How do you solve #log3(x+1) - log5(x-2)=2#?

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Answer 1 · 2016-05-06T17:36:45

#x=497/1003#

#log3(x+1)-log5(x-2)=2#

or #log(3(x+1))/(5(x-2))=log100#

or #(3(x+1))/(5(x-2))=100#

or #3x+3=100xx(5x-10)#

or #3x+3=500x-1000#

or #497x=1003#

or #x=497/1003#