How do you solve #log(x+1) = 2 + log x#?

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Answer 1 · 2015-12-01T22:03:31

#x=1/99#

Know that: #{(loga-logb=log (a/b)),(loga=b<=>a=10^b):}#

#log(x+1)=2+logx#

#log(x+1)-logx=2#

#log((x+1)/x)=2#

#(x+1)/x=10^2#

#x+1=100x#

#x=1/99#