Respuesta :

1.

[tex]2\cdot x^2+2\cdot x^1+1\cdot x^0=25\\2x^2+2x+1=25\\2x^2+2x-24=0\\x^2+x-12=0\\x^2+4x-3x-12=0\\x(x+4)-3(x+4)=0\\(x-3)(x+4)=0\\x=3 \vee x=-4[/tex]

The base can't be negative, therefore [tex]x=3[/tex].

2.

[tex]3\cdot x^0+3\cdot x^1+4\cdot x^0-4\cdot x^1+0\cdot x^0=0\\3+3x+4-4x=0\\x=7[/tex]

3.

[tex]2\cdot x^1+3\cdotx^0=1\cdot 2^3+1\cdot2^2+1\cdot2^1+1\cdot2^0\\2x+1=8+4+2+1\\2x=14\\x=7[/tex]