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Hola, Si claro
![log(x) + log(x + 3) = 2 log(x + 1) \\ log(x(x + 3)) = log( {(x + 1)}^{2} ) \\ log( {x}^{2} +3x) - log( {x}^{2} + 2x + 1) = 0 \\ log( \frac{ {x}^{2} +3 x}{ {x}^{2} + 2x + 1 } ) = 0 \\ \frac{ {x}^{2} + 3x}{ {x}^{2} + 2x + 1 } = {10}^{0} = 1 \\ {x}^{2} +3x = {x}^{2} + 2x + 1 \\ 3x -2x= 1 log(x) + log(x + 3) = 2 log(x + 1) \\ log(x(x + 3)) = log( {(x + 1)}^{2} ) \\ log( {x}^{2} +3x) - log( {x}^{2} + 2x + 1) = 0 \\ log( \frac{ {x}^{2} +3 x}{ {x}^{2} + 2x + 1 } ) = 0 \\ \frac{ {x}^{2} + 3x}{ {x}^{2} + 2x + 1 } = {10}^{0} = 1 \\ {x}^{2} +3x = {x}^{2} + 2x + 1 \\ 3x -2x= 1](https://tex.z-dn.net/?f=+log%28x%29+%2B+log%28x+%2B+3%29+%3D+2+log%28x+%2B+1%29+%5C%5C+log%28x%28x+%2B+3%29%29+%3D+log%28+%7B%28x+%2B+1%29%7D%5E%7B2%7D+%29+%5C%5C+log%28+%7Bx%7D%5E%7B2%7D+%2B3x%29+-+log%28+%7Bx%7D%5E%7B2%7D+%2B+2x+%2B+1%29+%3D+0+%5C%5C+log%28+%5Cfrac%7B+%7Bx%7D%5E%7B2%7D+%2B3+x%7D%7B+%7Bx%7D%5E%7B2%7D+%2B+2x+%2B+1+%7D+%29+%3D+0+%5C%5C+%5Cfrac%7B+%7Bx%7D%5E%7B2%7D+%2B+3x%7D%7B+%7Bx%7D%5E%7B2%7D+%2B+2x+%2B+1+%7D+%3D+%7B10%7D%5E%7B0%7D+%3D+1+%5C%5C+%7Bx%7D%5E%7B2%7D+%2B3x+%3D+%7Bx%7D%5E%7B2%7D+%2B+2x+%2B+1+%5C%5C+3x+-2x%3D+1+)
Se elimina el término cuadrática.
Se elimina el término cuadrática.
Eudoxo86:
Ya lo resolví, esta ya corregido
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