How do you solve $log_4(x+3) + log _4(2-x) =1$ ?

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Answer 1 · 2016-02-27T04:09:09

We know that $log_4 4=1$ hence

$log_4(x+3) + log _4(2-x) =log_4 4=> log_4[(x+3)*(2-x)]=log_4 4=> (x+3)*(2-x)=4=> -x^2-x+2=0=> (x-1)*(x+2)=0$

From the last equation the acceptable values are $x=1$ , $x=-2$

How do you solve log_4(x+3) + log _4(2-x) =1log_4(x+3) + log _4(2-x) =1?