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solve log2(4) - log2(x+3)=log2(8)
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are those "2"s the base of logarithms?
Yessir
yes
Then 4/(x+3) = 8 4 = 8(x+3) x+3 = 1/2 x = 1/2 - 3 x= -5/2
wait why would you put the (x+3) next to the 8
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you multiply both sides by (x+3) to remove it from denominator.
k, but how did you get t(x+3) by it self
\[\log_{2}4/(x+3)=\log_{2} 8\] \[4/(x+3)=8\]
You can remove the logarithms from both sides of the equation in such a fashion if they have the same base.
\[\log_2(4)-\log^2(x+3)=\log^2(8)\\\\ \log_2(2^2)-\log_2(x+3)=\log_2(8)\\\\ (2)-\log_2(x+3)-\log_2(8)=0\\2-\log_2(x+3)-(3)=0\\-1-\log_2(x+3)=0\\log_2(x+3)=1*-1\\x=\frac{-5}{2} \]
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