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Solve log2(x+2)-log2 3=6
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log a - log b = log (a/b)
take (e) for both side \[\log_2 (x+2)-\log_2(3)=6\] \[e ^{\log_2(x+2)}-e ^{\log_2(3)}=e ^{6}\] \[x+2-3=e ^{6}\]
Thank you for your assistance.
@Aj01, that is incorrect... the base is 2 not "e"
\[\large 2 ^{\log_2 (\frac{x+2}{3})} = 2^6\] \[\frac{x+2}{3} = 2^6\]
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