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how do you solve log base4 (x) + log base4 (x-12)=3
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\[\log_{4} (x)+\log_{4} (x-12)=3 \]
log a*b = log a + log b so log a + log b = log ab \[\log_{4} x +\log_{4} (x-12)=\log_{4} (x*(x-12)=3) \]
\[x*(x-12)=4^{3}=64\]
x^2-12x-64=0 (x-16)(x+4)=0 now do little bit by urself @soccer106
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