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Mathematics 10 Online
OpenStudy (anonymous):

solve log2(4) - log2(x+3)=log2(8)

OpenStudy (anonymous):

are those "2"s the base of logarithms?

OpenStudy (anonymous):

Yessir

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

Then 4/(x+3) = 8 4 = 8(x+3) x+3 = 1/2 x = 1/2 - 3 x= -5/2

OpenStudy (anonymous):

wait why would you put the (x+3) next to the 8

OpenStudy (anonymous):

you multiply both sides by (x+3) to remove it from denominator.

OpenStudy (anonymous):

k, but how did you get t(x+3) by it self

OpenStudy (anonymous):

\[\log_{2}4/(x+3)=\log_{2} 8\] \[4/(x+3)=8\]

OpenStudy (anonymous):

You can remove the logarithms from both sides of the equation in such a fashion if they have the same base.

OpenStudy (anonymous):

\[\log_2(4)-\log^2(x+3)=\log^2(8)\\\\ \log_2(2^2)-\log_2(x+3)=\log_2(8)\\\\ (2)-\log_2(x+3)-\log_2(8)=0\\2-\log_2(x+3)-(3)=0\\-1-\log_2(x+3)=0\\log_2(x+3)=1*-1\\x=\frac{-5}{2} \]

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