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log_{2}(2-x)-2=log_{2} (x+8)
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\[\log_{2}(2-x)-2=\log_{2} (x+8)\]
step by step please
\[\large \log_{2}(2-x)-2=\log_{2} (x+8)\] \[\large \log_{2}(2-x)-\log_{2}(4)=\log_{2} (x+8)\] \[\large \log_{2}\left(\frac{2-x}{4}\right)=\log_{2} (x+8)\] \[\large \frac{2-x}{4}=x+8\] \[\large 2-x=4(x+8)\] \[\large 2-x=4x+32\] \[\large 2-32=4x+x\] \[\large -30=5x\] \[\large \frac{-30}{5}=x\] \[\large -6=x\] \[\large x=-6\]
notes: I converted 2 to \[\log_2(4)\] because I wanted everything inside a log with base 2
so 2 = \[\log_{2}4 \] ?
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\[\large \log_{2}(4)=\log_{2}(2^2)=2\log_{2}(2)=2(1)=2\] So \[\large \log_{2}(4)=2\]
thank you again!
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