How do you solve #log(x-1) - log2 = 12 + log3x #?

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Answer 1 · 2018-08-03T13:04:29

#x=1/(1-3c)# where #c=e^(12+log(2))#

We will using that #log(a)-log(b)=log(a/b)# writing

#log(x-1)-log(3x)=12+log(2)# so

#log((x-1)/(3x))=12+log(2)# and

#(x-1)/(3x)=e^(12+log(2))#

let

#c=e^(12+log(2))#

then we get

#(x-1)/(3x)=c#

Multiplying by #3x#

#x-1=3xc # so # x-3xc=1#

so #x=1/(1-3c)#

but the is negative, so no solutions!