Respuesta :

[tex]\bf log_2(5)-log_2(x+2)=3\implies log_2\left( \cfrac{5}{x+2} \right)=3\\\\ -------------------------------\\\\ \textit{Logarithm Cancellation Rules}\\\\ log_a a^x\implies x\qquad \qquad \boxed{a^{log_ax}=x}\\\\ -------------------------------\\\\ 2^{log_2\left( \frac{5}{x+2} \right)}=2^3\implies \cfrac{5}{x+2}=2^3\implies \cfrac{5}{x+2}=8\implies 5=8x+16 \\\\\\ -11=8x\implies -\cfrac{11}{8}=x[/tex]
Banabanana
[tex]\log_25-\log_2(x+2)=3 \ \ \ \ \ \ [x\ \textgreater \ -2] \\ \\ \log_2 \left( \frac{5}{x+2}\right) =3 \\ \\ \frac{5}{x+2}=2^3 \\\\ \frac{5}{x+2}=8 \\ \\ 5=8(x+2) \\ \\ 5=8x+16 \\ \\ 5-16=8x \\ \\ -11=8x \\ \\ x=- \frac{11}{8} [/tex]

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