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want some medals?.. log2(2m+4)−log2(m−1)=3
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\[\log_{2}(2m+4) - \log_{2}(m-1)=3\]
this time it is a subtraction so write \[\log_2(\frac{2m+4}{m-1})\]=3\] then solve \[\frac{2m+4}{m-1}=2^3\]
thats what it was
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lol@satllite : )
\[\log_{2}(2m+4)-\log_{2}(m-1)= 3\] \[\frac{2m+4}{m-1}=8\] 2m+4=8m-8 6m=12 m=12/6=2
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