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Rainbow_Dash

  • 3 years ago

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

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

  2. ProgramGuru
    • 3 years ago
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    sorry it will be 12b-21

  3. Wired
    • 3 years ago
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    Log Quotient Rule: \[\Large \log_{b}({A/C}) = \log_{b}A − \log_{b}C\] Applying that gets you: \[\Large \log_{2}{(12b-21)}-\log_{2}{(b^{2}-3)}=2\] \[\Large \log_{2}{\frac{(12b-21)}{(b^{2}-3)}}=2\] \[\Large x=c^{y}\] \[\Large log_{c}{x}=y\] \[\Large X = \frac{(12b-21)}{(b^{2}-3)}\] \[\Large Y = 2\] \[\Large C = 2\] \[\Large 2^{2} = \frac{(12b-21)}{(b^{2}-3)}\] \[\Large 4b^{2}-12= 12b-21\] \[\Large 4b^{2}-12= 12b-21\] \[\Large 4b^{2}-12b+9=0\]

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