=x^3-3x-2. Then f'(x)=3x^2-3. Therefore f(x) has a local maximum at x=1 and a local minimum at x=-1. Since f(-1)=0 x=-1 is a double root. f(x)/(x+1)^2=x-2, therefore x=2 is a third solution. Of course, the two solutions x=-1 and x=2 are immediately obvious...
Hello again...
The 'usual' transformations lead to the cubic equation x^3-3x-2=0.
Let f(x)=x^3-3x-2. Then f'(x)=3x^2-3. Therefore f(x) has a local maximum at x=1 and a local minimum at x=-1. Since f(-1)=0 x=-1 is a double root. f(x)/(x+1)^2=x-2, therefore x=2 is a third solution.
Of course, the two solutions x=-1 and x=2 are immediately obvious...
Very nice! ❤
[3^(x³)]/(27^x)=9
[3^(x³)]/(3^3x)=3²
3^(x³-3x)=3²
x³-3x=2
x³-3x-2=0
x³-x-2x-2=0
x(x²-1)-2(x+1)=0
x(x+1)(x-1)-2(x+1)=0
(x+1)[x(x-1)-2]=0
(x+1)(x²-x-2)=0
(x+1)(x+1)(x-2)=0
x+1=0
x=-1 ❤❤ multiplicity of 2
x-2=0
x=2 ❤
x = { -1 , 2 }
(3)^x3/3^3x=9 ; i,e 3^(x^3-3x)=3^2 ; i,e x^3-3x=2 ; i,e x^3-3x-2=0 ; x^3+1-3x-3=0 ; i,e (x+1)(x^2-x+1)-3(x+1)=0 ; i,e (x+1)(x^2-x+1-3)=0 ; i,e (x+1)(x^2-x-2)=0 ; i,e (x+1)(x^2-2x+x-2)=0 i,e (x+1)(x-2)(x+1)=0 ; So, x=-1 or x=2 or x=-1.