Question

log(2) (12b-21) - log(2) (b^2 - 3) = 2

Answer 1

\[\log_{2}((12b-1)/ (b ^{2}-3))=2\log_{2}2 \]
\[(12b-3)/(b ^{2}-3)=4\]
Now solve the quadratic to get value of b!

Answer 2

sorry it will be 12b-21

Answer 3

Log Quotient Rule:
\[\Large \log_{b}({A/C}) = \log_{b}A − \log_{b}C\]

Applying that gets you:
\[\Large \log_{2}{(12b-21)}-\log_{2}{(b^{2}-3)}=2\]
\[\Large \log_{2}{\frac{(12b-21)}{(b^{2}-3)}}=2\]

\[\Large x=c^{y}\]
\[\Large log_{c}{x}=y\]

\[\Large X = \frac{(12b-21)}{(b^{2}-3)}\]
\[\Large Y = 2\]
\[\Large C = 2\]
\[\Large 2^{2} = \frac{(12b-21)}{(b^{2}-3)}\]
\[\Large 4b^{2}-12= 12b-21\]
\[\Large 4b^{2}-12= 12b-21\]
\[\Large 4b^{2}-12b+9=0\]