How do you solve #log(2+x)-log(x-5)=log 2#?
1 Answer
Dec 3, 2015
Explanation:
Re-write as single logarithmic expression
Note:
#log((2+x)/(x-5))= log 2#
#10^log((2+x)/(x-5)) = 10^(log2)#
#(2+x)/(x-5) = 2#
#(2+x)/(x-5)*color(red)((x-5)) = 2*color(red)((x-5))#
#(2+x)/cancel(x-5)*cancel((x-5)) = 2(x-5)#
# 2+x " " "= 2x- 10#
#+10 - x = -x +10# ===============
#color(red)(12 " " " = x)#
Check :
Yes, answer is