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

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Answer 1 · 2018-07-17T15:03:12

#x=9#

Given equation:

#log(5+x)-\log(x-2)=\log2\quad (\forall \ \ x>2)#

#\log({5+x}/{x-2})=\log 2#

Comparing the numbers on both the sides, we get

#{5+x}/{x-2}=2#

#5+x=2x-4#

#x=5+4#

#x=9#