Respuesta :

[tex]\bf log_{12}(x-3)-log_{12}(x+7)=log_{12}(3)\\\\ -----------------------------\\\\ log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)\qquad thus\\\\ -----------------------------\\\\ log_{12}\left( \cfrac{x-3}{x+7} \right)=log_{12}(3)\impliedby \begin{array}{llll} \textit{removing the log from}\\ \textit{both sides} \end{array} \\\\\\ \cfrac{x-3}{x+7}=3\implies x-3=3x+21\implies -24=2x\implies -12=x[/tex]

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