Question

log2(12b-21)-log2(b2-3)=2

Answer 1

is it base "2"? or a number "2"?

Answer 2

\[\log_{2} (12b-21) -\log_{2}(b ^{2}-3) = 2\]

Answer 3

\[\log_{2} \frac{ 12b-21 }{ b ^{2}-3 }=2\]

Answer 4

Apply the "Laws of Logarithm" and express in Exponential Form to find the value of "b".
:)
continuation of "caozeyuan's solution":

exponential form
\[\frac{ 12b-21 }{ b ^{2}-3 }=2^{2}\]

cross-multiply
\[12b-21=4b ^{2}-12\]

combine like terms and equate to zero
\[4b ^{2}-12b+9=0\]

look for factors of the trinomial. :)

Answer 5

It's a perfect square trinomial:
\[(2b-3)^{2}\]
zero property
\[2b-3=0\]\[2b=3\]\[b=\frac{ 3 }{ 2 }\]that's the answer.
:)