Question

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

Answer 1

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`