How do you solve #log(2+x)-log(x-5)=log 2#?

1 Answer
Vikki
Dec 3, 2015

#x= 12#

Explanation:

Re-write as single logarithmic expression

Note: #log(a) - log(b) = log(a/b)#

#log(2+x) - log(x-5) = log2#

#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 :
#log(12+2) - log(12-5) = log 2# ?
#log(14) - log(7) #
#log(14/7)#
#log 2 = log 2#

Yes, answer is #x= 12#

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