How do you solve #log x - log(8-5x) = 2#?

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Answer 1 · 2018-05-24T01:52:44

#logx-log(8-5x)=2#

#=>log_10(x/(8-5x))=2#

#=>x/(8-5x)=100#

#=>x=(8-5x)*100#

#=>x=800-500x#

#=>501x=800#

#=>x=800/501#