How do you solve #log (x+3)=log 8-log 2#?

1 Answer
Feb 28, 2016

#x=1#

Explanation:

What you need to know:
#color(white)("XXX")log(p) - log(q) = log (p/q)#

Therefore:
#color(white)("XXX")log(x+3)=log(8)-log(2)#

#rarrcolor(white)("XXX")log(x+3) = log(4)#

#rarrcolor(white)("XXX")x+3 = 4#

#rarrcolor(white)("XXX")x=1#