Respuesta :

bcalle
I solved it on your other posting of it. x = 12
All of the steps are there
[tex]\bf log_2(x-2)-log_2(5)=1\\\\ -----------------------------\\\\ log_{{ a}}\left( \frac{x}{y}\right)\implies log_{{ a}}(x)-log_{{ a}}(y)\qquad thus\\\\ -----------------------------\\\\ log_2\left( \cfrac{x-2}{5} \right)=1\\\\ -----------------------------\\\\ {{ a}}^{log_{{ a}}x}=x\impliedby \textit{log cancellation rule}\\\\ -----------------------------\\\\ 2^{\cfrac{}{}log_2\left( \frac{x-2}{5} \right)}=2^1\implies \cfrac{x-2}{5}=2^1 \\\\\\ \cfrac{x-2}{5}=2\implies x-2=10\implies x=12[/tex]

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