How do you solve #log (2x + 1) - log (x - 2) = 1#?
1 Answer
Mar 16, 2016
Explanation:
#log(2x+1)-log(x-2)=1#
#log((2x+1)/(x-2))=1#
#log((2x+1)/(x-2))=log(10)#
#(2x+1)/(x-2)=10#
#2x+1=10(x-2)#
#2x+1=10x-20#
#8x=21#
#color(green)(|bar(ul(color(white)(a/a)x=21/8color(white)(a/a)|)))#